## Find the square pyramid volume as a function of \(a\) and \(H\) by slicing method

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