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 lemniscate is symmetric about both the x -axis and y -axis. Also, the line $\theta =\pi$ passes through the origin. Therefore, it suffices to find the surface area generated by rotating the part of the lemniscate in the first quadrant about the x -axis, and then multiply by 4.

The surface area is given by the formula:

$$2\pi \int_{0}^{a}y\sqrt{1+\left(\frac{dy}{dx}\right)^2}dx$$

where $y=\sqrt{10\cos 2\theta}$, $x=\frac{1}{2}\sin 2\theta$, and $a=\frac{\pi}{4}$.

So we have:

$$\begin{aligned} \frac{dy}{dx}&=\frac{dy}{d\theta}\cdot \frac{d\theta}{dx}\ &=\frac{-5\sin 2\theta}{\sqrt{\cos 2\theta}}\ \end{aligned}$$

Therefore, we can simplify the integral as follows:

$$\begin{aligned} &2\pi\int_{0}^{a}y\sqrt{1+\left(\frac{dy}{dx}\right)^2}dx\ =&2\pi\int_{0}^{a}y\sqrt{1+\left(\frac{-5\sin 2\theta}{\sqrt{\cos 2\theta}}\right)^2}\cdot\frac{1}{2}\cos 2\theta d\theta\ =&\pi\int_{0}^{a}2\sqrt{10}\cos \theta \sqrt{1+\frac{25\sin^2 2\theta}{\cos 2\theta}}d\theta\ =&\pi\int_{0}^{a}2\sqrt{10}\cos \theta \sqrt{\frac{10\cos 2\theta +5}{\cos 2\theta}}d\theta\ =&2\pi\sqrt{10}\int_{0}^{a}\sqrt{2+\frac{5}{\cos 2\theta}}\cos \theta d\theta\ =&2\pi\sqrt{10}\int_{0}^{a}\sqrt{2}\sqrt{1+\tan ^2 \theta}\cdot \frac{\cos \theta}{\sqrt{1-\tan^2 \theta}}d\theta\ =&2\pi\sqrt{10}\int_{0}^{a}{\sqrt{2}}\cdot \frac{\cos \theta}{\cos \theta}d\theta\ =&2\pi\sqrt{10}\int_{0}^{a}{\sqrt{2}}d\theta\ =&2\pi\sqrt{10}\cdot a\cdot {\sqrt{2}}\ =&\pi\sqrt{10}(2-\sqrt {2}) \end{aligned}$$

Thus, the answer is $\boxed{\textbf{(B)}}$.

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 $r ^ { 2 } = 10 \cos 2 \theta$