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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
 modified 6 months ago by awari.swati831 • 140 written 8 months ago by
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$\text{ The point of intersection r =} 2sin\theta \text{ and r = } 2cos\theta \\$

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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