# Polar Coordinants and Double Integrals

1. Oct 17, 2006

### Menisto

The problem is to find the area of the piece enclosed by the intersection of the circles r = sin t and r = cos t.

I tried to set up the integral:

Integral[0 to Pi/4]Integral[Sin[t] to Cos[t]] r dr d@

but this doesn't seem work, I get out .25, and just by eyeballing it, I can tell it is less. It seems tricky because the upper bound r = cos [t] is being traced out between the angles Pi/4 and Pi/2, while the lower bound r = sin [t] is being traced out between the angles 0 and Pi/4.

Last edited: Oct 17, 2006
2. Oct 17, 2006

### Hurkyl

Staff Emeritus
Don't forget you're working in polar coordinates, not rectangular ones. You shouldn't be thinking of vertical lines; you should be thinking of radial lines.

3. Oct 17, 2006

### Menisto

I'm struggling to see how I would describe this region using only a single double integral, or would I need two?

4. Oct 17, 2006

### Hurkyl

Staff Emeritus
Yah; in polar coordinates, I would expect you either to use two integrals or to take advantage of symmetry.

5. Oct 17, 2006

### Menisto

Symmetry: Say if I were to describe the region from the center of the cosine circle. The angle would be from Pi/4 to Pi/2. Now, if I flipped the portion of the sin circle corresponding to these angles, it would again look like the region. Would the radius then go from 0 to Cos[t] - Sin[t]? The answer is much more reasonable, but it still seems like I'm doing something that seems wrong.

6. Oct 17, 2006

### Hurkyl

Staff Emeritus
Nope. Remember, you're looking at radial lines. What are the endpoints of a radial line lying in your shape?

7. Oct 17, 2006

### Menisto

I don't get what you're asking. The radial line of the cosine circle would start at [.5, 0] and go to cos[t]?

Last edited: Oct 17, 2006
8. Oct 17, 2006

### Menisto

I feel so stupid, It's just the double integral of with angles 0 to pi/4 and radius 0 to sin[t], times 2 (symmetry).

9. Oct 17, 2006

### Hurkyl

Staff Emeritus
That sounds right!

10. Oct 17, 2006

### Menisto

Thanks for the help!