Area and Volume Revisited — Question 6

Question 6 – Solution

Strategy. In cylindrical coordinates the torus is obtained by rotating a radius-$1$ disk centered at $(r,z)=(3,0)$.

Step 1: Bounds

The radial variable ranges from $2$ to $4$, and $$-\sqrt{1-(r-3)^2}\le z\le\sqrt{1-(r-3)^2},\qquad 0\le\theta\le 2\pi.$$

Step 2: Evaluate With $u=r-3$, $$\begin{align*} V&=2\pi\int_2^4 2r\sqrt{1-(r-3)^2}\,dr\\ &=4\pi\int_{-1}^1(u+3)\sqrt{1-u^2}\,du. \end{align*}$$ The $u$-term is odd, while $\int_{-1}^1\sqrt{1-u^2}\,du=\pi/2$. Therefore $$V=4\pi(3)\frac{\pi}{2}=\boxed{6\pi^2}.$$

Verification The generating disk has area $\pi$, and its center travels a circle of circumference $2\pi(3)=6\pi$. Their product is $6\pi^2$, agreeing with the integral.