double integrals in polar coordinates

**demode** · June 21st, 2010, 09:14 PM

And this is the answer they have given us:

They converted to polar coordinates. I've seen this type of problem being solved without conversion to polar coordinates. I tried to evaluate the integral without converting to polar and I ended up with a different answer:

$\int^1_0 \int^{-y+1}_{x=0} 2x-2y dx dy$

$\int^1_0 y^2-2y+1dy = \frac{1}{3}-1+1=\frac{1}{3}$

But my answer does not agree with the model answer! Shouldn't we end up with the same result regardless of whether we use polar or rectangular coordinates? I appreciate any help because I have an exam tomorrow...

**autumn** · June 21st, 2010, 10:57 PM

You left out the r with the jacobian or the r associated with the x or y.
In either case you need two r's in the integral

$x=r\cos\theta$ and $y=r\sin\theta$ while $dxdy=rdrd\theta$

Furthermore, you shoudn't go anywhere near polar, since this is a triangle and your polar bounds are of a circle.

the line x+y=1 converted to polar is hideous.... $r={1\over \sin\theta+\cos\theta}$

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