# Where am I going Wrong in this Polar Coordinate Conversion?

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?

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.
the integral should be $$\int_{0}^{\pi /2}\int_{0}^{2}r^4drd\theta$$