Double integrals with an absolute value

[spanishmaths]

Homework Statement

Im getting very confused with working out how to integrate the following double integral with an absolute value:
\int^{2}_{-2}\int^{2}_{-2}\left|x^{2}+y^{2}-1\right|dxdy

Homework Equations

The Attempt at a Solution

I know you have to split it down into where it is positive and negative, so when evaluating the inside one (ie with respect to x), we would have three sections, between -2 and 1 , between -1 and 1 (here we would multiply the function by -1) and then between 1 and 2.
This should give a function of y , which we would then integrate with respect to y and evaluate at exactly the same limits.
Is this correct? I have done it and it gives a very funny looking answer, so I thought I'd check here.

Thanks in advance.

 

[HallsofIvy]

No. x^2+ y^2- 1 changes sign when x^2+ y^2= 1, the unit circle. You need to divide this into two sections, yes, but not individual values of x and y.

You want
\int_A\int (1- x^2-y^2) dxdy+ \int_B\int (x^2+ y^2- 1) dxdy
where A is the interior of the unit circle and B is the portion of the square with vertices (2,2), (-2,2), (2,-2), and (2,2) that is outside that circle. The first integral is easy in polar coordinates. Because the second integral has one boundary that is a circle and another that is a square, that is going to be complicated in any coordinate system. I think I would do it in four separate sections:
\int_{x=-2}^{-1}\int_{y= -2}^{2} (x^2+ y^2+ 1) dydx
\int_{x= -1}^1 \int_{y= \sqrt{1- x^2}^2} (x^2+ y^2+ 1)dydx
\int_{x= -1}^1 \int_{y= -2}^{-\sqrt{1- x^2}}(x^2+ y^2+ 1)dydx
\int_{x= 1}^2\int_{y= -2}^2(x^2+ y^2+ 1)dydx

 

[SammyS]

...
You want
\int_A\int (1- x^2-y^2) dxdy+ \int_B\int (x^2+ y^2- 1) dxdy
where A is the interior of the unit circle and B is the portion of the square with vertices
...

\int_{x=-2}^{-1}\int_{y= -2}^{2} (x^2+ y^2+ 1) dydx
\int_{x= -1}^1 \int_{y= \sqrt{1- x^2}^2} (x^2+ y^2+ 1)dydx
\int_{x= -1}^1 \int_{y= -2}^{-\sqrt{1- x^2}}(x^2+ y^2+ 1)dydx
\int_{x= 1}^2\int_{y= -2}^2(x^2+ y^2+ 1)dydx

Those last 4 integrands appear to be in error.

I would use symmetry then integrate only in the first quadrant.

\int^{2}_{-2}\int^{2}_{-2}\left|x^{2}+y^{2}-1\right|dxdy=4\,\int_0^{\,2}\int^{\,2}_{0}\left|x^{2}+y^{2}-1\right|dxdy

You can then split the integral up into the sum of three integrals.

There are a couple of "cute" ways it can be done, also.