Triple Integrals in Cylindrical Coordinates — Question 4

Question 4 – Solution

Strategy. Convert the footprint to $r\le 4\cos\theta$ and use the conical height $z=r$.

Step 1: Footprint From $r^2\le 4r\cos\theta$, nonnegative radii satisfy $0\le r\le 4\cos\theta$. This requires $-\pi/2\le\theta\le\pi/2$ and describes the disk centered at $(2,0)$ with radius $2$.

Step 2: Evaluate $$\begin{align*} V&=\int_{-\pi/2}^{\pi/2}\int_0^{4\cos\theta}\int_0^r r\,dz\,dr\,d\theta\\ &=\frac{64}{3}\int_{-\pi/2}^{\pi/2}\cos^3\theta\,d\theta =\frac{64}{3}\left(\frac 43\right) =\boxed{\frac{256}{9}}. \end{align*}$$

Verification At $\theta=\pm\pi/2$ the radial interval collapses to $r=0$; at $\theta=0$ it reaches $r=4$. These are exactly the leftmost and rightmost points of the footprint, so no rays are omitted or duplicated.