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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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1  
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

2 Answers 2

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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This looked like an interesting academic exercise to work through, even though making the variable substitution increases the amount of effort required. The original region of integration looks like this:

enter image description here

and the integration to be performed works pretty easily over it:

$$ \int_0^{\pi/2}\int_0^{\pi/2} \cos\ \theta \ \cos \ \phi \ \ d\theta \ d\phi \ \ = \ \ \int_0^{\pi/2} \sin\ \theta \ \cos \ \phi \ \vert_{\theta = 0}^{\theta = \pi/2} \ \ d\phi \ $$

$$ = \ \ \int_0^{\pi/2} \ ( \ 1 \ - \ 0 \ ) \ \cos \ \phi \ \ \ d\phi \ \ = \ \ \sin \ \phi \ \vert_0^{\pi/2} \ = \ 1 \ - \ 0 \ = \ 1 \ \ . $$

(I've carried out this calculation for comparison against the alternate method.)

Under the variable substitution PanAkry proposes, $ \ \alpha \ = \ \theta+\phi \ \ $ and $ \ \ \beta \ = \ \theta-\phi \ $ , we can transform the lines along which the sides of the original square lie in the $ \ \theta - \phi \ $ plane to obtain

$$ \mathbf{\overline{AB}} \ : \quad \phi \ = \ 0 \ \ \rightarrow \ \ \alpha \ = \ \theta \ + \ 0 \ \ , \ \ \beta \ = \ \theta \ - \ 0 \ \ \Rightarrow \ \ \beta \ = \ \alpha \ \ ; $$

$$ \mathbf{\overline{BC}} \ : \quad \theta \ = \ \frac{\pi}{2} \ \ \rightarrow \ \ \alpha \ = \ \frac{\pi}{2} \ + \ \phi \ \ , \ \ \beta \ = \ \frac{\pi}{2} \ - \ \phi \ \ \Rightarrow \ \ \beta \ = \ \pi \ - \ \alpha \ \ ; $$

$$ \mathbf{\overline{CD}} \ : \quad \phi \ = \ \frac{\pi}{2} \ \ \rightarrow \ \ \alpha \ = \ \theta \ + \ \frac{\pi}{2} \ \ , \ \ \beta \ = \ \theta \ - \ \frac{\pi}{2} \ \ \Rightarrow \ \ \beta \ = \ \alpha \ - \pi \ ; $$

$$ \mathbf{\overline{DA}} \ : \quad \theta \ = \ 0 \ \ \rightarrow \ \ \alpha \ = \ 0 \ + \ \phi \ \ , \ \ \beta \ = \ 0 \ - \ \phi \ \ \Rightarrow \ \ \beta \ = \ -\alpha \ \ . $$

We find that the sides of the transformed square in the $ \ \alpha - \beta \ $ plane indeed fall along straight lines; solving for the intersections of these lines gives the four transformed vertices described by Michael Hardy in his answer. The graph for this region is shown below.

enter image description here

The transformation has the effect of changing the area of the square (which is what its non-unit Jacobian tells us) and also rotates it, which complicates the integration slightly, as we must divide the region in half (this will be the case regardless of the order of integration):

$$ \rightarrow \ \ \int_0^{\pi / 2} \int_{-\alpha}^{\alpha} \ \frac{1}{2} \ ( \ \cos \alpha \ + \ \cos \beta \ ) \ \vert \ \mathfrak{J} \ \vert \ \ d\beta \ d\alpha $$ $$ \ \ + \ \ \int_{\pi / 2}^{\pi} \ \int_{\alpha - \pi}^{\pi - \alpha} \ \frac{1}{2} \ ( \ \cos \alpha \ + \ \cos \beta \ ) \ \vert \ \mathfrak{J} \ \vert\ \ d\beta \ d\alpha \ \ . $$

We can compute the Jacobian from (for example)

$$ \mathfrak{J}^{-1} \ \ = \ \ \left[\begin{array}{cc}\frac{\partial \alpha}{\partial \theta}&\frac{\partial \alpha}{\partial \phi}\\\frac{\partial \beta}{\partial \theta}&\frac{\partial \beta}{\partial \phi}\end{array}\right] \ \ = \ \ \left[\begin{array}{cc}1& \ 1\\1&-1\end{array}\right] \ \ , \ \ \det \ \mathfrak{J}^{-1} \ = \ -2 $$

$$ \mathfrak{J} \ \ = \ \ \frac{1}{\det \ \mathfrak{J}^{-1}} \ \left[\begin{array}{cc}-1& \ -1\\-1& \ 1\end{array}\right] \ \ = \ \ \left[\begin{array}{cc}\frac{1}{2}& \ \frac{1}{2}\\\frac{1}{2}&-\frac{1}{2}\end{array}\right] \ \ \Rightarrow \ \ \det \ \mathfrak{J} \ = \ -\frac{1}{2} \ \ . $$

(It may be noted that since the determinant of the Jacobian is negative, the orientation of the transformed square is reversed: the circulation around the transformed vertices $ \ A \ 'B \ 'C \ 'D \ ' \ $ is now clockwise.)

Carrying out the integration produces

$$ \frac{1}{2} \cdot \frac{1}{2} \ \int_0^{\pi / 2} \int_{-\alpha}^{\alpha} \ ( \ \cos \alpha \ + \ \cos \beta \ ) \ \ d\beta \ d\alpha $$ $$\ \ + \ \ \frac{1}{2} \cdot \frac{1}{2} \ \int_{\pi / 2}^{\pi} \ \int_{\alpha - \pi}^{\pi - \alpha} \ ( \ \cos \alpha \ + \ \cos \beta \ ) \ \ d\beta \ d\alpha \ \ $$

$$ = \ \ \frac{1}{4} \ \left[ \ \int_0^{\pi / 2} \ ( \ \beta \ \cos \alpha \ + \ \sin \beta \ ) \ \vert_{-\alpha}^{\alpha} \ \ d\alpha \ \ + \ \ \int_{\pi / 2}^{\pi} \ ( \ \beta \ \cos \alpha \ + \ \sin \beta \ ) \ \vert_{\alpha - \pi}^{\pi - \alpha} \ \ d\alpha \ \right] \ $$

$$ = \ \ \frac{1}{4} \ \left[ \ \int_0^{\pi / 2} \ 2 \ ( \ \alpha \ \cos \alpha \ + \ \sin \alpha \ ) \ \ d\alpha \ \ + \ \ \int_{\pi / 2}^{\pi} \ 2 \ ( \ [\pi - \alpha] \ \cos \alpha \ + \ \sin [\pi - \alpha] \ ) \ \ d\alpha \ \right] $$

$$ = \ \ \frac{1}{2} \ \left[ \ ( \ \alpha \ \sin \alpha \ + \ \cos \alpha \ + \ [- \cos \alpha ] \ ) \ \vert_0^{\pi / 2} \quad + \ ( \ \pi \ \sin \alpha \ - \ [ \ \alpha \ \sin \alpha \ + \ \cos \alpha \ ] \ + \ [ -\cos \alpha ] \ ) \ \vert_{\pi / 2}^{\pi} \ \right] $$

$$ = \ \ \frac{1}{2} \ \left[ \ ( \ \alpha \ \sin \alpha \ ) \ \vert_0^{\pi / 2} \quad + \ ( \ \pi \ \sin \alpha \ - \ \alpha \ \sin \alpha \ - \ 2 \ \cos \alpha \ ] \ ) \ \vert_{\pi / 2}^{\pi} \ \right] $$

$$ = \ \ \frac{1}{2} \ \left[ \ \left( \ \frac{\pi}{2} \ \sin \frac{\pi}{2} \ - \ 0 \ \right) \ + \ \left( \ 0 \ - \ 0 \ - \ 2 \ \cos \pi \ - \ \pi \ \sin \frac{\pi}{2} \ + \ \frac{\pi}{2} \ \sin \frac{\pi}{2} \ + \ 0 \ \right) \ \right] $$

$$ = \ \ \frac{1}{2} \ \left[ \ \frac{\pi}{2} \cdot 1 \ - \ 2 \ (-1) \ - \ \pi \cdot 1 \ + \ \frac{\pi}{2} \cdot 1 \ \right] \ = \ 1 \ \ . $$

So we recover the value of the original integral, having made use of a "product-to-sum" trigonometric identity, albeit with somewhat more struggle (I wouldn't recommend taking this route on an exam...).

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