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Find the volume of a cone

Learn how to prove the area of a cone step by step with integral

Math

Cone volume

A cone with height hh and bottom radius rr

R=h−xhrV=∫0hπR2 dx=∫0hπ(h−xhr)2dx=∫0hπr2(1−xh)2dx=∫0hπr2(1−2xh+x2h2)dx=πr2∫0h(1−2hx+1h2x2)dx=πr2∫0h1 dx−πr2∫0h2hx dx+πr2∫0h1h2x2 dx=πr2∫0hdx−πr2h∫0h2x dx+πr2h2∫0hx2 dx=πr2(h)−πr2h(h2)+πr2h2(h33)=πr2h−πr2h+πr2h3=πr2h3.\begin{aligned} R &= \frac{h-x}{h}r \\[6pt] V &= \int_{0}^{h} \pi R^2 \, dx \\[6pt] &= \int_{0}^{h} \pi \left(\frac{h-x}{h}r\right)^2 dx \\[6pt] &= \int_{0}^{h} \pi r^2 \left(1-\frac{x}{h}\right)^2 dx \\[6pt] &= \int_{0}^{h} \pi r^2 \left( 1-\frac{2x}{h}+\frac{x^2}{h^2} \right) dx \\[6pt] &= \pi r^2\int_{0}^{h} \left( 1 - \frac{2}{h}x + \frac{1}{h^2}x^2 \right) dx \\[6pt] &= \pi r^2\int_{0}^{h}1\,dx - \pi r^2\int_{0}^{h}\frac{2}{h}x\,dx + \pi r^2\int_{0}^{h}\frac{1}{h^2}x^2\,dx \\[6pt] &= \pi r^2\int_{0}^{h}dx - \pi\frac{r^2}{h}\int_{0}^{h}2x\,dx + \pi\frac{r^2}{h^2}\int_{0}^{h}x^2\,dx \\[6pt] &= \pi r^2(h) - \pi\frac{r^2}{h}(h^2) + \pi\frac{r^2}{h^2}\left(\frac{h^3}{3}\right) \\[6pt] &= \pi r^2h - \pi r^2h + \frac{\pi r^2h}{3} \\[6pt] &= \frac{\pi r^2h}{3}. \end{aligned}
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