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Given in my homework I have to compute (by hand) $$\iint\limits_{x^2+y^2\leq 1}(x^2+y^2)\,\mathrm dx\,\,\mathrm dy.$$ My solution so far: Let $f(x,y)=x^2+y^2$ and $K=\{(x,y)\in\mathbb{R}^2:x^2+y^2\leq 1\}$. With the transformation of polar coordinates $\varphi:(r,\phi)\mapsto(r\cdot\cos\phi,r\cdot\sin\phi)=(x,y)$ and the determinant of the Jacobian matrix $|\det D_\varphi|=r$ we can rewrite the set as $K=\{(r\cdot\cos\phi,r\cdot\sin\phi):r\in[0,1],\phi\in[0,2\pi]\}$ and our integrand too as $f(x,y)=x^2+y^2=r^2$. $$\iint\limits_K f(x,y)\,\mathrm dx\,\mathrm dy=\int\limits_0^1\int\limits_0^{2\pi}r^2\cdot |\det D_\varphi|\,\mathrm d\phi\,\mathrm dr=2\pi\int\limits_0^1r^3\,\mathrm dr=\left.2\pi\cdot\frac{1}{4}r^4\right|_0^1=\frac{\pi}{2}.$$

I now want to check the result with a CAS. I am pretty new to Mathematica so I just didn't found any clue on how to enter such an integral for computation. Any help out there?

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In Mathematica:

Integrate[Integrate[x^2 + y^2, {x, -Sqrt[1 - y^2], Sqrt[1 - y^2]}], {y, -1, 1}]

Or, shorter:

Integrate[x^2 + y^2, {y, -1, 1}, {x, -Sqrt[1 - y^2], Sqrt[1 - y^2]}]

The main trick is to calculate the bound on $x$ based on the current value of $y$, which is what you need to make the integration bounds explicit. Indeed, $x_{max}=\sqrt{1-y^2}$. This is something you can do in most integral-calculating math software.

You can also define the region implicitly, see this.

For this specific problem, that would give:

Integrate[(x^2 + y^2) Boole[x^2 + y^2 <= 1], {y, -100, 100}, {x, -100, 100}]

Where the "100" bounds are just to limit the computation. However, Mathematica is even smart enough to calculate:

Integrate[(x^2 + y^2) Boole[x^2 + y^2 <= 1], {y, -Infinity, 
  Infinity}, {x, -Infinity, Infinity}]
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  • $\begingroup$ Rather than writing this particular problem as an iterated integral, I wonder if it could be written as a "pure" double integral. Not all double integrals can be simply written as iterated integrals. $\endgroup$
    – Ragib Zaman
    Jun 7, 2012 at 12:27
  • $\begingroup$ The last version with Integrate[(x^2 + y^2) Boole[x^2 + y^2 <= 1], {y, -Infinity, Infinity}, {x, -Infinity, Infinity}] is awesome. I will stick to the Boole function. $\endgroup$
    – Christian Ivicevic
    Jun 7, 2012 at 12:34
  • $\begingroup$ You're indeed supposed to be using infinite limits when using the Boole[] form of the integral; the assumption is that the integrand is zero outside the boundary described within the Boole[] function, so it all works out. $\endgroup$
    – J. M.'s eventual burnout ♦
    Jun 10, 2012 at 11:48
  • $\begingroup$ For checking a paper-and-pencil evaluation, definitely the method using Boole is the way to go: that way, if you incorrectly described the limits on x in terms of y (or vice versa) by hand, you wouldn't repeat the same error when doing the Mathematica evaluation. $\endgroup$
    – murray
    Jun 10, 2012 at 20:26
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As noted, Boole[] is the go-to method for older versions of Mathematica. Now, thanks to version 10's support for regions, one can easily enter

Integrate[x^2 + y^2, {x, y} ∈ ImplicitRegion[x^2 + y^2 <= 1, {x, y}]]

to obtain the expected answer.

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    $\begingroup$ Or in this case you could use a special region: Integrate[x^2 + y^2, {x, y} ∈ Disk[]]. $\endgroup$
    – Greg Hurst
    Oct 15, 2015 at 22:37

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