I am trying to convert the integral below into Cartesian form. However, I'm having trouble finding what the limits should be. The integral should evaluate to $\frac12\pi(b^4-a^4)$. For background, it's using Green's theorem to relate $\frac12\pi(b^4-a^4)$ to an area integral over the region $a^2\le x^2 + y^2\le b^2$



  • $\begingroup$ If $r$ is ranging from one constant to another, that means the distance from the origin is fixed as the angle varies, indicating that $r=a$ is the equation for a circle of radius $a$. So if $r$ is varying over $[a,b]$, the region is an annulus (ring region between two circles of differing radii with the same center). $\endgroup$ – user170231 Feb 10 '15 at 15:52
  • $\begingroup$ The region is not convex or even simply connected, so you will probably have to break it up into several integrals. The region is an annulus. By symmetry of the integrand, you could first rewrite it as $4\cdot\int_0^{\pi/2}\int_a^b r^3\;dr\;d\theta$. Then you could rewrite as a single integral in Cartesian form. $\endgroup$ – MPW Feb 10 '15 at 15:53
  • $\begingroup$ You're halfway to determining that with the inequality you've provided. Solve for $y$. $\endgroup$ – user170231 Feb 10 '15 at 15:53
  • $\begingroup$ user170231, I don't think that works out right $\endgroup$ – ThanksABundle Feb 10 '15 at 15:54
  • $\begingroup$ $a^2\le x^2+y^2\le b^2~~\implies~~\sqrt{a^2-x^2}\le |y|\le\sqrt{b^2-x^2}$. In the top half of the annulus, $y$ is bounded between the "greater" half-circle, $\sqrt{b^2-x^2}$ and the "lesser" half-circle $\sqrt{a^2-x^2}$ (assuming $a<b$). Similarly, in the lower half of the annulus, you have $-\sqrt{b^2-x^2}\le y\le-\sqrt{a^2-x^2}$. The order of the square roots changes because the $a$-half-circle lies above the $b$-half-circle. $\endgroup$ – user170231 Feb 10 '15 at 15:58

In other words, you want something of the form $$ \int_p^q \int_u^v f(x,y) \;dy\;dx,$$ and the question is what you should put for $p$, $q$, $u$, and $v$.

One problem you will immediately have is that there is a "hole" in your region of integration, namely, the disk of radius $a$ around the origin. To convert this into a problem you can integrate over $x$ and $y$, you have to find some way to exclude that region.

One approach is to break the integral up into multiple parts and take their sum. You would then need at least one region "above" the hole and one "below", and you will make your life easier then if you don't try to extend those regions all the way to the left or right edges of the whole region but rather integrate yet two more regions, one to the "left" of the hole and one to the "right".

Alternatively, you can integrate over the entire disk of radius $b$, and then subtract the integral over the disk of radius $a$.

To integrate over a disk of radius $R$, you can let $x$ vary from $-R$ to $R$. For each $x$, the limits of $y$ are a function of $x$. You can find that function by considering this condition: $$y^2 + x^2 = R^2.$$ You can confirm that the most extreme values of $y$ are the ones that solve $$y_\mathrm{ext}^2 + x^2 = R^2.$$ This equation has two solutions, and you can confirm that they are the correct limits. Don't forget that these limits are actually a function of $x$ when you evaluate the inner integral.


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