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The volume of a solid with known cross sections can be calculated by taking the definite integral of all the cross sections, with $ A(x) $ being equal to a single section.

$ V=\int\limits_a^b A(x)dx $

For example, suppose we want to find the volume of the solid with each cross section being a circle, with the diameter of each cross section being the distance between $ y=4-2x $ and the x-axis from 0 to 2. Since one cross section will be equal to $ \pi r^2 $ and $ r=2-x $ , a single cross section will have an area of $ \pi(2-x)^2 $ or $ \pi(4-4x+x^2) $ . The volume will therefore be equal to

$ \int\limits_0^2\pi(4-4x+x^2)dx=\pi\int\limits_0^2(4-4x+x^2)dx=\pi\left[4x-2x^2+\dfrac{x^3}{3}\right]_0^2 $
$ =\pi\left(\left(4(2)-2(2)^2+\frac{(2)^3}{3}\right)-\left((4(0)-4(0)^2+\frac{0^3}{3}\right)\right)=\pi\left(8-8+\frac83\right)-0=\frac{8}{3}\pi $

This method can be used to derive geometric formulas. Here, we will find the volume of a pyramid with sides of the base equal to $ b $ and height $ h $ . Since $ V=\int\limits_a^b A(y)dy $ , all that is needed is to find a function of $ y $ to describe the area of a cross section of the pyramid at height $ y $ . Using trigonometry, we can find that $ \frac{h}{\frac{b}{2}}=\frac{y}{\frac{x}{2}} $ . All that is needed is to isolate $ x $ so that $ x=\frac{by}{h} $ and square it (since this function describes the area of a cross section), then integrate $ \frac{b^2y^2}{h^2} $ from $ 0 $ to $ h $ . It is important to remember that $ b $ and $ h $ are constants and must be treated as such.

$ V=\int\limits_0^h\dfrac{b^2y^2}{h^2}dy=\frac{b^2}{h^2}\int\limits_0^h y^2dy=\frac{b^2}{h^2}\left[\frac{y^3}{3}\right]_0^h=\frac{b^2}{h^2}\left(\frac{h^3}{3}-\frac{0^3}{3}\right)=\frac{b^2h^3}{3h^2}-0 $
$ =\frac{b^2h}{3} $
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