Double integration: polar coordinates

I get really confused about how to get the limits and sketch the region for double integration when dealing with polar coordinates. For example:

\displaystyle \int_{y=0}^{y=1} \int_{x=\sqrt{1-y^2}}^{x=\sqrt{4-y^2}} xy^{\alpha} e^{\sqrt{x^2 + y^2}} \mathrm{d}x \mathrm{d}y

First point of confusion: when converting to polar coordinates, I always use this Jacobian:

\frac{\partial (x, y)}{\partial (r, \theta)}

. Does that mean when I replace the dxdy in the integral I need to replace it with

drd\theta

or can I choose to do

d\theta dr

?

Second point of confusion: am I right in saying r can't be negative? i.e. from the upper limit of the inner integral, we have x^2 + y^2 = 4, so r^2 = 4. So does that mean r = 2 and never -2?

Third point of confusion: how do you do the theta limits? I get that 2 <= r <= 4 in the inner integral. But then how do you do stuff for theta? I mean, I'd have thought since we have 0 <= y <= 1, it means the we should have the following:

i.e. theta running from 0 to pi and r running from 2 to 4 (which corresponds to 0 <= y <= 1 and 2 <= r <= 4 in my eyes).

But the solutions ignore the negative quadrant and only consider the positive one (the blue bit). Why do they do that?

Moreover, how do I get the limits for theta if it's not 0 to pi?

(edited 12 years ago)

For the first 2 points: yes, you can swap dr d0 with d0 dr (although obviously this may change the form of the region limits). And yes, r >=0 always.

For the theta limits: to be honest I don't see a very nice solution. You need y <=1, so you need

r \sin \theta <=1

. So you need simultaneously 1<=r<=2 and

r \sin \theta <=1

, which I think will require you to divide the integral into two regions.

Are you sure the question is correct (and that you're supposed to use polar coordinates)? I haven't actually done the whole thing, but if you integrate in cartesians,

\int y^{\alpha} x e^{\sqrt{x^2+y^2}}\,dx

isn't too bad to find, and looking at the limits, it should cancel down quite nicely. Conversely, that integral looks like it will be pretty nasty in polar coordinates: y^alpha will become

r^\alpha \sin^\alpha \theta

, which doesn't look like fun.

The limits are going to be horrible,

Reply 2

Swayum

OP

18

Original post by DFranklin

For the first 2 points: yes, you can swap dr d0 with d0 dr (although obviously this may change the form of the region limits). And yes, r >=0 always.

So to double check, I don't need to adjust my Jacobian to

\frac{\partial (x, y)}{\partial (\theta, r)}

if I want to do a dtheta dr integral? I can continue using this Jacobian

\frac{\partial (x, y)}{\partial (r, \theta)}

and put

d\theta dr

?

For the theta limits: to be honest I don't see a very nice solution. You need y <=1, so you need

r \sin \theta <=1

. So you need simultaneously 1<=r<=2 and

r \sin \theta <=1

, which I think will require you to divide the integral into two regions.

Are you sure the question is correct (and that you're supposed to use polar coordinates)? I haven't actually done the whole thing, but if you integrate in cartesians,

\int y^{\alpha} x e^{\sqrt{x^2+y^2}}\,dx

isn't too bad to find, and looking at the limits, it should cancel down quite nicely. Conversely, that integral looks like it will be pretty nasty in polar coordinates: y^alpha will become

r^\alpha \sin^\alpha \theta

, which doesn't look like fun.

The limits are going to be horrible,

Well, the question specifically says to use polar coordinates, so doing it in Cartesian isn't an option. Double checked the question and it's correct as I've posted it. Just out of interest, how would you do this integral

\int y^{\alpha} x e^{\sqrt{x^2+y^2}}\,dx

?

As for the theta limits, the solutions say that 0 <= theta <= arcsin(1/r) and then they've sketched the region as just the positive quadrant in my sketch above (i.e. just the blue bit). I don't get how they're getting just the positive quadrant though.

Their double integral boils down to

\displaystyle \int_{r=1}^{r=2} \int_{\theta = 0}^{\theta = \arcsin(1/r)} (r\cos \theta )(r\sin \theta )^{\alpha}e^rr \mathrm{d}\theta \mathrm{d}r

which isn't too bad because the theta integral is in the form f'(x)(f(x))^n

(edited 12 years ago)

Reply 3

DFranklin

18

Original post by Swayum

So to double check, I don't need to adjust my Jacobian to

\frac{\partial (x, y)}{\partial (\theta, r)}

if I want to do a dtheta dr integral? I can continue using this Jacobian

\frac{\partial (x, y)}{\partial (r, \theta)}

and put

d\theta dr

?

Yes; it's just a double integral (in r, theta), and you can always(*) reverse the order of integration in a double integral.

(*) Well you can unless the integrands are somewhat pathological, and you're unlikely to get this in an applied question.

Just out of interest, how would you do this integral

\int y^{\alpha} x e^{\sqrt{x^2+y^2}}\,dx

?

I think you can make the substitution u = sqrt(x^2+y^2) and it will work out. y^alpha is a constant as far as integrating w.r.t. x.

As for the theta limits, the solutions say that 0 <= theta <= arcsin(1/r) and then they've sketched the region as just the positive quadrant in my sketch above (i.e. just the blue bit). I don't get how they're getting just the positive quadrant though.

Yes, I was thinking of r as a function of theta (which is nasty), but theta as a function of r is easier.

As far as the positive quadrant, in the original integral, x goes from sqrt(1-y^2) to sqrt(4-y^2), and sqrt is always positive (in these questions), so x and y are always positive, so you want the +ve quadrant.

Their double integral boils down to