# Help! polar integrals

1. Aug 22, 2006

### sarahisme

2. Aug 22, 2006

3. Aug 22, 2006

### Tom Mattson

Staff Emeritus
Yes, please show the working. If you don't then someone else has to do the problem to check your work, and that isn't the idea of the Homework Help section of the site.

4. Aug 22, 2006

### sarahisme

yep, ok sure thing guys. i'll post my working in a sec... :)

5. Aug 22, 2006

### sarahisme

its part (b) which i am really struggling with....

6. Aug 22, 2006

### Staff: Mentor

a. is correct, and the integrands in the others appear correct, but it would be useful to know how the limits were determined. Please do show your work. Thanks.

7. Aug 22, 2006

### 0rthodontist

Your revised working of b. looks correct. In c., the limits of integration are not correct.

8. Aug 23, 2006

### sarahisme

ok, but can the radius really go from -1/cos(theta) to 1/cos(theta)? (i.e. how can radius be negative?

9. Aug 23, 2006

### sarahisme

as for (c), here is my working...

10. Aug 23, 2006

### 0rthodontist

You're right--I was wrong. The region includes the area where theta goes from -pi/4 to pi/4 and also the area where theta goes from 3pi/4 to 5pi/4. You should really just split the two parts of the region up into two integrals, which can be condensed into one.
theta should only go from -pi/2 to pi/2. If r <= cos theta then cos theta >= 0.

11. Aug 23, 2006

### sarahisme

would you please be able to show me how to do this, i can't seem to do it...what graph should i be looking at to do this (to be able see the bowtie?)

-=sarah

12. Aug 23, 2006

### 0rthodontist

If |y| <= |x| then you should see this as the region that is vertically between the lines y = x and y = -x. If |x| <= 1 this is the vertical stripe going from -1 to 1 on the x axis. The intersection of these regions is the "bowtie." (if you describe it that way then you must know what it is). In the step where you find |tan theta| <= 1, this can also be satisfied when theta is from 3pi/4 to 5pi/4, though personally I'd trust the diagram more.

13. Aug 23, 2006

### sarahisme

ok, i am still a little confused (sorry :()

this is the graph i am looking at in the x-y plane....the regions enclosing the star shapes are the regions we want ot integrate over, right?

Last edited: Aug 23, 2006
14. Aug 23, 2006

### 0rthodontist

Last edited: Aug 23, 2006
15. Aug 23, 2006

### sarahisme

in which case can't we just do this for part (b):

Last edited: Aug 23, 2006
16. Aug 23, 2006

### 0rthodontist

No, because that only covers the right half of the bowtie.

17. Aug 23, 2006

### sarahisme

for part (c) i get a graph like this (which is y = +/- sqrt(x-x^2)

and so we want to integrate

0<=x<=1
0<=y<=sqrt(x-x^2)

i think thats almost right? :S

18. Aug 23, 2006

### sarahisme

yeah, but i put a 2 out the front and so double the area. because the two halfs of the bowtie have the same area?

19. Aug 23, 2006

### 0rthodontist

Oh, I didn't see the 2 before. Yes, that's correct, but not because the halves have the same area--because the function inside depends only on the radius, not the angle.

Now that you've drawn the diagram for part c, what can you say about the limits of integration for the angle?

Last edited: Aug 23, 2006
20. Aug 23, 2006

### sarahisme

ok well here is the limits for the region we want to integrate over in that diagram...

0<=x<=1
0<=y<=sqrt(x-x^2)

0<=x<=1 converts to 0<=r<=1/cos(theta) in polar coords

right?

then

0<=y<=sqrt(x-x^2) converts to .... i am not sure, this is where i get stuck :yuck: