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Evaluate the integral $\int\int_R(x^2-2y^2) \, dA$ where $R$ is the first quadrant region between the circles of radius $4$ and radius $5$.

I am having troubles with finding the limits of the integral. Thank you!

Michael Hardy
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user131040
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  • You could use the fact that for a circle of radius $r$, the equation is (centered at the origin) $x^2+y^2=r^2$, and then change to polar coordinates. –  Mar 07 '14 at 00:59
  • Hint: $\int\int_R(x^2-2y^2)dA= \int_0^{2\pi}\int_4^5(r^2(1+rsin\theta))rdrd\theta$ and you can do $r^4sin\theta$ by parts. – William Chang Mar 07 '14 at 01:25

1 Answers1

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Just use polar coordinates.

$x^2=r^2\cos^2{\theta};y^2=r^2\sin^2{\theta}$

$x^2-2y^2=r^2(\cos^2{\theta}-2\sin^2{\theta})$

$x^2-2y^2=r^2(1-\sin^2{\theta}-2\sin^2{\theta})=r^2(1-3\sin^2{\theta})$

$x^2-2y^2=r^2(1-\frac{3}{2}+\frac{3}{2}cos2{\theta})=r^2(-\frac{1}{2}+\frac{3}{2}cos2{\theta}))$

$\iint_Rx^2-2y^2\,dx\,dy=-\frac{1}{2}\int_{0}^{\pi/2}\int_{4}^5r^3\,dr\,d\theta +\frac{3}{2}\int_{0}^{\pi/2}\int_{4}^5r^3\cos2{\theta}\,dr\,d\theta$

Note that "first cuadrant" means $\theta$ goes from $0$ to $\pi/2$

Juanjo
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