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Q16E

Expert-verifiedFound in: Page 528

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 \(\frac{1}{2}\).

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

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

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 \(\theta = \frac{\pi }{2}\).

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 \(\frac{1}{2}\).

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