Quick question on Polar Coordinates.

When evaluating the double integral and changing variables, I'm not sure if the limits are correct.

The question is as follows:

Evaluate $$\int\!\!\!\int_D xy\sqrt{x^2 + y^2}\,dxdy $$where $D = \{(x,y) \mid 1 \leq x^2 + y^2 \leq 4,\ x \geq 0,\ y \geq 0\}$

So my question is when I change to polar coordinates, is the limit for the integral with respect to r from 1 to 2 or 1 to 4?

Instinctively, I would say it's 1 to 4 but the answer given out by the lecturer (which does not have all the steps) has the limts at 1 to 2.

Is it maybe because $x^2 + y^2 = a^2$?

Note: I have the new integral, in terms of r and $\theta$ as:


  • 2
    $\begingroup$ Since $r=\sqrt{x^2+y^2}$, if $1\leq x^2+y^2\leq 4$, then $1\leq r^2\leq 4$, so $1\leq r\leq 2$ (with appropriate bounds for $\theta$. $\endgroup$ Apr 1, 2012 at 2:07
  • $\begingroup$ Thanks for that. It makes perfect sense now. Should of seen it. $\endgroup$ Apr 1, 2012 at 2:17

1 Answer 1


Look at the actual variables in your integral. You have $\sqrt{x^2+y^2}=\sqrt{r^2}=|r|$, $x=r\cos\theta$, and $y=r\sin\theta$, and you have $dxdy=r dr d\theta$, so your integrand must be $|r|r^3 \sin\theta\cos\theta dr d\theta$. Now look at the region over which you’re integrating:

$$\begin{align*}&1\le x^2+y^2\le 4\;,\tag{1}\\ &x\ge 0\;,\tag{2}\\ &y\ge 0\;.\tag{e} \end{align*}$$

What limits on $r$ and $\theta$ describe this region? $(1)$ says that $1\le r^2\le 4$, so $1\le |r|\le 2$. $(2)$ and $(3)$ say that the region is limited to the first quadrant. If you take $0\le\theta\le\pi/2$, you stay in the first quadrant provided that you also keep $r\ge 0$. Thus, $(1)-(3)$ translate to

$$\begin{align*} &1\le r\le 2\;,\\\\ &0\le\theta\le\frac{\pi}2\;. \end{align*}$$

Your integral should therefore be $$\int_0^{\pi/2}\int_1^2 r^4\sin\theta \cos\theta dr d\theta\;.$$

  • $\begingroup$ Thanks, I got the limits for the outer integral, but my brain was not conceptualising the limits w.r.t r properly. $\endgroup$ Apr 1, 2012 at 2:19
  • $\begingroup$ If x is positive and y is negative, isn't that 4th quadrant? $\endgroup$
    – Mike
    Apr 1, 2012 at 6:12
  • $\begingroup$ Just edited the question, I think I must of changed it by mistake in an earlier edit. Y should be +ive. $\endgroup$ Apr 1, 2012 at 10:58

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