Find the surface area generated by rotating the lemniscate $r ^ { 2 } = 10 \cos 2 \theta$ about the line $\theta = \pi$.
A) $2 \pi \sqrt { 10 }$
B) $2 \pi \sqrt { 10 } ( 2 - \sqrt { 2 } )$
C) $\pi \sqrt { 10 }$
D) $\frac { \pi \sqrt { 10 } } { 10 }$
E) $\frac { \pi } { 10 }$

Question

Quizplus · Accepted Answer

The surface area generated by rotating a polar curve $r(\theta)$ about a line $\theta = \pi$ can be found using the formula for surface area of revolution in polar coordinates. For the lemniscate $r^2 = 10 \cos 2\theta$, the limits of integration are from $-\pi/4$ to $\pi/4$. The correct calculation leads to the surface area $2 \pi \sqrt{10} (2 - \sqrt{2})$.

Find the Surface Area Generated by Rotating the Lemniscate r2=10cos⁡2θr ^ { 2 } = 10 \cos 2 \thetar2=10cos2θ

Find the Surface Area Generated by Rotating the Lemniscate $r ^ { 2 } = 10 \cos 2 \theta$