Area inside r=6sin(theta) but outside r=3

[RKOwens4]

Homework Statement

The title really says it all. Find the area of the region inside r=6sin(theta) but outside r=3.

Homework Equations

The Attempt at a Solution

I first found the x values where the two curves intersect and came up with .52359877 and 2.61799388. I then integrated 6sin(theta) (or, -6cos(theta)) from .52359877 to 2.61799388. The answer I got was 10.3923, but that's incorrect.

 

[RKOwens4]

Anyone?

 

[LCKurtz]

Homework Statement

The title really says it all. Find the area of the region inside r=6sin(theta) but outside r=3.

Homework Equations

The Attempt at a Solution

I first found the x values where the two curves intersect and came up with .52359877 and 2.61799388. I then integrated 6sin(theta) (or, -6cos(theta)) from .52359877 to 2.61799388. The answer I got was 10.3923, but that's incorrect.

What's with all the decimals? Leave things in terms of pi. I don't believe those are x values. This is a polar coordinate problem, in terms of r and theta. You might start by stating the correct formula for the area between two polar curves expressed in terms of polar coordinates.

 

[RKOwens4]

Okay, I found the formula for the area between two polar curves, which is:

1/2 Integral from B to A of (f2(theta)^2 - f1(theta)^2)

But what do I do about the r=3? And how do I find out what B and A are?

 

[Mark44]

The two polar curves are symmetric about the y-axis, so you can find the area in the first quadrant and double it to get the entire area.

For b, you can use pi/2. For a, find the point of intersection in quadrant 1 of the two curves - use the exact value, not a decimal approximation.

For the outer curve, r = 6 sin(theta). For the inner curve, r = 3. These are the functions you're calling f1 and f2 (not necessarily in order).

This problem is very similar to the other problem you posted.