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5.5 volumes

The document describes how to calculate the volume of a solid object using Cavalieri's principle. It involves partitioning the solid into thin cross-sectional slices and approximating the volume of each slice as a cylinder with the slice's cross-sectional area and thickness. The total volume is then approximated as the sum of the cylindrical slice volumes. As the number of slices approaches infinity, this sum approaches the actual volume calculated as the integral of the cross-sectional area function over the solid's distance range.

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Volumes and Cavalieri’s Principle
Volumes and Cavalieri’s Principle In the last section, to find the area A of a region R, we summed the areas of more and more but thinner and thinner rectangular strips that pave over R with better and better precision.
Volumes and Cavalieri’s Principle In the last section, to find the area A of a region R, we summed the areas of more and more but thinner and thinner rectangular strips that pave over R with better and better precision. R a=x b=x x
Volumes and Cavalieri’s Principle In the last section, to find the area A of a region R, we summed the areas of more and more but thinner and thinner rectangular strips that pave over R with better and better precision. By the Fund. Theorem of Calculus, n Σ L(xi)Δx i=1 where L(x) is the cross–sectional length function. R a=x b=x x
Volumes and Cavalieri’s Principle In the last section, to find the area A of a region R, we summed the areas of more and more but thinner and thinner rectangular strips that pave over R with better and better precision. By the Fund. Theorem of Calculus, n Σ L(xi)Δx where L(x) is the cross–sectional length function. L(x) x R a=x b=x x i=1
Volumes and Cavalieri’s Principle In the last section, to find the area A of a region R, we summed the areas of more and more but thinner and thinner rectangular strips that pave over R with better and better precision. By the Fund. Theorem of Calculus, n Σ L(xi)Δx where L(x) is the cross–sectional length function. L(x) x R Δx L(x1) a=x0 b=xn x1 x2 x i=1 We use the right end point in each subinterval since the choice of the points does not matter.
Volumes and Cavalieri’s Principle In the last section, to find the area A of a region R, we summed the areas of more and more but thinner and thinner rectangular strips that pave over R with better and better precision. By the Fund. Theorem of Calculus, n Σ L(xi)Δx where L(x) is the cross–sectional length function. x R Δx Δx L(x1) L(x2) a=x0 b=xn x1 x2 x i=1 L(x) We use the right end point in each subinterval since the choice of the points does not matter.
Volumes and Cavalieri’s Principle In the last section, to find the area A of a region R, we summed the areas of more and more but thinner and thinner rectangular strips that pave over R with better and better precision. By the Fund. Theorem of Calculus, n Σ L(xi)Δx where L(x) is the cross–sectional length function. Δx Δx a=x0 x1 x2 L(x) x R L(x1) L(x2) x i=1 L(x3) We use the right end point in each subinterval since the choice of the points does not matter. b=xn
Volumes and Cavalieri’s Principle In the last section, to find the area A of a region R, we summed the areas of more and more but thinner and thinner rectangular strips that pave over R with better and better precision. By the Fund. Theorem of Calculus, n Σ L(xi)Δx where L(x) is the cross–sectional length function. Δx L(x) x R Δx Δx L(x1) L(x2) a=x0 x1 x2 xi–1 xi Δx L(xn) x i=1 L(x3) We use the right end point in each subinterval since the choice of the points does not matter. b=xn
Volumes and Cavalieri’s Principle In the last section, to find the area A of a region R, we summed the areas of more and more but thinner and thinner rectangular strips that pave over R with better and better precision. By the Fund. Theorem of Calculus, n Σ L(xi)Δx where L(x) is the cross–sectional length function. Δx L(x) x R Δx Δx L(x1) L(x2) a=x0 x1 x2 xi–1 xi Δx L(xn) n Σ L(xi)Δx = Sum of the areas of all the rectangles x i=1 L(x3) i=1 b=xn
Volumes and Cavalieri’s Principle In the last section, to find the area A of a region R, we summed the areas of more and more but thinner and thinner rectangular strips that pave over R with better and better precision. By the Fund. Theorem of Calculus, as n 0, n Σ L(xi)Δx where L(x) is the cross–sectional length function. Δx L(x) x R Δx Δx L(x1) L(x2) a=x0 x1 x2 xi–1 xi Δx L(xn) n Σ L(xi)Δx = Sum of the areas of all the rectangles x i=1 L(x3) i=1 b=xn
Volumes and Cavalieri’s Principle In the last section, to find the area A of a region R, we summed the areas of more and more but thinner and thinner rectangular strips that pave over R with better and better precision. By the Fund. Theorem of b Calculus, as n 0, ∫ x=a area of R = L(x) dx n Σ L(xi)Δx where L(x) is the cross–sectional length function. Δx L(x) x R Δx Δx L(x1) L(x2) a=x0 x1 x2 xi–1 xi Δx L(xn) n Σ L(xi)Δx = Sum of the areas of all the rectangles x i=1 L(x3) i=1 b=xn
Volumes and Cavalieri’s Principle In the last section, to find the area A of a region R, we summed the areas of more and more but thinner and thinner rectangular strips that pave over R with better and better precision. By the Fund. Theorem of b Calculus, as n 0, ∫ x=a area of R = L(x) dx n Σ L(xi)Δx where L(x) is the cross–sectional length function. So Δx L(x) x R Δx Δx L(x1) L(x2) a=x0 x1 x2 xi–1 xi Δx L(xn) n Σ L(xi)Δx = Sum of the areas of all the rectangles x i=1 “area is the integral of the cross–sectional lengths”. L(x3) i=1 b=xn
Volumes and Cavalieri’s Principle In this section, we find the volume V of a solid S by summing more and more but thinner and thinner slices that layer S with better and better precision.
Volumes and Cavalieri’s Principle In this section, we find the volume V of a solid S by summing more and more but thinner and thinner slices that layer S with better and better precision. The intersection of a plane and a solid is a cross– sectional area.
Volumes and Cavalieri’s Principle In this section, we find the volume V of a solid S by summing more and more but thinner and thinner slices that layer S with better and better precision. The intersection of a plane and a solid is a cross– sectional area. For example, the cross–sectional areas of a box are rectangles of the same dimensions.
Volumes and Cavalieri’s Principle In this section, we find the volume V of a solid S by summing more and more but thinner and thinner slices that layer S with better and better precision. The intersection of a plane and a solid is a cross– sectional area. For example, the cross–sectional areas of a box are rectangles of the same dimensions. However we have solids that are not boxes with exactly the same cross–sectional areas. We will show these solids have the same volume. Two solids with the same cross–sectional areas
Volumes and Cavalieri’s Principle Let S be a solid and we want to find its volume. S
Let S be a solid and we want to find its volume. S x Volumes and Cavalieri’s Principle Set a ruler X in a chosen direction and measure S from end to end along X.
Volumes and Cavalieri’s Principle Let S be a solid and we want to find its volume. Set a ruler X in a chosen direction and measure S from end to end along X. Assume that S measures from x = a to x = b. S a b x
Volumes and Cavalieri’s Principle Let S be a solid and we want to find its volume. Set a ruler X in a chosen direction and measure S from end to end along X. Assume that S measures from x = a to x = b. At a generic point x, let A(x) be its cross–sectional area at x as shown. A(x) S a b x x
Volumes and Cavalieri’s Principle A(x) a=x0 b=xn x xi–1 xi 1 x2 x S Partition [a, b] into n equal size subintervals as {x0=a, x1, x2, ..xi, .. xn=b}.
Volumes and Cavalieri’s Principle A(x) A(xi) a=x0 b=xn x xi–1 xi 1 x2 x S Partition [a, b] into n equal size subintervals as {x0=a, x1, x2, ..xi, .. xn=b}. Select the right end point xi of each subinterval [xi–1, xi] and then the cross– sectional area at xi is A(xi).
Volumes and Cavalieri’s Principle A(x) A(xi) a=x0 b=xn x xi–1 xi 1 x2 x S Partition [a, b] into n equal size subintervals as {x0=a, x1, x2, ..xi, .. xn=b}. Select the right end point xi of each subinterval [xi–1, xi] and then the cross– sectional area at xi is A(xi). (Again, the choice of the point in each interval does not matter as n ∞).
Volumes and Cavalieri’s Principle A(x) Δx A(xi) a=x0 b=xn x xi–1 xi 1 x2 x S Partition [a, b] into n equal size subintervals as {x0=a, x1, x2, ..xi, .. xn=b}. Select the right end point xi of each subinterval [xi–1, xi] and then the cross– sectional area at xi is A(xi). (Again, the choice of the point in each interval does not matter as n ∞). Let Δx be the length of each subinterval.
Volumes and Cavalieri’s Principle A(xi) A(x) Δx a=x0 b=xn x xi–1 xi 1 x2 x S Δx A(xi) Ci The cylinder Ci with cross–sectional area A(xi) and thickness Δx approximates the slice of the solid between xi–1 and xi.
Volumes and Cavalieri’s Principle A(xi) A(x) Δx a=x0 b=xn x xi–1 xi 1 x2 x S The cylinder Ci with cross–sectional area A(xi) and thickness Δx approximates the slice of the solid between xi–1 and xi. Let the volume of Ci be vi then vi = A(xi)Δx. Δx A(xi) Ci volume of Ci is vi = A(xi)Δx
Volumes and Cavalieri’s Principle A(xi) A(x) Δx a=x0 b=xn x xi–1 xi 1 x2 x S The cylinder Ci with cross–sectional area A(xi) and thickness Δx approximates the slice of the solid between xi–1 and xi. Let the volume of Ci be vi then vi = A(xi)Δx. Δx A(xi) Ci volume of Ci is vi = A(xi)Δx The sum of the volumes from all the cylinders, i.e. all the vi‘s approximates the volume V of the solid S.
Volumes and Cavalieri’s Principle a=x0 b=xn xi–1 xx i 1 x2 x ≈ n = Σ A(xi)Δx, i=1 S n Σ vi Hence V A(x) Δx A(xi) Δx A(xi) Ci volume of Ci is vi = A(xi)Δx
Volumes and Cavalieri’s Principle a=x0 b=xn xi–1 xx i 1 x2 x Hence n V ≈ Σ vi n = Σ A(xi)Δx, i=1 i=1 S A(x) Δx A(xi) Δx A(xi) Ci volume of Ci is vi = A(xi)Δx n Σ A(xi)Δx = n ∞ ∫ as n ∞ by the FTC, V = lim A(x) dx i=1 b x=a
Volumes and Cavalieri’s Principle a=x0 b=xn xi–1 xx i 1 x2 x Hence n V ≈ Σ vi n = Σ A(xi)Δx, i=1 i=1 S A(x) Δx A(xi) Δx A(xi) Ci volume of Ci is vi = A(xi)Δx n Σ A(xi)Δx = n ∞ ∫ as n ∞ by the FTC, V = lim A(x) dx i=1 b x=a Hence, similar to the relation between the area and the cross–sectional lengths, we have that the “volume is the integral of the cross–sectional areas”.
Volumes and Cavalieri’s Principle The fact that Area = ∫ b L(x) dx x=a lead us to the following conclusion.
Volumes and Cavalieri’s Principle The fact that Area = ∫ b L(x) dx x=a lead us to the following conclusion. Two 2D regions that have identical cross–sectional lengths L(x) have the same areas.
Volumes and Cavalieri’s Principle The fact that Area = ∫ b L(x) dx x=a lead us to the following conclusion. Two 2D regions that have identical cross–sectional lengths L(x) have the same areas. This is known as the Cavalieri’s Principle.
Volumes and Cavalieri’s Principle The fact that Area = ∫ b L(x) dx x=a lead us to the following conclusion. Two 2D regions that have identical cross–sectional lengths L(x) have the same areas. This is known as the Cavalieri’s Principle. For example, two triangles with the same base B and height H have the same area H B
Volumes and Cavalieri’s Principle The fact that Area = ∫ b L(x) dx x=a lead us to the following conclusion. Two 2D regions that have identical cross–sectional lengths L(x) have the same areas. This is known as the Cavalieri’s Principle. For example, two triangles with the same base B and height H have the same area because their cross– sectional lengths u and v are the same everywhere. H B x u v
Volumes and Cavalieri’s Principle The fact that Area = ∫ b L(x) dx x=a lead us to the following conclusion. Two 2D regions that have identical cross–sectional lengths L(x) have the same areas. This is known as the Cavalieri’s Principle. For example, two triangles with the same base B and height H have the same area because their cross– sectional lengths u and v are the same everywhere. This is true because H B x u v u / x = B / H = v / x for all x’s
Volumes and Cavalieri’s Principle The fact that Area = ∫ b L(x) dx x=a lead us to the following conclusion. Two 2D regions that have identical cross–sectional lengths L(x) have the same areas. This is known as the Cavalieri’s Principle. For example, two triangles with the same base B and height H have the same area because their cross– sectional lengths u and v are the same everywhere. This is true because H B x u v u / x = B / H = v / x for all x’s so u = v.
Volumes and Cavalieri’s Principle The fact that b Volume = A(x) dx ∫ x=a gives us to the following.
Volumes and Cavalieri’s Principle The fact that b Volume = A(x) dx ∫ x=a gives us to the following. Two 3D solids that have the identical cross–sectional areas A(x) have the same volumes.
Volumes and Cavalieri’s Principle The fact that b Volume = A(x) dx ∫ x=a gives us to the following. Two 3D solids that have the identical cross–sectional areas A(x) have the same volumes. This is the 3D version of Cavalieri’s Principle.
Volumes and Cavalieri’s Principle The fact that b Volume = A(x) dx ∫ x=a gives us to the following. Two 3D solids that have the identical cross–sectional areas A(x) have the same volumes. This is the 3D version of Cavalieri’s Principle. Hence both solids below have the volume l w l x h w
Volumes and Cavalieri’s Principle The fact that b Volume = A(x) dx ∫ x=a gives us to the following. Two 3D solids that have the identical cross–sectional areas A(x) have the same volumes. This is the 3D version of Cavalieri’s Principle. Hence both solids below have the volume l w l h w h l * w dx ∫ x=0 x V =
Volumes and Cavalieri’s Principle The fact that b Volume = A(x) dx ∫ x=a gives us to the following. Two 3D solids that have the identical cross–sectional areas A(x) have the same volumes. This is the 3D version of Cavalieri’s Principle. Hence both solids below have the volume l w l h w h l * w dx ∫ x=0 x = h (lw) x| x=0 V =
Volumes and Cavalieri’s Principle The fact that b Volume = A(x) dx ∫ x=a gives us to the following. Two 3D solids that have the identical cross–sectional areas A(x) have the same volumes. This is the 3D version of Cavalieri’s Principle. Hence both solids below have the volume l w l h w h l * w dx ∫ x=0 x V = = h (lw) x| x=0 = lwh
Volumes and Cavalieri’s Principle The fact that b Volume = A(x) dx ∫ x=a gives us to the following. Two 3D solids that have the identical cross–sectional areas A(x) have the same volumes. This is the 3D version of Cavalieri’s Principle. Hence both solids below have the volume l w l h w h l * w dx ∫ x=0 x V = = h (lw) x| x=0 = lwh Here are more examples of volumes.
Volumes and Cavalieri’s Principle Example A. For each of the following solids, set a direction for x. Find the span of the solid along x, the cross–sectional area function A(x) and the volume V.
Volumes and Cavalieri’s Principle Example A. For each of the following solids, set a direction for x. Find the span of the solid along x, the cross–sectional area function A(x) and the volume V. a. (Cylinders) r h
Volumes and Cavalieri’s Principle Example A. For each of the following solids, set a direction for x. Find the span of the solid along x, the cross–sectional area function A(x) and the volume V. a. (Cylinders) Set the x direction vertically upward as shown. The range is from x = 0 to x = h. r x=h h x=0
Volumes and Cavalieri’s Principle Example A. For each of the following solids, set a direction for x. Find the span of the solid along x, the cross–sectional area function A(x) and the volume V. a. (Cylinders) Set the x direction vertically upward as shown. The range is from x = 0 to x = h. The cross–sections are circles of constant radius r so A(x) = πr2. r x=h h x=0
Volumes and Cavalieri’s Principle Example A. For each of the following solids, set a direction for x. Find the span of the solid along x, the cross–sectional area function A(x) and the volume V. a. (Cylinders) Set the x direction vertically upward as shown. The range is from x = 0 to x = h. The cross–sections are circles of constant radius r so A(x) = πr2. Hence the volume of the cylinder is r x=h h x=0 V = ∫ b A(x) dx x=a h π r2 dx ∫ x=0 =
Volumes and Cavalieri’s Principle Example A. For each of the following solids, set a direction for x. Find the span of the solid along x, the cross–sectional area function A(x) and the volume V. a. (Cylinders) Set the x direction vertically upward as shown. The range is from x = 0 to x = h. The cross–sections are circles of constant radius r so A(x) = πr2. Hence the volume of the cylinder is r x=h h x=0 b A(x) dx h π r2 dx ∫ x=0 h | = π r2h x=0 V = = π r2x ∫ x=a =
Volumes and Cavalieri’s Principle b. (Cones) h R
Volumes and Cavalieri’s Principle b. (Cones) Set the x direction vertically downward as shown. The range is from x = 0 to x = h. x=0 h x=h R
Volumes and Cavalieri’s Principle b. (Cones) Set the x direction vertically downward as shown. The range is from x = 0 to x = h. The cross–sections are circles of whose radius r varies according x. x=0 h x=h R x r
Volumes and Cavalieri’s Principle b. (Cones) Set the x direction vertically downward as shown. The range is from x = 0 to x = h. The cross–sections are circles of whose radius r varies according x. r / R = x / h x=0 h x=h R x r By a similar triangles argument we have that
Volumes and Cavalieri’s Principle b. (Cones) Set the x direction vertically downward as shown. The range is from x = 0 to x = h. The cross–sections are circles of whose radius r varies according x. By a similar triangles argument we have that r / R = x / h or that r = xR h x=0 h x=h R x r
Volumes and Cavalieri’s Principle b. (Cones) Set the x direction vertically downward as shown. The range is from x = 0 to x = h. The cross–sections are circles of whose radius r varies according x. By a similar triangles argument we have that r / R = x / h or that r = xR h A(x) = π ( 2 therefore xR h ) x=0 h x=h R x r
Volumes and Cavalieri’s Principle b. (Cones) Set the x direction vertically downward as shown. The range is from x = 0 to x = h. The cross–sections are circles of whose radius r varies according x. By a similar triangles argument we have that r / R = x / h or that r = h ∫ x=0 V = dx xR h A(x) = π ( 2 therefore xR h ) and the volume of the cone is π ( xR 2 h ) x=0 h x=h R x r
Volumes and Cavalieri’s Principle b. (Cones) Set the x direction vertically downward as shown. The range is from x = 0 to x = h. The cross–sections are circles of whose radius r varies according x. By a similar triangles argument we have that r / R = x / h or that r = h ∫ x=0 V = dx xR h A(x) = π ( 2 therefore xR h ) and the volume of the cone is π ( xR 2 h ) ∫ h = dx x=0 πR2 h2 x2 x=0 h x=h R x r
Volumes and Cavalieri’s Principle b. (Cones) Set the x direction vertically downward as shown. The range is from x = 0 to x = h. The cross–sections are circles of whose radius r varies according x. By a similar triangles argument we have that r / R = x / h or that r = h ∫ x=0 V = dx xR h A(x) = π ( 2 therefore xR h ) and the volume of the cone is π ( xR 2 h ) ∫ h = dx x=0 πR2 h2 h x2 | x=0 = πR2 h2 x3 3 x=0 h x=h R x r
Volumes and Cavalieri’s Principle b. (Cones) Set the x direction vertically downward as shown. The range is from x = 0 to x = h. The cross–sections are circles of whose radius r varies according x. By a similar triangles argument we have that r / R = x / h or that r = x=0 h x=h h ∫ x=0 V = dx R x r xR h A(x) = π ( 2 therefore xR h ) and the volume of the cone is π ( xR 2 h ) ∫ h = dx x=0 πR2 h2 h x2 | = = x=0 πR2 h2 x3 3 π R2h 3
Volumes and Cavalieri’s Principle In mathematics, a solid P which is formed by moving a 2D base B in the 3D space in a straight line is called a prism. B A prism P with base B h
Volumes and Cavalieri’s Principle In mathematics, a solid P which is formed by moving a 2D base B in the 3D space in a straight line is called a prism. A prism formed by moving the base B in the perpendicular direction to the base B is called a right (straight) prism. B A prism P with base B h
Volumes and Cavalieri’s Principle In mathematics, a solid P which is formed by moving a 2D base B in the 3D space in a straight line is called a prism. A prism formed by moving the base B in the perpendicular direction to the base B is called a right (straight) prism. Let h be the height of the prisms as shown, then both prisms have the same volume V = Bh by the same argument for cylinders. h h B B A prism P with base B A right prism P
Volumes and Cavalieri’s Principle In fact, any solid formed by rotating and moving a base B upward to the height of h,
Volumes and Cavalieri’s Principle In fact, any solid formed by rotating and moving a base B upward to the height of h, such as the ones shown here, http://www.iroonie.com/modern-modular-indoor- spiral-staircase-designs/modern-wooden- spiral-staircase-layouts/
Volumes and Cavalieri’s Principle In fact, any solid formed by rotating and moving a base B upward to the height of h, such as the ones shown here, http://www.iroonie.com/modern-modular-indoor- spiral-staircase-designs/modern-wooden- spiral-staircase-layouts/ http://www.screen-dream. de/images/Sharp_ Spiral.jpg
Volumes and Cavalieri’s Principle In fact, any solid formed by rotating and moving a base B upward to the height of h, such as the ones shown here, also have volumes V = Bh. http://www.iroonie.com/modern-modular-indoor- spiral-staircase-designs/modern-wooden- spiral-staircase-layouts/ http://www.screen-dream. de/images/Sharp_ Spiral.jpg
Volumes and Cavalieri’s Principle In fact, any solid formed by rotating and moving a base B upward to the height of h, such as the ones shown here, also have volumes V = Bh. If the staircase here is a continuous ramp, let h be the height, B be the landing–surface area of the step, then the volume of the ramp would be V = Bh. http://www.iroonie.com/modern-modular-indoor- spiral-staircase-designs/modern-wooden- spiral-staircase-layouts/ http://www.screen-dream. de/images/Sharp_ Spiral.jpg
Volumes and Cavalieri’s Principle A solid C which consists of exactly all the straight line segments that connect a given point V (the vertex) and to another end point in the base B, is a cone
Volumes and Cavalieri’s Principle A solid C which consists of exactly all the straight line segments that connect a given point V (the vertex) and to another end point in the base B, is a cone B A cone V
Volumes and Cavalieri’s Principle A solid C which consists of exactly all the straight line segments that connect a given point V (the vertex) and to another end point in the base B, is a cone. If the base B is a circle, a rectangle, a regular polygon, or any 2D region with a “center” c, then the cone is a right (straight) cone if the vertex V is right above c. B A cone V
Volumes and Cavalieri’s Principle A solid C which consists of exactly all the straight line segments that connect a given point V (the vertex) and to another end point in the base B, is a cone. If the base B is a circle, a rectangle, a regular polygon, or any 2D region with a “center” c, then the cone is a right (straight) cone if the vertex V is right above c. V V c B B A cone A straight cone
Volumes and Cavalieri’s Principle A solid C which consists of exactly all the straight line segments that connect a given point V (the vertex) and to another end point in the base B, is a cone. If the base B is a circle, a rectangle, a regular polygon, or any 2D region with a “center” c, then the cone is a right (straight) cone if the vertex V is right above c. The volume V of a cone is = Bh/3 where h is the height of the cone. V V h h c B B A cone A straight cone
Volumes and Cavalieri’s Principle A solid C which consists of exactly all the straight line segments that connect a given point V (the vertex) and to another end point in the base B, is a cone. If the base B is a circle, a rectangle, a regular polygon, or any 2D region with a “center” c, then the cone is a right (straight) cone if the vertex V is right above c. The volume V of a cone is = Bh/3 where h is the height of the cone. This is true because the cross–sectional area V V function is A(x) = Bx2/h2 where the x is as shown h h c x A(x) B B A cone A straight cone
Volumes and Cavalieri’s Principle A solid C which consists of exactly all the straight line segments that connect a given point V (the vertex) and to another end point in the base B, is a cone. If the base B is a circle, a rectangle, a regular polygon, or any 2D region with a “center” c, then the cone is a right (straight) cone if the vertex V is right above c. The volume V of a cone is = Bh/3 where h is the height of the cone. This is true because the cross–sectional area V V function is A(x) = Bx2/h2 A(x) x where the x is as shown h h and this integrates to V = Bh/3 by the c Cavalerie’s Principle. B B A cone A straight cone
The Volume of the Intersection of Two Pipes Example B. Find the volume of the intersection of two pipes with radius r that are joined at a right angle as shown. Intersection of two pipes http://mathworld.wolf ram.com/SteinmetzS olid.html
The Volume of the Intersection of Two Pipes Example B. Find the volume of the intersection of two pipes with radius r that are joined at a right angle as shown. Intersection of two pipes We’ll utilize Cavaliere’s Principle which states that the volume of a solid is the http://mathworld.wolf ram.com/SteinmetzS olid.html integral of the cross-sectional area function A(x)
The Volume of the Intersection of Two Pipes Example B. Find the volume of the intersection of two pipes with radius r that are joined at a right angle as shown. Intersection of two pipes We’ll utilize Cavaliere’s Principle which states that the volume of a solid is the http://mathworld.wolf ram.com/SteinmetzS olid.html integral of the cross-sectional area function A(x), i.e. b Volume = ∫ cross-sectional areas dx = ∫ A(x) dx. x=a x=a b
The Volume of the Intersection of Two Pipes Example B. Find the volume of the intersection of two pipes with radius r that are joined at a right angle as shown. We’ll utilize Cavaliere’s Principle which states that the volume of a solid is the Intersection of two pipes http://mathworld.wolf ram.com/SteinmetzS olid.html integral of the cross-sectional area function A(x), i.e. b b Volume = ∫ cross-sectional areas dx = ∫ A(x) dx. x=a x=a Here is a better view at the solid and it’s cross– sections in question (see pipesvol.svr).
The Volume of the Intersection of Two Pipes Example B. Find the volume of the intersection of two pipes with radius r that are joined at a right angle as shown. Intersection of two pipes We’ll utilize Cavaliere’s Principle which states that the volume of a solid is the http://mathworld.wolf ram.com/SteinmetzS olid.html integral of the cross-sectional area function A(x), i.e. b b Volume = ∫ cross-sectional areas dx = ∫ A(x) dx. x=a x=a Here is a better view at the solid and it’s cross– sections in question (see pipesvol.svr). We see that the cross–sectional areas are squares.
The Volume of the Intersection of Two Pipes We will find the volume of the top half and double it for the final answer.
The Volume of the Intersection of Two Pipes We will find the volume of the top half and double it for the final answer. As we observed, the horizontal cross–sections are squares
The Volume of the Intersection of Two Pipes We will find the volume of the top half and double it for the final answer. s As we observed, the horizontal cross–sections are squares whose sides s are cross–sectional lengths of the circle. s
The Volume of the Intersection of Two Pipes We will find the volume of the top half and double it for the final answer. As we observed, the horizontal cross–sections are squares whose sides s are cross–sectional lengths of the circle. Let x and r be as shown. s s r x
The Volume of the Intersection of Two Pipes We will find the volume of the top half and double it for the final answer. As we observed, the horizontal cross–sections are squares whose sides s are cross–sectional lengths of the circle. Let x and r be as shown. s s r x r x
The Volume of the Intersection of Two Pipes We will find the volume of the top half and double it for the final answer. As we observed, the horizontal cross–sections are squares whose sides s are cross–sectional lengths of the circle. Let x and r be as shown. Hence s/2 = √r2 – x2 s s r x r x
The Volume of the Intersection of Two Pipes We will find the volume of the top half and double it for the final answer. As we observed, the horizontal cross–sections are squares whose sides s are cross–sectional lengths of the circle. Let x and r be as shown. Hence s/2 = √r2 – x2 or that s s r x s = 2√r2 – x2 where x goes from 0 to r. r x
The Volume of the Intersection of Two Pipes We will find the volume of the top half and double it for the final answer. As we observed, the horizontal cross–sections are squares whose sides s are cross–sectional lengths of the circle. Let x and r be as shown. Hence s/2 = √r2 – x2 or that A(x) = 4(r2 – x2) s s r x r x s = 2√r2 – x2 where x goes from 0 to r. Therefore the cross–sectional area function is A(x) = s2 = 4(r2 – x2).
The Volume of the Intersection of Two Pipes So the volume of the top is r ∫ 4(r2 – x2) dx 0 b s r x x=a s A(x) = 4(r2 – x2) s = 2√r2 – x2 r x
The Volume of the Intersection of Two Pipes So the volume of the top is r ∫ 4(r2 – x2) dx 0 b x=a s s s = 2√r2 – x2 r x r = 4∫ (r2 – x2) dx 0 A(x) = 4(r2 – x2) r x
The Volume of the Intersection of Two Pipes So the volume of the top is r ∫ 4(r2 – x2) dx 0 b x=a s s s = 2√r2 – x2 r x r = 4∫ (r2 – x2) dx 0 = 4(r2x – x3/3) | r x=0 A(x) = 4(r2 – x2) r x
The Volume of the Intersection of Two Pipes So the volume of the top is r ∫ 4(r2 – x2) dx 0 b x=a s s s = 2√r2 – x2 r x r = 4∫ (r2 – x2) dx 0 = 4(r2x – x3/3) | r x=0 = 4(2r3/3) = 8r3/3 A(x) = 4(r2 – x2) r x
The Volume of the Intersection of Two Pipes So the volume of the top is r ∫ 4(r2 – x2) dx 0 b x=a s s s = 2√r2 – x2 r x r = 4∫ (r2 – x2) dx 0 = 4(r2x – x3/3) | r x=0 = 4(2r3/3) = 8r3/3 Therefore the entire volume of the intersection is twice the above or 16r3/3. A(x) = 4(r2 – x2) r x
The Volume of the Intersection of Two Pipes So the volume of the top is r ∫ 4(r2 – x2) dx 0 b x=a s s s = 2√r2 – x2 r x r = 4∫ (r2 – x2) dx 0 = 4(r2x – x3/3) | r x=0 = 4(2r3/3) = 8r3/3 Therefore the entire volume of the intersection is twice the above or 16r3/3. A(x) = 4(r2 – x2) r x Cavalieri’s Principle gives us the mathematical meaning of the term “volumes”. Starting with 1D objects – line segments, we define their “volumes” to be just the lengths of the segments.
The Volume of the Intersection of Two Pipes By Cavalerie’s Principle, the “volumes” of 2D regions, i.e. areas, are the integrals of the cross– sectional–length functions L(x) which are “volumes” of 1D line segments.
The Volume of the Intersection of Two Pipes By Cavalerie’s Principle, the “volumes” of 2D regions, i.e. areas, are the integrals of the cross– sectional–length functions L(x) which are “volumes” of 1D line segments. Similarly, the volumes of 3D solids are the integrals of the cross–sectional–area functions A(x) which are “volumes” of 2D regions.
The Volume of the Intersection of Two Pipes By Cavalerie’s Principle, the “volumes” of 2D regions, i.e. areas, are the integrals of the cross– sectional–length functions L(x) which are “volumes” of 1D line segments. Similarly, the volumes of 3D solids are the integrals of the cross–sectional–area functions A(x) which are “volumes” of 2D regions. Continuing in this manner, we define algebraically the term “volume” in the “higher dimensional spaces” via Cavalerie’s Principle,
The Volume of the Intersection of Two Pipes By Cavalerie’s Principle, the “volumes” of 2D regions, i.e. areas, are the integrals of the cross– sectional–length functions L(x) which are “volumes” of 1D line segments. Similarly, the volumes of 3D solids are the integrals of the cross–sectional–area functions A(x) which are “volumes” of 2D regions. Continuing in this manner, we define algebraically the term “volume” in the “higher dimensional spaces” via Cavalerie’s Principle, i.e. volumes are the integrals of the cross–sectional functions which are “volumes” in one less dimensional space.

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5.5 volumes

  • 1.
  • 2.
    Volumes and Cavalieri’sPrinciple In the last section, to find the area A of a region R, we summed the areas of more and more but thinner and thinner rectangular strips that pave over R with better and better precision.
  • 3.
    Volumes and Cavalieri’sPrinciple In the last section, to find the area A of a region R, we summed the areas of more and more but thinner and thinner rectangular strips that pave over R with better and better precision. R a=x b=x x
  • 4.
    Volumes and Cavalieri’sPrinciple In the last section, to find the area A of a region R, we summed the areas of more and more but thinner and thinner rectangular strips that pave over R with better and better precision. By the Fund. Theorem of Calculus, n Σ L(xi)Δx i=1 where L(x) is the cross–sectional length function. R a=x b=x x
  • 5.
    Volumes and Cavalieri’sPrinciple In the last section, to find the area A of a region R, we summed the areas of more and more but thinner and thinner rectangular strips that pave over R with better and better precision. By the Fund. Theorem of Calculus, n Σ L(xi)Δx where L(x) is the cross–sectional length function. L(x) x R a=x b=x x i=1
  • 6.
    Volumes and Cavalieri’sPrinciple In the last section, to find the area A of a region R, we summed the areas of more and more but thinner and thinner rectangular strips that pave over R with better and better precision. By the Fund. Theorem of Calculus, n Σ L(xi)Δx where L(x) is the cross–sectional length function. L(x) x R Δx L(x1) a=x0 b=xn x1 x2 x i=1 We use the right end point in each subinterval since the choice of the points does not matter.
  • 7.
    Volumes and Cavalieri’sPrinciple In the last section, to find the area A of a region R, we summed the areas of more and more but thinner and thinner rectangular strips that pave over R with better and better precision. By the Fund. Theorem of Calculus, n Σ L(xi)Δx where L(x) is the cross–sectional length function. x R Δx Δx L(x1) L(x2) a=x0 b=xn x1 x2 x i=1 L(x) We use the right end point in each subinterval since the choice of the points does not matter.
  • 8.
    Volumes and Cavalieri’sPrinciple In the last section, to find the area A of a region R, we summed the areas of more and more but thinner and thinner rectangular strips that pave over R with better and better precision. By the Fund. Theorem of Calculus, n Σ L(xi)Δx where L(x) is the cross–sectional length function. Δx Δx a=x0 x1 x2 L(x) x R L(x1) L(x2) x i=1 L(x3) We use the right end point in each subinterval since the choice of the points does not matter. b=xn
  • 9.
    Volumes and Cavalieri’sPrinciple In the last section, to find the area A of a region R, we summed the areas of more and more but thinner and thinner rectangular strips that pave over R with better and better precision. By the Fund. Theorem of Calculus, n Σ L(xi)Δx where L(x) is the cross–sectional length function. Δx L(x) x R Δx Δx L(x1) L(x2) a=x0 x1 x2 xi–1 xi Δx L(xn) x i=1 L(x3) We use the right end point in each subinterval since the choice of the points does not matter. b=xn
  • 10.
    Volumes and Cavalieri’sPrinciple In the last section, to find the area A of a region R, we summed the areas of more and more but thinner and thinner rectangular strips that pave over R with better and better precision. By the Fund. Theorem of Calculus, n Σ L(xi)Δx where L(x) is the cross–sectional length function. Δx L(x) x R Δx Δx L(x1) L(x2) a=x0 x1 x2 xi–1 xi Δx L(xn) n Σ L(xi)Δx = Sum of the areas of all the rectangles x i=1 L(x3) i=1 b=xn
  • 11.
    Volumes and Cavalieri’sPrinciple In the last section, to find the area A of a region R, we summed the areas of more and more but thinner and thinner rectangular strips that pave over R with better and better precision. By the Fund. Theorem of Calculus, as n 0, n Σ L(xi)Δx where L(x) is the cross–sectional length function. Δx L(x) x R Δx Δx L(x1) L(x2) a=x0 x1 x2 xi–1 xi Δx L(xn) n Σ L(xi)Δx = Sum of the areas of all the rectangles x i=1 L(x3) i=1 b=xn
  • 12.
    Volumes and Cavalieri’sPrinciple In the last section, to find the area A of a region R, we summed the areas of more and more but thinner and thinner rectangular strips that pave over R with better and better precision. By the Fund. Theorem of b Calculus, as n 0, ∫ x=a area of R = L(x) dx n Σ L(xi)Δx where L(x) is the cross–sectional length function. Δx L(x) x R Δx Δx L(x1) L(x2) a=x0 x1 x2 xi–1 xi Δx L(xn) n Σ L(xi)Δx = Sum of the areas of all the rectangles x i=1 L(x3) i=1 b=xn
  • 13.
    Volumes and Cavalieri’sPrinciple In the last section, to find the area A of a region R, we summed the areas of more and more but thinner and thinner rectangular strips that pave over R with better and better precision. By the Fund. Theorem of b Calculus, as n 0, ∫ x=a area of R = L(x) dx n Σ L(xi)Δx where L(x) is the cross–sectional length function. So Δx L(x) x R Δx Δx L(x1) L(x2) a=x0 x1 x2 xi–1 xi Δx L(xn) n Σ L(xi)Δx = Sum of the areas of all the rectangles x i=1 “area is the integral of the cross–sectional lengths”. L(x3) i=1 b=xn
  • 14.
    Volumes and Cavalieri’sPrinciple In this section, we find the volume V of a solid S by summing more and more but thinner and thinner slices that layer S with better and better precision.
  • 15.
    Volumes and Cavalieri’sPrinciple In this section, we find the volume V of a solid S by summing more and more but thinner and thinner slices that layer S with better and better precision. The intersection of a plane and a solid is a cross– sectional area.
  • 16.
    Volumes and Cavalieri’sPrinciple In this section, we find the volume V of a solid S by summing more and more but thinner and thinner slices that layer S with better and better precision. The intersection of a plane and a solid is a cross– sectional area. For example, the cross–sectional areas of a box are rectangles of the same dimensions.
  • 17.
    Volumes and Cavalieri’sPrinciple In this section, we find the volume V of a solid S by summing more and more but thinner and thinner slices that layer S with better and better precision. The intersection of a plane and a solid is a cross– sectional area. For example, the cross–sectional areas of a box are rectangles of the same dimensions. However we have solids that are not boxes with exactly the same cross–sectional areas. We will show these solids have the same volume. Two solids with the same cross–sectional areas
  • 18.
    Volumes and Cavalieri’sPrinciple Let S be a solid and we want to find its volume. S
  • 19.
    Let S bea solid and we want to find its volume. S x Volumes and Cavalieri’s Principle Set a ruler X in a chosen direction and measure S from end to end along X.
  • 20.
    Volumes and Cavalieri’sPrinciple Let S be a solid and we want to find its volume. Set a ruler X in a chosen direction and measure S from end to end along X. Assume that S measures from x = a to x = b. S a b x
  • 21.
    Volumes and Cavalieri’sPrinciple Let S be a solid and we want to find its volume. Set a ruler X in a chosen direction and measure S from end to end along X. Assume that S measures from x = a to x = b. At a generic point x, let A(x) be its cross–sectional area at x as shown. A(x) S a b x x
  • 22.
    Volumes and Cavalieri’sPrinciple A(x) a=x0 b=xn x xi–1 xi 1 x2 x S Partition [a, b] into n equal size subintervals as {x0=a, x1, x2, ..xi, .. xn=b}.
  • 23.
    Volumes and Cavalieri’sPrinciple A(x) A(xi) a=x0 b=xn x xi–1 xi 1 x2 x S Partition [a, b] into n equal size subintervals as {x0=a, x1, x2, ..xi, .. xn=b}. Select the right end point xi of each subinterval [xi–1, xi] and then the cross– sectional area at xi is A(xi).
  • 24.
    Volumes and Cavalieri’sPrinciple A(x) A(xi) a=x0 b=xn x xi–1 xi 1 x2 x S Partition [a, b] into n equal size subintervals as {x0=a, x1, x2, ..xi, .. xn=b}. Select the right end point xi of each subinterval [xi–1, xi] and then the cross– sectional area at xi is A(xi). (Again, the choice of the point in each interval does not matter as n ∞).
  • 25.
    Volumes and Cavalieri’sPrinciple A(x) Δx A(xi) a=x0 b=xn x xi–1 xi 1 x2 x S Partition [a, b] into n equal size subintervals as {x0=a, x1, x2, ..xi, .. xn=b}. Select the right end point xi of each subinterval [xi–1, xi] and then the cross– sectional area at xi is A(xi). (Again, the choice of the point in each interval does not matter as n ∞). Let Δx be the length of each subinterval.
  • 26.
    Volumes and Cavalieri’sPrinciple A(xi) A(x) Δx a=x0 b=xn x xi–1 xi 1 x2 x S Δx A(xi) Ci The cylinder Ci with cross–sectional area A(xi) and thickness Δx approximates the slice of the solid between xi–1 and xi.
  • 27.
    Volumes and Cavalieri’sPrinciple A(xi) A(x) Δx a=x0 b=xn x xi–1 xi 1 x2 x S The cylinder Ci with cross–sectional area A(xi) and thickness Δx approximates the slice of the solid between xi–1 and xi. Let the volume of Ci be vi then vi = A(xi)Δx. Δx A(xi) Ci volume of Ci is vi = A(xi)Δx
  • 28.
    Volumes and Cavalieri’sPrinciple A(xi) A(x) Δx a=x0 b=xn x xi–1 xi 1 x2 x S The cylinder Ci with cross–sectional area A(xi) and thickness Δx approximates the slice of the solid between xi–1 and xi. Let the volume of Ci be vi then vi = A(xi)Δx. Δx A(xi) Ci volume of Ci is vi = A(xi)Δx The sum of the volumes from all the cylinders, i.e. all the vi‘s approximates the volume V of the solid S.
  • 29.
    Volumes and Cavalieri’sPrinciple a=x0 b=xn xi–1 xx i 1 x2 x ≈ n = Σ A(xi)Δx, i=1 S n Σ vi Hence V A(x) Δx A(xi) Δx A(xi) Ci volume of Ci is vi = A(xi)Δx
  • 30.
    Volumes and Cavalieri’sPrinciple a=x0 b=xn xi–1 xx i 1 x2 x Hence n V ≈ Σ vi n = Σ A(xi)Δx, i=1 i=1 S A(x) Δx A(xi) Δx A(xi) Ci volume of Ci is vi = A(xi)Δx n Σ A(xi)Δx = n ∞ ∫ as n ∞ by the FTC, V = lim A(x) dx i=1 b x=a
  • 31.
    Volumes and Cavalieri’sPrinciple a=x0 b=xn xi–1 xx i 1 x2 x Hence n V ≈ Σ vi n = Σ A(xi)Δx, i=1 i=1 S A(x) Δx A(xi) Δx A(xi) Ci volume of Ci is vi = A(xi)Δx n Σ A(xi)Δx = n ∞ ∫ as n ∞ by the FTC, V = lim A(x) dx i=1 b x=a Hence, similar to the relation between the area and the cross–sectional lengths, we have that the “volume is the integral of the cross–sectional areas”.
  • 32.
    Volumes and Cavalieri’sPrinciple The fact that Area = ∫ b L(x) dx x=a lead us to the following conclusion.
  • 33.
    Volumes and Cavalieri’sPrinciple The fact that Area = ∫ b L(x) dx x=a lead us to the following conclusion. Two 2D regions that have identical cross–sectional lengths L(x) have the same areas.
  • 34.
    Volumes and Cavalieri’sPrinciple The fact that Area = ∫ b L(x) dx x=a lead us to the following conclusion. Two 2D regions that have identical cross–sectional lengths L(x) have the same areas. This is known as the Cavalieri’s Principle.
  • 35.
    Volumes and Cavalieri’sPrinciple The fact that Area = ∫ b L(x) dx x=a lead us to the following conclusion. Two 2D regions that have identical cross–sectional lengths L(x) have the same areas. This is known as the Cavalieri’s Principle. For example, two triangles with the same base B and height H have the same area H B
  • 36.
    Volumes and Cavalieri’sPrinciple The fact that Area = ∫ b L(x) dx x=a lead us to the following conclusion. Two 2D regions that have identical cross–sectional lengths L(x) have the same areas. This is known as the Cavalieri’s Principle. For example, two triangles with the same base B and height H have the same area because their cross– sectional lengths u and v are the same everywhere. H B x u v
  • 37.
    Volumes and Cavalieri’sPrinciple The fact that Area = ∫ b L(x) dx x=a lead us to the following conclusion. Two 2D regions that have identical cross–sectional lengths L(x) have the same areas. This is known as the Cavalieri’s Principle. For example, two triangles with the same base B and height H have the same area because their cross– sectional lengths u and v are the same everywhere. This is true because H B x u v u / x = B / H = v / x for all x’s
  • 38.
    Volumes and Cavalieri’sPrinciple The fact that Area = ∫ b L(x) dx x=a lead us to the following conclusion. Two 2D regions that have identical cross–sectional lengths L(x) have the same areas. This is known as the Cavalieri’s Principle. For example, two triangles with the same base B and height H have the same area because their cross– sectional lengths u and v are the same everywhere. This is true because H B x u v u / x = B / H = v / x for all x’s so u = v.
  • 39.
    Volumes and Cavalieri’sPrinciple The fact that b Volume = A(x) dx ∫ x=a gives us to the following.
  • 40.
    Volumes and Cavalieri’sPrinciple The fact that b Volume = A(x) dx ∫ x=a gives us to the following. Two 3D solids that have the identical cross–sectional areas A(x) have the same volumes.
  • 41.
    Volumes and Cavalieri’sPrinciple The fact that b Volume = A(x) dx ∫ x=a gives us to the following. Two 3D solids that have the identical cross–sectional areas A(x) have the same volumes. This is the 3D version of Cavalieri’s Principle.
  • 42.
    Volumes and Cavalieri’sPrinciple The fact that b Volume = A(x) dx ∫ x=a gives us to the following. Two 3D solids that have the identical cross–sectional areas A(x) have the same volumes. This is the 3D version of Cavalieri’s Principle. Hence both solids below have the volume l w l x h w
  • 43.
    Volumes and Cavalieri’sPrinciple The fact that b Volume = A(x) dx ∫ x=a gives us to the following. Two 3D solids that have the identical cross–sectional areas A(x) have the same volumes. This is the 3D version of Cavalieri’s Principle. Hence both solids below have the volume l w l h w h l * w dx ∫ x=0 x V =
  • 44.
    Volumes and Cavalieri’sPrinciple The fact that b Volume = A(x) dx ∫ x=a gives us to the following. Two 3D solids that have the identical cross–sectional areas A(x) have the same volumes. This is the 3D version of Cavalieri’s Principle. Hence both solids below have the volume l w l h w h l * w dx ∫ x=0 x = h (lw) x| x=0 V =
  • 45.
    Volumes and Cavalieri’sPrinciple The fact that b Volume = A(x) dx ∫ x=a gives us to the following. Two 3D solids that have the identical cross–sectional areas A(x) have the same volumes. This is the 3D version of Cavalieri’s Principle. Hence both solids below have the volume l w l h w h l * w dx ∫ x=0 x V = = h (lw) x| x=0 = lwh
  • 46.
    Volumes and Cavalieri’sPrinciple The fact that b Volume = A(x) dx ∫ x=a gives us to the following. Two 3D solids that have the identical cross–sectional areas A(x) have the same volumes. This is the 3D version of Cavalieri’s Principle. Hence both solids below have the volume l w l h w h l * w dx ∫ x=0 x V = = h (lw) x| x=0 = lwh Here are more examples of volumes.
  • 47.
    Volumes and Cavalieri’sPrinciple Example A. For each of the following solids, set a direction for x. Find the span of the solid along x, the cross–sectional area function A(x) and the volume V.
  • 48.
    Volumes and Cavalieri’sPrinciple Example A. For each of the following solids, set a direction for x. Find the span of the solid along x, the cross–sectional area function A(x) and the volume V. a. (Cylinders) r h
  • 49.
    Volumes and Cavalieri’sPrinciple Example A. For each of the following solids, set a direction for x. Find the span of the solid along x, the cross–sectional area function A(x) and the volume V. a. (Cylinders) Set the x direction vertically upward as shown. The range is from x = 0 to x = h. r x=h h x=0
  • 50.
    Volumes and Cavalieri’sPrinciple Example A. For each of the following solids, set a direction for x. Find the span of the solid along x, the cross–sectional area function A(x) and the volume V. a. (Cylinders) Set the x direction vertically upward as shown. The range is from x = 0 to x = h. The cross–sections are circles of constant radius r so A(x) = πr2. r x=h h x=0
  • 51.
    Volumes and Cavalieri’sPrinciple Example A. For each of the following solids, set a direction for x. Find the span of the solid along x, the cross–sectional area function A(x) and the volume V. a. (Cylinders) Set the x direction vertically upward as shown. The range is from x = 0 to x = h. The cross–sections are circles of constant radius r so A(x) = πr2. Hence the volume of the cylinder is r x=h h x=0 V = ∫ b A(x) dx x=a h π r2 dx ∫ x=0 =
  • 52.
    Volumes and Cavalieri’sPrinciple Example A. For each of the following solids, set a direction for x. Find the span of the solid along x, the cross–sectional area function A(x) and the volume V. a. (Cylinders) Set the x direction vertically upward as shown. The range is from x = 0 to x = h. The cross–sections are circles of constant radius r so A(x) = πr2. Hence the volume of the cylinder is r x=h h x=0 b A(x) dx h π r2 dx ∫ x=0 h | = π r2h x=0 V = = π r2x ∫ x=a =
  • 53.
    Volumes and Cavalieri’sPrinciple b. (Cones) h R
  • 54.
    Volumes and Cavalieri’sPrinciple b. (Cones) Set the x direction vertically downward as shown. The range is from x = 0 to x = h. x=0 h x=h R
  • 55.
    Volumes and Cavalieri’sPrinciple b. (Cones) Set the x direction vertically downward as shown. The range is from x = 0 to x = h. The cross–sections are circles of whose radius r varies according x. x=0 h x=h R x r
  • 56.
    Volumes and Cavalieri’sPrinciple b. (Cones) Set the x direction vertically downward as shown. The range is from x = 0 to x = h. The cross–sections are circles of whose radius r varies according x. r / R = x / h x=0 h x=h R x r By a similar triangles argument we have that
  • 57.
    Volumes and Cavalieri’sPrinciple b. (Cones) Set the x direction vertically downward as shown. The range is from x = 0 to x = h. The cross–sections are circles of whose radius r varies according x. By a similar triangles argument we have that r / R = x / h or that r = xR h x=0 h x=h R x r
  • 58.
    Volumes and Cavalieri’sPrinciple b. (Cones) Set the x direction vertically downward as shown. The range is from x = 0 to x = h. The cross–sections are circles of whose radius r varies according x. By a similar triangles argument we have that r / R = x / h or that r = xR h A(x) = π ( 2 therefore xR h ) x=0 h x=h R x r
  • 59.
    Volumes and Cavalieri’sPrinciple b. (Cones) Set the x direction vertically downward as shown. The range is from x = 0 to x = h. The cross–sections are circles of whose radius r varies according x. By a similar triangles argument we have that r / R = x / h or that r = h ∫ x=0 V = dx xR h A(x) = π ( 2 therefore xR h ) and the volume of the cone is π ( xR 2 h ) x=0 h x=h R x r
  • 60.
    Volumes and Cavalieri’sPrinciple b. (Cones) Set the x direction vertically downward as shown. The range is from x = 0 to x = h. The cross–sections are circles of whose radius r varies according x. By a similar triangles argument we have that r / R = x / h or that r = h ∫ x=0 V = dx xR h A(x) = π ( 2 therefore xR h ) and the volume of the cone is π ( xR 2 h ) ∫ h = dx x=0 πR2 h2 x2 x=0 h x=h R x r
  • 61.
    Volumes and Cavalieri’sPrinciple b. (Cones) Set the x direction vertically downward as shown. The range is from x = 0 to x = h. The cross–sections are circles of whose radius r varies according x. By a similar triangles argument we have that r / R = x / h or that r = h ∫ x=0 V = dx xR h A(x) = π ( 2 therefore xR h ) and the volume of the cone is π ( xR 2 h ) ∫ h = dx x=0 πR2 h2 h x2 | x=0 = πR2 h2 x3 3 x=0 h x=h R x r
  • 62.
    Volumes and Cavalieri’sPrinciple b. (Cones) Set the x direction vertically downward as shown. The range is from x = 0 to x = h. The cross–sections are circles of whose radius r varies according x. By a similar triangles argument we have that r / R = x / h or that r = x=0 h x=h h ∫ x=0 V = dx R x r xR h A(x) = π ( 2 therefore xR h ) and the volume of the cone is π ( xR 2 h ) ∫ h = dx x=0 πR2 h2 h x2 | = = x=0 πR2 h2 x3 3 π R2h 3
  • 63.
    Volumes and Cavalieri’sPrinciple In mathematics, a solid P which is formed by moving a 2D base B in the 3D space in a straight line is called a prism. B A prism P with base B h
  • 64.
    Volumes and Cavalieri’sPrinciple In mathematics, a solid P which is formed by moving a 2D base B in the 3D space in a straight line is called a prism. A prism formed by moving the base B in the perpendicular direction to the base B is called a right (straight) prism. B A prism P with base B h
  • 65.
    Volumes and Cavalieri’sPrinciple In mathematics, a solid P which is formed by moving a 2D base B in the 3D space in a straight line is called a prism. A prism formed by moving the base B in the perpendicular direction to the base B is called a right (straight) prism. Let h be the height of the prisms as shown, then both prisms have the same volume V = Bh by the same argument for cylinders. h h B B A prism P with base B A right prism P
  • 66.
    Volumes and Cavalieri’sPrinciple In fact, any solid formed by rotating and moving a base B upward to the height of h,
  • 67.
    Volumes and Cavalieri’sPrinciple In fact, any solid formed by rotating and moving a base B upward to the height of h, such as the ones shown here, http://www.iroonie.com/modern-modular-indoor- spiral-staircase-designs/modern-wooden- spiral-staircase-layouts/
  • 68.
    Volumes and Cavalieri’sPrinciple In fact, any solid formed by rotating and moving a base B upward to the height of h, such as the ones shown here, http://www.iroonie.com/modern-modular-indoor- spiral-staircase-designs/modern-wooden- spiral-staircase-layouts/ http://www.screen-dream. de/images/Sharp_ Spiral.jpg
  • 69.
    Volumes and Cavalieri’sPrinciple In fact, any solid formed by rotating and moving a base B upward to the height of h, such as the ones shown here, also have volumes V = Bh. http://www.iroonie.com/modern-modular-indoor- spiral-staircase-designs/modern-wooden- spiral-staircase-layouts/ http://www.screen-dream. de/images/Sharp_ Spiral.jpg
  • 70.
    Volumes and Cavalieri’sPrinciple In fact, any solid formed by rotating and moving a base B upward to the height of h, such as the ones shown here, also have volumes V = Bh. If the staircase here is a continuous ramp, let h be the height, B be the landing–surface area of the step, then the volume of the ramp would be V = Bh. http://www.iroonie.com/modern-modular-indoor- spiral-staircase-designs/modern-wooden- spiral-staircase-layouts/ http://www.screen-dream. de/images/Sharp_ Spiral.jpg
  • 71.
    Volumes and Cavalieri’sPrinciple A solid C which consists of exactly all the straight line segments that connect a given point V (the vertex) and to another end point in the base B, is a cone
  • 72.
    Volumes and Cavalieri’sPrinciple A solid C which consists of exactly all the straight line segments that connect a given point V (the vertex) and to another end point in the base B, is a cone B A cone V
  • 73.
    Volumes and Cavalieri’sPrinciple A solid C which consists of exactly all the straight line segments that connect a given point V (the vertex) and to another end point in the base B, is a cone. If the base B is a circle, a rectangle, a regular polygon, or any 2D region with a “center” c, then the cone is a right (straight) cone if the vertex V is right above c. B A cone V
  • 74.
    Volumes and Cavalieri’sPrinciple A solid C which consists of exactly all the straight line segments that connect a given point V (the vertex) and to another end point in the base B, is a cone. If the base B is a circle, a rectangle, a regular polygon, or any 2D region with a “center” c, then the cone is a right (straight) cone if the vertex V is right above c. V V c B B A cone A straight cone
  • 75.
    Volumes and Cavalieri’sPrinciple A solid C which consists of exactly all the straight line segments that connect a given point V (the vertex) and to another end point in the base B, is a cone. If the base B is a circle, a rectangle, a regular polygon, or any 2D region with a “center” c, then the cone is a right (straight) cone if the vertex V is right above c. The volume V of a cone is = Bh/3 where h is the height of the cone. V V h h c B B A cone A straight cone
  • 76.
    Volumes and Cavalieri’sPrinciple A solid C which consists of exactly all the straight line segments that connect a given point V (the vertex) and to another end point in the base B, is a cone. If the base B is a circle, a rectangle, a regular polygon, or any 2D region with a “center” c, then the cone is a right (straight) cone if the vertex V is right above c. The volume V of a cone is = Bh/3 where h is the height of the cone. This is true because the cross–sectional area V V function is A(x) = Bx2/h2 where the x is as shown h h c x A(x) B B A cone A straight cone
  • 77.
    Volumes and Cavalieri’sPrinciple A solid C which consists of exactly all the straight line segments that connect a given point V (the vertex) and to another end point in the base B, is a cone. If the base B is a circle, a rectangle, a regular polygon, or any 2D region with a “center” c, then the cone is a right (straight) cone if the vertex V is right above c. The volume V of a cone is = Bh/3 where h is the height of the cone. This is true because the cross–sectional area V V function is A(x) = Bx2/h2 A(x) x where the x is as shown h h and this integrates to V = Bh/3 by the c Cavalerie’s Principle. B B A cone A straight cone
  • 78.
    The Volume ofthe Intersection of Two Pipes Example B. Find the volume of the intersection of two pipes with radius r that are joined at a right angle as shown. Intersection of two pipes http://mathworld.wolf ram.com/SteinmetzS olid.html
  • 79.
    The Volume ofthe Intersection of Two Pipes Example B. Find the volume of the intersection of two pipes with radius r that are joined at a right angle as shown. Intersection of two pipes We’ll utilize Cavaliere’s Principle which states that the volume of a solid is the http://mathworld.wolf ram.com/SteinmetzS olid.html integral of the cross-sectional area function A(x)
  • 80.
    The Volume ofthe Intersection of Two Pipes Example B. Find the volume of the intersection of two pipes with radius r that are joined at a right angle as shown. Intersection of two pipes We’ll utilize Cavaliere’s Principle which states that the volume of a solid is the http://mathworld.wolf ram.com/SteinmetzS olid.html integral of the cross-sectional area function A(x), i.e. b Volume = ∫ cross-sectional areas dx = ∫ A(x) dx. x=a x=a b
  • 81.
    The Volume ofthe Intersection of Two Pipes Example B. Find the volume of the intersection of two pipes with radius r that are joined at a right angle as shown. We’ll utilize Cavaliere’s Principle which states that the volume of a solid is the Intersection of two pipes http://mathworld.wolf ram.com/SteinmetzS olid.html integral of the cross-sectional area function A(x), i.e. b b Volume = ∫ cross-sectional areas dx = ∫ A(x) dx. x=a x=a Here is a better view at the solid and it’s cross– sections in question (see pipesvol.svr).
  • 82.
    The Volume ofthe Intersection of Two Pipes Example B. Find the volume of the intersection of two pipes with radius r that are joined at a right angle as shown. Intersection of two pipes We’ll utilize Cavaliere’s Principle which states that the volume of a solid is the http://mathworld.wolf ram.com/SteinmetzS olid.html integral of the cross-sectional area function A(x), i.e. b b Volume = ∫ cross-sectional areas dx = ∫ A(x) dx. x=a x=a Here is a better view at the solid and it’s cross– sections in question (see pipesvol.svr). We see that the cross–sectional areas are squares.
  • 83.
    The Volume ofthe Intersection of Two Pipes We will find the volume of the top half and double it for the final answer.
  • 84.
    The Volume ofthe Intersection of Two Pipes We will find the volume of the top half and double it for the final answer. As we observed, the horizontal cross–sections are squares
  • 85.
    The Volume ofthe Intersection of Two Pipes We will find the volume of the top half and double it for the final answer. s As we observed, the horizontal cross–sections are squares whose sides s are cross–sectional lengths of the circle. s
  • 86.
    The Volume ofthe Intersection of Two Pipes We will find the volume of the top half and double it for the final answer. As we observed, the horizontal cross–sections are squares whose sides s are cross–sectional lengths of the circle. Let x and r be as shown. s s r x
  • 87.
    The Volume ofthe Intersection of Two Pipes We will find the volume of the top half and double it for the final answer. As we observed, the horizontal cross–sections are squares whose sides s are cross–sectional lengths of the circle. Let x and r be as shown. s s r x r x
  • 88.
    The Volume ofthe Intersection of Two Pipes We will find the volume of the top half and double it for the final answer. As we observed, the horizontal cross–sections are squares whose sides s are cross–sectional lengths of the circle. Let x and r be as shown. Hence s/2 = √r2 – x2 s s r x r x
  • 89.
    The Volume ofthe Intersection of Two Pipes We will find the volume of the top half and double it for the final answer. As we observed, the horizontal cross–sections are squares whose sides s are cross–sectional lengths of the circle. Let x and r be as shown. Hence s/2 = √r2 – x2 or that s s r x s = 2√r2 – x2 where x goes from 0 to r. r x
  • 90.
    The Volume ofthe Intersection of Two Pipes We will find the volume of the top half and double it for the final answer. As we observed, the horizontal cross–sections are squares whose sides s are cross–sectional lengths of the circle. Let x and r be as shown. Hence s/2 = √r2 – x2 or that A(x) = 4(r2 – x2) s s r x r x s = 2√r2 – x2 where x goes from 0 to r. Therefore the cross–sectional area function is A(x) = s2 = 4(r2 – x2).
  • 91.
    The Volume ofthe Intersection of Two Pipes So the volume of the top is r ∫ 4(r2 – x2) dx 0 b s r x x=a s A(x) = 4(r2 – x2) s = 2√r2 – x2 r x
  • 92.
    The Volume ofthe Intersection of Two Pipes So the volume of the top is r ∫ 4(r2 – x2) dx 0 b x=a s s s = 2√r2 – x2 r x r = 4∫ (r2 – x2) dx 0 A(x) = 4(r2 – x2) r x
  • 93.
    The Volume ofthe Intersection of Two Pipes So the volume of the top is r ∫ 4(r2 – x2) dx 0 b x=a s s s = 2√r2 – x2 r x r = 4∫ (r2 – x2) dx 0 = 4(r2x – x3/3) | r x=0 A(x) = 4(r2 – x2) r x
  • 94.
    The Volume ofthe Intersection of Two Pipes So the volume of the top is r ∫ 4(r2 – x2) dx 0 b x=a s s s = 2√r2 – x2 r x r = 4∫ (r2 – x2) dx 0 = 4(r2x – x3/3) | r x=0 = 4(2r3/3) = 8r3/3 A(x) = 4(r2 – x2) r x
  • 95.
    The Volume ofthe Intersection of Two Pipes So the volume of the top is r ∫ 4(r2 – x2) dx 0 b x=a s s s = 2√r2 – x2 r x r = 4∫ (r2 – x2) dx 0 = 4(r2x – x3/3) | r x=0 = 4(2r3/3) = 8r3/3 Therefore the entire volume of the intersection is twice the above or 16r3/3. A(x) = 4(r2 – x2) r x
  • 96.
    The Volume ofthe Intersection of Two Pipes So the volume of the top is r ∫ 4(r2 – x2) dx 0 b x=a s s s = 2√r2 – x2 r x r = 4∫ (r2 – x2) dx 0 = 4(r2x – x3/3) | r x=0 = 4(2r3/3) = 8r3/3 Therefore the entire volume of the intersection is twice the above or 16r3/3. A(x) = 4(r2 – x2) r x Cavalieri’s Principle gives us the mathematical meaning of the term “volumes”. Starting with 1D objects – line segments, we define their “volumes” to be just the lengths of the segments.
  • 97.
    The Volume ofthe Intersection of Two Pipes By Cavalerie’s Principle, the “volumes” of 2D regions, i.e. areas, are the integrals of the cross– sectional–length functions L(x) which are “volumes” of 1D line segments.
  • 98.
    The Volume ofthe Intersection of Two Pipes By Cavalerie’s Principle, the “volumes” of 2D regions, i.e. areas, are the integrals of the cross– sectional–length functions L(x) which are “volumes” of 1D line segments. Similarly, the volumes of 3D solids are the integrals of the cross–sectional–area functions A(x) which are “volumes” of 2D regions.
  • 99.
    The Volume ofthe Intersection of Two Pipes By Cavalerie’s Principle, the “volumes” of 2D regions, i.e. areas, are the integrals of the cross– sectional–length functions L(x) which are “volumes” of 1D line segments. Similarly, the volumes of 3D solids are the integrals of the cross–sectional–area functions A(x) which are “volumes” of 2D regions. Continuing in this manner, we define algebraically the term “volume” in the “higher dimensional spaces” via Cavalerie’s Principle,
  • 100.
    The Volume ofthe Intersection of Two Pipes By Cavalerie’s Principle, the “volumes” of 2D regions, i.e. areas, are the integrals of the cross– sectional–length functions L(x) which are “volumes” of 1D line segments. Similarly, the volumes of 3D solids are the integrals of the cross–sectional–area functions A(x) which are “volumes” of 2D regions. Continuing in this manner, we define algebraically the term “volume” in the “higher dimensional spaces” via Cavalerie’s Principle, i.e. volumes are the integrals of the cross–sectional functions which are “volumes” in one less dimensional space.