Homework Help: 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!

Attached Files:

File size:
10.4 KB
Views:
110
• variable2.jpg
File size:
39.2 KB
Views:
104
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

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
$$\int_0^{\arctan(b/a)} d\phi \int_0^{a/\cos(\phi)} dr \frac{r^2 \cos(\phi)}{r^2} = a \int_0^{\arctan(b/a)} d\phi = a \arctan(b/a)$$