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Q16E

Expert-verified
Found in: Page 528

Essential Calculus: Early Transcendentals

Book edition 2nd
Author(s) James Stewart
Pages 830 pages
ISBN 9781133112280

To determine the area enclosed by the one loop of the curve $${r^2} = \sin 2\theta$$.

The area enclosed by one loop of the curve $${r^2} = \sin 2\theta$$ is $$\frac{1}{2}$$.

See the step by step solution

Step 1: Given data

The one loop of the curve is $${r^2} = \sin 2\theta$$.

Step 2: Formula of Area of the Polar region

The area $$A$$ of the polar region $$R$$ is $$A = \int_a^b {\frac{1}{2}} {r^2}d\theta$$.

Step 3: Sketch the polar region enclosed by the curve

The given curve $${r^2} = \sin 2\theta$$ becomes $$r = \pm \sqrt {\sin 2\theta }$$.

Here, the graph of the curve $$r = - \sqrt {\sin 2\theta }$$ is the reflection of the positive part of $$r = \sqrt {\sin 2\theta }$$.

Sketch the polar region enclosed by the polar curve $$r = \sqrt {\sin 2\theta }$$ as shown in the Figure below.

From figure, it is observed that upper loop of the curve is completed from, $$\theta = 0$$ to $$\theta = \frac{\pi }{2}$$.

Step 4: Calculation for the area enclosed by one loop of curve

Since, $$\sin 2\theta$$ cannot negative, it can be concluded that $$\theta$$ varies from $$\theta = 0$$ to $$\theta = \frac{\pi }{2}$$.

Obtain the area of the polar region.

Substitute $$r = \sqrt {\sin 2\theta } ,a = 0$$ and $$b = \frac{\pi }{2}$$ in $$A = \int_a^b {\frac{1}{2}} {r^2}d\theta$$.

\begin{aligned}{l}A = \int_0^{\frac{\pi }{2}} {\frac{1}{2}} {(\sqrt {\sin 2\theta } )^2}d\theta \\A = \frac{1}{2}\int_0^{\frac{\pi }{2}} {(\sin 2\theta )} d\theta = \frac{1}{2}\left( { - \frac{{\cos 2\theta }}{2}} \right)_0^{\frac{\pi }{2}}\\A = - \frac{1}{4}\left( {\cos 2\left( {\frac{\pi }{2}} \right) - \cos 2(0)} \right)\end{aligned}

Simplify further as shown below.

\begin{aligned}{l}A = - \frac{1}{4}( - 1 - 1)\\A = \frac{1}{2}\end{aligned}

Hence, the area enclosed by one loop of the curve $${r^2} = \sin 2\theta$$ is $$\frac{1}{2}$$.