# Homework Help: Finding the area, two polar curves given

1. Jul 6, 2010

### RKOwens4

1. The problem statement, all variables and given/known data

"Find the area of the region which is inside the polar curve r=5cos(theta) and outside the curve r=4-3cos(theta)."

2. Relevant equations

3. The attempt at a solution

I keep coming up with 18.708, but it says that's incorrect. I don't know what I'm doing wrong. Can someone please help?

2. Jul 6, 2010

### Staff: Mentor

We can't help you if you don't show us what you've done.

3. Jul 6, 2010

### RKOwens4

? I did it in my calculus notebook. I don't know how to post a picture of my notebook but I'll try to rewrite my mistakes (if I can remember what was going through my mind at the time).

I set the r's equal to each other and got:

5cos(theta) = 4-3cos(theta)
8cos(theta) = 4
cos(theta) = 1/2
theta=pi/3

area is pi(5/2)^2 or 25pi/4

area = 1/2 integral from 0 to pi/3 of (5cos(theta))^2 - (4-3cos(theta)^2) d(theta)

4. Jul 6, 2010

### RKOwens4

Can someone please help? This is my first time posting here but I thought you could post questions that you needed help on in the homework and coursework questions section.

5. Jul 6, 2010

### Staff: Mentor

The integral you show will give you half the area. There is another intersection point at theta = -pi/3. Because of the symmetry, the area above the horizontal axis is the same as that below the hor. axis, so if you multiply your integral by 2, that should give you what you're looking for.

6. Jul 6, 2010

### RKOwens4

I think I'm having trouble integrating the integral. I get 2sin(2*theta). Does that look right to you? When I calculate it from 0 to pi/3 it's 1.732, which is incorrect, and when I double the answer it's still incorrect. Can you at least tell me if I've integrated it correctly?

7. Jul 7, 2010

### Staff: Mentor

How did you go from
$$\int_0^{\pi/3}\left((5cos(\theta))^2 - (4 - 3cos(\theta))^2\right)d\theta$$

to 2sin(2*theta)?

You can see my LaTeX code by clicking the integral expression above.

8. Jul 7, 2010

### RKOwens4

I'm not sure what I was thinking at the time. I probably just subtracted 3cos(theta)^2 from 5cos(theta)^2 and ended up with 2cos(theta)^2 - 4, then integrated. I honestly don't remember. I know I did something wrong, but can you show me how to correctly integrate it? I'm pretty sure I have the beginning part figured out correctly and can figure out the rest of the problem if I just know how to integrate it.

9. Jul 7, 2010

### Staff: Mentor

OK, I see what your mistake was. You misinterpreted the integral to be
$$\int_0^{\pi/3}\left(5cos^2(\theta) - (4 - 3cos^2(\theta))\right)d\theta$$

That is incorrect since you need to square the 4 - 3cos(theta) part, as I show in the integral in post 7. Also, you should have gotten 8cos^2(theta) - 4, not 2cos^2(theta) - 4.

The only tricky part of the integration is that you need to use a trig identity when working with cos^2(theta), which is cos^2(theta) = (cos(2theta) + 1)/2.

10. Jul 7, 2010

### RKOwens4

Hmmm, so is 8cos^2(theta)-4 the integral? When I use that for 0 to pi/3, I get -6. When I multiply it by 2 I get -12 (or 12), but it says that's incorrect.

11. Jul 7, 2010

### Staff: Mentor

No, it is not. Try to read what I write more carefully. The right integral is in post #7.
12 and -12 are entirely different numbers.

12. Jul 7, 2010

### RKOwens4

Maybe I'm using the word integral incorrectly. What I meant is, the function in post #7... don't you have to take the anti-derivative of that, before you can plug in pi/3 and then zero? What I'm trying to figure out is, what is the anti-derivative? The one I came up with was incorrect.

And yes I know -12 and 12 are two completely different numbers, but if you're talking about area I thought you take the absolute value.

13. Jul 7, 2010

### Staff: Mentor

The integral in post #7 - yes, you need to evaluate this integral, but the whole point of the last several posts is that you have to expand the part with (4 - 3cos(theta))^2, and it doesn't appear that you have done that. After you expand the integrand in post 7, you should be able to find the antiderivative. Keep in mind what I said about the trig identity.
If you get a negative value for an area, then you have set up the integral incorrectly. Area is always nonnegative.

14. Jul 7, 2010

### RKOwens4

Ah, I gotcha. Thanks.