## Find the square pyramid volume as a function of $$a$$ and $$H$$ by slicing method

Home -> Solved problems -> Volume square pyramid ### Solution

The first step is to find the volume of one slice of the pyramid. The slice is considered as a square cuboid with side $$x$$ and infinitesimally small height $$\text{d}h$$ as shown in blue in the figure below, so we can affirm that the slice’s volume is $$\text{d}v=x^{2}\text{d}h$$. Now to continue we need to find a relationship between $$x$$ and $$h$$ before applying integration . $$h$$ is the distance between the cuboid and the square base of the pyramid. So to do that, we use Thales‘s theorem in the triangle with the orange line segments. We get,$\frac{\frac{x}{\sqrt{2}}}{\frac{a}{\sqrt{2}}}=\frac{H-h}{H}$$\frac{x}{a}=\frac{H-h}{H}$$x=a(\frac{H-h}{H})$ After finding a relationship between $$x$$ and $$h$$ we can calculate the pyramid volume by applying integral between $$0$$ and $$H$$. $$a$$ and $$H$$ are constants. $\text{d}v=x^{2}\text{d}h$$=(a(\frac{H-h}{H}))^{2}\text{d}h$$v=\int_{0}^{H} (a(\frac{H-h}{H}))^{2}\text{d}h$$=(\frac{a}{H})^{2}\int_{0}^{H} (H-h)^{2}\text{d}h$$=(\frac{a}{H})^{2}(-\frac{(H-h)^{3}}{3})|_0^H$$=(\frac{a}{H})^{2}(-\frac{(H-H)^{3}}{3})-(\frac{a}{H})^{2}(-\frac{(H-0)^{3}}{3})$$=-(\frac{a}{H})^{2}(-\frac{H^{3}}{3})$$=\frac{a^{2}H^{3}}{3H^{2}}$$=\frac{1}{3}a^{2}H$
Home -> Solved problems -> Volume square pyramid

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