# Thread: Double integral help (possible polar coordinates)

1. ## Double integral help (possible polar coordinates)

So im trying to solve this problem, and im having serious issues setting up the integral. What is given is:
SS (1/(x^2 + y^2)^2) dxdy, where the region is determined by the conditions x^2 + y^2<=1 and x+y>=1.
Im guessing I have to use polar coordinates for this, when I try this I get :
SS (1/r^2cos^2 + r^2sin^2)^2) r drd(theta) since x=rcos and y=rsin and JAcobian is r. and since sin^2+cos^=1 I get:
SS 1/r^3 Which I dont think is right under the boundaries I am using, any help?
(and yes I have tried to draw this unit circle out with the restriction but it does not help)

2. ## Re: Double integral help (possible polar coordinates)

$\displaystyle \int \int \frac{1}{(x^2 + y^2)^2} \,dx \,dy$ over that domain is equal to

$\displaystyle \int \int \frac{1}{r^4} r \, dr \, d \theta$, or $\displaystyle \int \int \frac{1}{r^3} \, dr \, d \theta$ (you're right so far).

Now you need to determine bounds on r and $\displaystyle \theta$. If you draw the picture, you should see that $\displaystyle 0 \le \theta \le \frac{\pi}{2}$. Now what's left is to determine the bounds on r. Clearly, the upper bound for r is 1.

Finding the lower bound is a little tricky, and uses a little trigonometry (unless I overlooked a simpler method). I'll leave that up to you.

3. ## Re: Double integral help (possible polar coordinates)

hah alright, that is my problem then, I know the bound is not 0 because When you integrate you get 1/0, and form the picture it obvious its at the line y=-x+1 but im not sure how no put that in terms of r, maybe just by substituting x=rcos into that equation? but then lower bound for r would be -rcos +1...?

4. ## Re: Double integral help (possible polar coordinates)

Here is a picture:

Consider the triangle formed by the origin, (1,0), and the point where the radius meets the line y = 1 - x. How do you find r (the lower bound) in terms of $\displaystyle \theta$?

5. ## Re: Double integral help (possible polar coordinates)

haha I was somewhat close, r=1/sin+cos
Thanks for the help!

6. ## Re: Double integral help (possible polar coordinates)

Originally Posted by calculuskid1
haha I was somewhat close, r=1/sin+cos
Sorry, I don't think the expression for the lower bound is that simple. Note that the lower bound is the radius measured from the origin (0,0) to the point where r meets the line.

7. ## Re: Double integral help (possible polar coordinates)

So.. looking at your graph, the only thing I can think of it using the Sine law, So r/Sin45 = -x+1/ sin (theta)
Substituting x=rcos (theta) and isolating for r i get r=sin 45/ (sin(theta) +sin(45)cos(theta))

8. ## Re: Double integral help (possible polar coordinates)

Where did you get -x+1 from? I was thinking $\displaystyle \frac{r}{\sin 45^{\circ}} = \frac{1}{\sin(135^{\circ} - \theta)}$ but if your solution works, then go for it.

(Sorry I should probably be using radians, but for this case it doesn't matter too much)

9. ## Re: Double integral help (possible polar coordinates)

Wow that is much simpler! My -x+1 was the equation of the line associated with angle theta. Thank you very much for all your help!