Evaluate this surface Integral

[unscientific]

Homework Statement

The problem is attached in the picture.

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!

 

[clamtrox]

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

 

[unscientific]

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

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..

 

[HallsofIvy]

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

 

[clamtrox]

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..

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.

 

[unscientific]

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.

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..

 

[clamtrox]

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..

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

 

[unscientific]

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

If tan ∅ < (b/a) then p = a/cos ∅.

If tan ∅ > (b/a) then p = b/sin ∅.


If tan ∅ = (b/a) then p = √a²+b²

 

[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.

 

[unscientific]

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.

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

 

[clamtrox]

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

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