Question

Can someone walk me through with evaluating, I've done it over and over and yet I am not getting the same answer as the solutions.


`\int\limits^r_{\sqrt{r^2- (h^2)/(4)} ^`
`4 \pi x √(r^2-x^2) dx`

Answer 1

`\displaystyle\int_{\sqrt{r^2-\frac{h^2}4}}^r4\pi x√(r^2-x^2)\,\mathrm dx`

Let
`u=r^2-x^2`, so that
`-\frac{\mathrm du}2=x\,\mathrm dx`. Then when
`x=\sqrt{r^2-\frac{h^2}4}`, you have
`u=\frac{h^2}4`; and when
`x=r`, you have
`u=0`.

So the integral is equivalent to


`\displaystyle-\frac12\int_{\frac{h^2}4}^04\pi\sqrt u\,\mathrm du`

`\displaystyle2\pi\int_0^{\frac{h^2}4}√(u)\,\mathrm du`

`2\pi*\frac23u^(3/2)\bigg|_(u=0)^(u=h^2/4)`

`\frac{4\pi}3\left(\frac{h^2}4\right)^(3/2)`

`\frac\pi6h^3`

assuming
`h>0`, which is probably the case since the integral appears to represent one that computes the volume of a solid of revolution.