# Evaluate this surface Integral

1. Aug 15, 2012

### unscientific

1. The problem statement, all variables and given/known data

The problem is attached in the picture.

3. The attempt at a solution

The suggested solution went straight into the hardcore integration. I was trying a different approach by changing the variables (x,y) into (u,v) which appear to make the integration much easier...

The Jacobian was found to be J = p

Not sure where I went wrong!

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2. Aug 15, 2012

### clamtrox

Where did you get the upper limit for ρ integral? Now you're integrating over a circle, not a rectangle.

3. Aug 15, 2012

### unscientific

Hmm if you draw a rectangle with lengths ab, the p ranges from 0 to maximum of √(a2+b2) which is at the corner.

The angle ranges from 0 to ∏/2..

4. Aug 15, 2012

### HallsofIvy

Staff Emeritus
You cannot set the $\rho$ and $\theta$ limits independently. The upper limit on $\rho$ will be a rather complicated function of $\theta$.

5. Aug 15, 2012

### clamtrox

This is true, but if you integrate ρ from 0 to √(a2+b2) for each value of the angle, you are integrating over a circle. Actually the calculation is probably a bit simpler anyway in polar coordinates, so you should definitely work it out. The easiest way is to split the angular integral into two parts: consider first $\tan(\theta) < b/a$. You should be able to find a simple enough expression for the upper bound of ρ in this region.

6. Aug 15, 2012

### unscientific

Not sure how to do it... I know the limit of ∅ is from 0 to ∏/2 but I don't even know the limits of p..

7. Aug 15, 2012

### clamtrox

You need to calculate the distance from the rectangle's corner to an arbitrary point on an opposing edge. Can you do that?

8. Aug 15, 2012

### unscientific

If tan ∅ < (b/a) then p = a/cos ∅.

If tan ∅ > (b/a) then p = b/sin ∅.

If tan ∅ = (b/a) then p = √a2+b2

9. Aug 15, 2012

### clamtrox

Perfect. Those are your integration limits for the radius: now you just need to split the angle integral into
$$\int_0^{\pi/2} d\phi = \int_0^{\arctan(b/a)} d\phi + \int_{\arctan(b/a)}^{\pi/2} d\phi$$ plug in the radius limits and integrate.

10. Aug 16, 2012

### unscientific

Nice. But this really looks more complex than doing it the 'hardcore' way..

11. Aug 16, 2012

### clamtrox

I'm sure it does, if you're not used to changing variables in multidimensional integrals. Did you actually do the calculation? I think you would have found that the integrals you need to compute in spherical coordinates are a lot easier than those in cartesian coordinates. For the first term, you'd get
$$\int_0^{\arctan(b/a)} d\phi \int_0^{a/\cos(\phi)} dr \frac{r^2 \cos(\phi)}{r^2} = a \int_0^{\arctan(b/a)} d\phi = a \arctan(b/a)$$