Triple Integrals in Cylindrical Coordinates — Question 7

Question 7 – Solution

Strategy. At fixed cylindrical radius $r$, the sphere supplies symmetric upper and lower $z$-bounds.

Step 1: Bounds Since $r^2+z^2\le 4$, $$0\le\theta\le 2\pi,\quad 0\le r\le 2,\quad -\sqrt{4-r^2}\le z\le\sqrt{4-r^2}.$$

Step 2: Evaluate $$\begin{align*} I&=\int_0^{2\pi}\int_0^2\int_{-\sqrt{4-r^2}}^{\sqrt{4-r^2}}r^2(r\,dz\,dr\,d\theta)\\ &=4\pi\int_0^2r^3\sqrt{4-r^2}\,dr. \end{align*}$$ With $u=4-r^2$, $r^3dr=-(4-u)du/2$, so $$\int_0^2r^3\sqrt{4-r^2}\,dr =\frac 12\int_0^4(4-u)u^{1/2}du=\frac{64}{15}.$$ Thus $\boxed{I=256\pi/15}$.

Verification The integrand is nonnegative and at most $4$. Since the ball volume is $32\pi/3$, the required bound $0<I<128\pi/3$ holds.