Triple Integrals in Cylindrical Coordinates — Question 10

Question 10 – Solution

Strategy. The horizontal plane meets the paraboloid at $r=a$; compute both volume and first $z$-moment before imposing the average.

Step 1: Bounds The solid lies above $z=r^2$ and below $z=a^2$: $$0\le\theta\le 2\pi,\qquad 0\le r\le a, \qquad r^2\le z\le a^2.$$

Step 2: Volume and moment $$\begin{align*} V(a)&=2\pi\int_0^a(a^2-r^2)r\,dr=\frac{\pi a^4}{2},\\ M_z(a)&=2\pi\int_0^a\int_{r^2}^{a^2}zr\,dz\,dr =\frac{\pi a^6}{3}. \end{align*}$$ Therefore $$z_{\mathrm{avg}}=\frac{M_z(a)}{V(a)}=\frac{2a^2}{3}.$$

Step 3: Recover the parameter The condition $2a^2/3=6$ gives $a^2=9$. Since $a>0$, $$\boxed{a=3},\qquad \boxed{V(3)=\frac{81\pi}{2}}.$$

Verification For $a=3$, the solid has $0\le z\le 9$, and the average $6$ lies inside that range. Substitution returns $2(3^2)/3=6$ exactly.