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I try to understand change of variables for double integrals.

For this reason I try to calculate $\int_0^{\pi/2}\int_0^{\pi/2}\cos\theta\cos{\phi}\,d\theta d\phi$ by using this change $\alpha=\theta+\phi$, $\beta=\theta-\phi$ and $\cos\theta\cos\phi=\frac{1}{2}(\cos\alpha+\cos\beta)$. I repeat I know how to calculate the first integral and this is just an exercise for me. I found that the Jacobian is $J=-\frac{1}{2}$ but I even don't found on which surface I have to integrate!

Does somebody have an idea?

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When $(\theta,\phi)$ is in the rectangle $[0,\pi/2] \times [0,\pi/2]$, where is $(\alpha,\beta)$? –  GEdgar Mar 9 '12 at 16:11
I up-voted this question, so someone must have down-voted it. –  Michael Hardy Mar 11 '12 at 3:21
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1 Answer

When $(\phi,\theta)=(0,0)$ then $(\alpha,\beta)=(0,0)$.

When $(\phi,\theta)=(0,\pi/2)$ then $(\alpha,\beta)=(\pi/2,\pi/2)$.

When $(\phi,\theta)=(\pi/2,0)$ then $(\alpha,\beta)=(\pi/2,-\pi/2)$.

When $(\phi,\theta)=(\pi/2,\pi/2)$ then $(\alpha,\beta)=(\pi,0)$.

Those are the four corners. The edges of the square transform to straight line segments because of the linearity of the transformation. So just draw straight lines between those corners.

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