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Question: By transforming to polar co-ordinates $\iint \frac{(x^2+y^2)}{(x^2+y^2)^{3/2}}dxdy$ evaluate over the region of the circle $x^2+y^2=2ax$ in the first quadrant.
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Subject : Applied Mathematics 2

Topic : Double integration

Difficulty : High

am2(82) • 167 views
 modified 6 months ago by awari.swati831 • 140 written 8 months ago by
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$\text{ The circle } x^2 + y^2 - 2ax = 0 \text { is a circle with center (a,0) and radius a. To change to polar coordinate.} \\$

$x= rcos\theta, y = rsin\theta \\$

$dxdy = rdrd\theta \\$

$\text {The equation of circle becomes r = acos}\theta \\$

$I = \int_{\theta=0}^{\frac{\pi}{2}} \int_{r=0}^{acos\theta} \frac {r^2cos^2\theta - r^2sin^2\theta } {(r^2)^\frac{3}{2} }r dr d\theta \\$

$I = \int_{\theta=0}^{\frac{\pi}{2}} \int_{r=0}^{acos\theta} \frac {r^2cos2\theta} {(r^3)} r dr d\theta \\$

$= \int_{\theta = 0}^{\frac{\pi}{2}} \left [r \right ]_0^{acos\theta} cos2\theta d\theta \\$

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

$= \frac{a}{2} \int_{\theta = 0}^{\frac{\pi}{2}} ( cos3\theta + cos\theta) d\theta \\$

$= \frac{a}{2} \left [\frac{sin3\theta}{3} + sin\theta \right ]_0^{\frac{\pi}{2}} \\$

$= \frac{a}{2} \int_{\theta = 0}^{\frac{\pi}{2}} \left [\frac{1}{3}\frac{sin3\pi}{2} + 1 \right ] \\$

$= \frac{a}{2} \left[ \frac{-1}{3} + 1 \right ] \\$ $= \frac{a}{2} \frac{2}{3} = \frac{a}{3} \$