I have this following triple integral:

$$\int^{2}_{2}\int^{\sqrt{4-x^2}}_{-\sqrt{4-x^2}}\int^2_{\sqrt{x^2+y^2}}(x^2 + y^2)\,dz\,dx\,dy=10.0531$$

I am required to convert this system into cylindrical coordinates.

So far, I have deduced that $\sqrt{x^2 + y^2}$ for the lower bounds of $dz$ can be represented as radius $r$ for $dr$, but I am having a hard time figuring out what the bounds of $dy$ will become when it is converted into $dTheta$. I know those bounds represent a half circle split by the y-axis.

Also, it seems that after converting $dz$ to $dr$ the function becomes $r^3$ instead of $r^2$. Why is this?

  • $\begingroup$ Start by drawing a picture of the region of integration in the $xy$ plane (i.e. for $-2\le x\le 2$ and $-\sqrt{4-x^2} \le y\le \sqrt{4-x^2}$. What is that region? It should become clear what the range of $\theta$ is. $\endgroup$ – rogerl Nov 2 '17 at 18:40
  • $\begingroup$ For your second question, remember that when converting to polar coordinates you must introduce a factor of $r$. $\endgroup$ – rogerl Nov 2 '17 at 18:47
  • $\begingroup$ @rogerl Sorry, I did not see your comments till I've posted my answer $\endgroup$ – Andrei Nov 2 '17 at 18:49

The first question is the integration range. You need to draw the region $-2\le x\le2$ and $-\sqrt{4-x^2}\le y \le\sqrt{4-x^2}$. This is a disk (hint: that's why you are asked to calculate in cylindrical coordinates). Then $0\le r\le2$ and $0\le\theta\le 2\pi$.

For the second question, note that $dx dy$ becomes $r dr d\theta$. You have an extra $r$ from the Jacobian of the coordinate transformation.


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