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![Solving Volume of a Cross-Sectional
Solid
•Base: y 1 2
x 2 x
y 2
1
y x2 2 x •Cross Sectional Shapes:
2
•Triangles perpendicular to the x-
axis.
•Interval: [0,4]
x
•Triangle Base: Triangle Height = 2:3
•Triangle Base= (2*(Triangle Height))/
3
1 3
•Area of the Whole Triangle: ( Base * 2)( )
2 4
•Function for Solving the Volume of the
Solid: 4
3 1 2
( x 2 x ) 2 dx
0
3 2](https://image.slidesharecdn.com/solvingvolumesusingcrosssectionalareas-130210213753-phpapp02/75/Solving-volumes-using-cross-sectional-areas-2-2048.jpg)
![Solutions
4
Volume ( Area) dx
0
y 4
3 1 2
1 ( x 2 x ) 2 dx
y x2 2 x 3 2
2 0
4 5
3 1 4
[( ( x 2 x 2 4 x)dx]
3 0 4
x 7 4
3 1 5 4
( x x 2
2x2 )
3 20 7 0
7 7
3 1 4 1 4
[( (4) 5 (4) 2
2(4) ) ( (0) 5
2
(0) 2 2(0) 2 )]
3 20 7 20 7
3 256 512 3 5472 1824 3
( 32 ) ( )
3 5 7 3 35 35
90 .2646 90 .26](https://image.slidesharecdn.com/solvingvolumesusingcrosssectionalareas-130210213753-phpapp02/75/Solving-volumes-using-cross-sectional-areas-3-2048.jpg)
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Solving volumes using cross sectional areas
- 1. SOLVING VOLUMES USING CROSS SECTIONALAREAS AP Calculus AB/ Christy Sohn (11)
- 2. Solving Volume ofa Cross-Sectional Solid •Base: y 1 2 x 2 x y 2 1 y x2 2 x •Cross Sectional Shapes: 2 •Triangles perpendicular to the x- axis. •Interval: [0,4] x •Triangle Base: Triangle Height = 2:3 •Triangle Base= (2*(Triangle Height))/ 3 1 3 •Area of the Whole Triangle: ( Base * 2)( ) 2 4 •Function for Solving the Volume of the Solid: 4 3 1 2 ( x 2 x ) 2 dx 0 3 2
- 3. Solutions 4 Volume ( Area) dx 0 y 4 3 1 2 1 ( x 2 x ) 2 dx y x2 2 x 3 2 2 0 4 5 3 1 4 [( ( x 2 x 2 4 x)dx] 3 0 4 x 7 4 3 1 5 4 ( x x 2 2x2 ) 3 20 7 0 7 7 3 1 4 1 4 [( (4) 5 (4) 2 2(4) ) ( (0) 5 2 (0) 2 2(0) 2 )] 3 20 7 20 7 3 256 512 3 5472 1824 3 ( 32 ) ( ) 3 5 7 3 35 35 90 .2646 90 .26