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Question: Evaluate $\iint r^3\, drd\theta$ over the area included between the circles $r=2sin\theta$ and $r=2cos\theta$
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Subject : Applied Mathematics 2

Topic : Double integration

Difficulty : Medium

am2(82) • 210 views
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modified 6 months ago by gravatar for awari.swati831 awari.swati831140 written 8 months ago by gravatar for smitapn612 smitapn6120
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$ \text{ The point of intersection r =} 2sin\theta \text{ and r = } 2cos\theta \\ $

$ tan\theta = 1 \Longrightarrow \theta = \frac{\pi}{4} \\ $

$ Area_1 = \int_{\theta = 0}^\frac{\pi}{4} \int_{r= 0}^{ 2sin\theta} r dr d\theta\\ $

$ = \int_{\theta = 0}^\frac{\pi}{4} \left[\frac{r^2}{2} \right]_0^{2sin\theta} d\theta \\ $

$ = \frac {1}{2} \int_{\theta = 0}^ {\frac{\pi}{4}} 4 sin^2\theta d\theta \\ $

$ = 2 \int_{\theta = 0}^ {\frac{\pi}{4}} \frac {(1 - cos2\theta)} {2} d\theta \\ $

$ = \frac{2}{2} \left[\theta - \frac {sin2\theta} {2} \right ]_0^\frac{\pi}{4} \\ $

$ = \frac{\pi}{4} - \frac{1}{2} \\ $

$ Area_2 = \int_{\theta = \frac{\pi}{4}}^\frac{\pi}{2} \int_{r= 0}^{acos\theta} r dr d\theta \\ $

$ = \int_{\theta = \frac{\pi}{4}}^\frac{\pi}{2} \left[\frac{r^2}{2} \right]_0^{acos\theta} d\theta \\ $

$ = \frac{1}{2} \int_{\theta = \frac{\pi}{4}}^\frac{\pi}{2} 4 cos^2 \theta d\theta \\ $

$ = 2 \int_{\theta =\frac{\pi}{4} }^ {\frac{\pi}{4}} \frac {(1 + cos2\theta)} {2} d\theta \\ $

$ = \left[\theta + \frac {sin2\theta} {2} \right ]_\frac{\pi}{4}^\frac{\pi}{2} \\ $

$ = \frac{\pi}{2} - \frac{\pi}{4} - \frac{1}{2} \\ $

$ = \frac{\pi}{4} - \frac{1}{2} \\ $

$ \text {Total Area =} Area_1 + Area_2 \\ $

$ = \frac{\pi}{4} - \frac{1}{2} + \frac{\pi}{4} - \frac{1}{2} \\ $

$ = \frac{\pi}{2} - 1 \\ $

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modified 6 months ago by gravatar for awari.swati831 awari.swati831140 written 7 months ago by gravatar for smitapn612 smitapn6120
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