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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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