Triple Integrals in Cylindrical Coordinates — Question 2

Question 2 – Solution

Strategy. In cylindrical coordinates the paraboloid becomes $z=9-r^2$; its intersection with $z=0$ determines the radial endpoint.

Step 1: Geometry Setting $z=0$ gives $r=3$. A meridian half-plane shows the height above each radius.

Thus $0\le\theta\le 2\pi$, $0\le r\le 3$, and $0\le z\le 9-r^2$.

Step 2: Evaluate $$\begin{align*} V&=\int_0^{2\pi}\int_0^3\int_0^{9-r^2}r\,dz\,dr\,d\theta\\ &=2\pi\int_0^3(9r-r^3)\,dr =2\pi\left[\frac 92r^2-\frac 14r^4\right]_0^3 =\boxed{\frac{81\pi}{2}}. \end{align*}$$

Verification The solid lies inside the cylinder of radius $3$ and height $9$, whose volume is $81\pi$. The computed volume is positive and exactly half that containing-cylinder volume.