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.
$\int\limits^r_{\sqrt{r^2- \frac{h^2}{4}} ^$ $4 \pi x \sqrt{r^2-x^2} dx$

Answer 1

$\displaystyle\int_{\sqrt{r^2-\frac{h^2}4}}^r4\pi x\sqrt{r^2-x^2}\,\mathrm dx$

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

So the integral is equivalent to

$\displaystyle-\dfrac12\int_{\frac{h^2}4}^04\pi\sqrt u\,\mathrm du$
$\displaystyle2\pi\int_0^{\frac{h^2}4}\sqrt{u}\,\mathrm du$
$2\pi\times\dfrac23u^{3/2}\bigg|_{u=0}^{u=h^2/4}$
$\dfrac{4\pi}3\left(\dfrac{h^2}4\right)^{3/2}$
$\dfrac\pi6h^3$

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