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Solve the following double integral by converting to polar coordinates first:

$\int_{0}^{2}\int_{0}^{\sqrt{4-x^2}}(x^2+y^2)^{3/2}dydx$

My attempt at a solution:

$\int\int_{R}dydx$(Cartesian) = $\int\int_{R}rdrd\theta$(Polar)

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

$y=\sqrt{4-x^2} $ ---> $x^2 + y^2 = 4 $ ∴ $\int_{r=0}^{2}$

Because we can only use the top half of the circle, $\int_{\theta=0}^{\pi}$

Therefore, the overall integral I arrive at is:

$\int_{0}^{\pi}\int_{0}^{2}(r^2)^{3/2}rdrd\theta$,

Which simplifies down to:

$\int_{0}^{\pi}\int_{0}^{2}r^4drd\theta$

Solving this, I get an answer of $\frac{32\pi}{5}$. The answer in the book, however, is $\frac{16\pi}{5}$. What am I doing wrong? Where am I ending up with an answer twice as big as it should be?

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2 Answers 2

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You should only be using the upper left quadrant of the circle. $x$ ranges from $0$ to $2$; to get the full top half, it would need to range from $-2$ to $2$. I find it usually helps to draw the full region before starting any coordinate transforms.

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  • $\begingroup$ Ohhhh, of course, I didn't take into account the restrictions that the Cartesian coordinates also place on the Polar integration limits. Thank you, this has been a recurring mistake that has made these kinds of problems a nightmare for me! $\endgroup$
    – Mock
    Apr 5, 2015 at 22:34
  • $\begingroup$ You're very welcome. If you're satisfied with the answer, would you mind accepting it? This will mark the question closed and let other uses know it doesn't need their attention. (You're welcome not to if you would like it to get more attention) $\endgroup$
    – Zach Effman
    Apr 5, 2015 at 22:42
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the integral should be $$\int_{0}^{\pi /2}\int_{0}^{2}r^4drd\theta $$

because the region is quarter circle

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