# Why won't this double integral integrate to 0?

If $-1 \leq x = y \leq 1$, and I am supposed to calculate double integral of $x^2 + y^2$, why won't this equal $0$? I know the answer is $0$ but when I calculate the integral using the limits $-1 \le x \le y$ and$-1 \le y <\le 1$, I do not get $0$.

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This integral isn't equal zero. You have function that more than 0 in every point exept (0,0). It is positive. – Evgeny Savinov Jan 23 '12 at 22:52
Well, he is integrating over a region where $x=y$. – Byron Schmuland Jan 23 '12 at 22:57
Do you mean to be integrating over $-1 \leq x = y \leq 1$ and not $-1 \leq x,y \leq 1$? – mixedmath Jan 23 '12 at 23:13
Yes, the former – lord12 Jan 23 '12 at 23:17
Note the one-dimensional version of the problem: For real $x$'s let $f(x)= 2700000$, what is $\int_E f(x)dx$, where $E=\{0,3/5,1\}$? – AD. Jan 24 '12 at 6:59

As usual, evaluate the double integral by using an iterated integral, integrating first with respect to $x$. Then $x$ goes from $y$ all the way to $\dots y$.

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You are integrating over the set $A=\left\{(x,y)|-1\leq x=y \leq 1\right\}=\left\{(x,x)|-1\leq x \leq 1\right\}$. The integral of any function over $A$ is trivially 0, as $A$ is a set of measure 0 in $\mathbb{R}^2$.

If you are not familiar with measure theory: An easy way to see this is replacing the function you are integrating over with the constant function $f(x,y)=2$, which is $\geq x^2+y^2$ . Therefore also the integral of this function will be $\geq$ the integral you are looking for. The integral of $f$ over any region will give you twice its area of the region and the area of $A$ is 0.

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you're right. edited. – Michalis Jan 24 '12 at 9:45

Here is a way to get zero by actually evaluating the double integral. Note that you are integrating (using a two dimensional notion of area) a function over a one dimensional space, and so as Michalis says, you are integrating over a measure zero space, hence the answer is zero.

The integral is $\int_{-1}^1 \int_x^x x^2+y^2 dy \;dx=\int_{-1}^1 0\; dx = 0$.

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Ok, let's go.

The first variant: let $x=r\cos(w)$ and $y=r\sin(w)$, so $\int\int (x^2+y^2)dxdy=\int\int r^2 r drdw=8\int_0^{\frac{\pi}{4}}dw\int_0^{\frac{1}{\cos(w)}}r^3dr=8\int_0^{\frac{\pi}{4}}\frac{dw}{4\cos^4(w)}=(t=\tan(w))$

=$2\int_0^1(t^2+1)dt=2(\frac{1}{3}+1)=8/3$

The second: if $x=y$, it is really is equal to zero $\int\int (x^2+y^2)dxdy=\int\int r^2 r drdw=\int_{\pi/4}^{\pi/4}dw\int_{2^{1/2}}^{2^{1/2}} r^3 dr+\int_{5\pi/4}^{5\pi/4}dw\int_{2^{1/2}}^{2^{1/2}} r^3 dr=0$

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