Triple Integrals in Cylindrical Coordinates — Question 1

Question 1 – Solution

Strategy. Treat the solid as a quarter-cylinder; radial symmetry makes the integrand $r^2$.

Step 1: Convert Cylindrical coordinates give $$x=r\cos\theta,\quad y=r\sin\theta,\quad x^2+y^2=r^2, \quad dV=r\,dz\,dr\,d\theta.$$ The first-quadrant quarter-cylinder has bounds $$0\le\theta\le\frac\pi 2,\qquad 0\le r\le 2, \qquad -1\le z\le 2.$$

Step 2: Evaluate $$\begin{align*} I&=\int_0^{\pi/2}\int_0^2\int_{-1}^2 r^2(r\,dz\,dr\,d\theta)\\ &=\left(\frac\pi 2\right)(3)\left[\frac{r^4}{4}\right]_0^2 =\boxed{6\pi}. \end{align*}$$

Verification The volume is $(\pi 2^2/4)(3)=3\pi$. On a radius-$2$ disk, the average of $r^2$ is $2^2/2=2$; restricting to one quadrant does not change that radial average. Thus $I=(3\pi)(2)=6\pi$.