$\displaystyle \int_0^2 r^3 \sqrt{4-r^2} dr$
Not sure how to integrate this... I tried a u substitution for $\displaystyle 4-r^2$ but couldn't find anything to replace $\displaystyle r^3$ with.
$\displaystyle \displaystyle \begin{align*} \int{r^3\sqrt{4 - r^2}\,dr} = -\frac{1}{2}\int{-2r\cdot r^2\sqrt{4 - r^2}\,dr} \end{align*} $
Let $\displaystyle \displaystyle \begin{align*} u = 4 - r^2 \implies du = -2r\,dr \end{align*} $ and the integral becomes
$\displaystyle \displaystyle \begin{align*} -\frac{1}{2}\int{-2r\cdot r^2\sqrt{4 - r^2}\,dr} &= -\frac{1}{2}\int{\left(4 - u\right)\sqrt{u}\,du} \\ &= -\frac{1}{2}\int{4u^{\frac{1}{2}} - u^{\frac{3}{2}}\,du} \\ &= -\frac{1}{2}\left(\frac{4u^{\frac{3}{2}}}{\frac{3}{ 2}} - \frac{u^{\frac{5}{2}}}{\frac{5}{2}}\right) + C \end{align*} $
Now after you change the terminals to u terminals and insert them, you will have the answer for the definite integral.