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


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

Ad blocker interference detected!

Wikia is a free-to-use site that makes money from advertising. We have a modified experience for viewers using ad blockers

Wikia is not accessible if you’ve made further modifications. Remove the custom ad blocker rule(s) and the page will load as expected.