Question: Change to polar co-ordinates and evaluate $\int_{0}^{2}\int_{0}^{\sqrt{2x-x^2}}\frac{x\,dy\,dx}{\sqrt{x^2+y^2}}$ .
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Subject : Applied Mathematics 2

Topic : Double integration

Difficulty : Medium

am2(82) • 253 views
 modified 8 months ago by awari.swati831 • 150 written 10 months ago by
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$\text{ The region of integration is from y = 0 to y = } \sqrt{2x-x^2} \Longrightarrow x^2 + y^2 -2x = 0 \\$

$\text{x= 0 to x = 2} \\$

$\text{changing to polar coordinates x = } rcos\theta , y= rsin\theta, dxdy = rdrd\theta \\$

$I = \int_{\theta=0}^{\frac{\pi}{2}} \int_{r=0}^{2cos\theta} \frac {rcos\theta rdr d\theta} {r} \\$

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

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

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

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

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