Finding the area, two polar curves given

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Homework Help Overview

The problem involves finding the area of the region that is inside the polar curve r=5cos(θ) and outside the curve r=4-3cos(θ). This falls under the subject area of polar coordinates and calculus.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss setting the polar equations equal to find intersection points and express confusion over the area calculation. There are attempts to derive the integral needed for the area, with some participants questioning the correctness of their integration steps.

Discussion Status

There is ongoing exploration of the integral setup and evaluation, with participants providing guidance on correcting the integral expression and discussing the need for proper expansion of terms. Multiple interpretations of the integral and its evaluation are being explored, but no consensus has been reached on the correct approach yet.

Contextual Notes

Participants mention issues with integrating and evaluating the integral, as well as confusion over the signs of the area calculated. There is an emphasis on ensuring the integral is set up correctly to avoid negative values for area.

RKOwens4
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Homework Statement



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


Homework Equations





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?
 
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RKOwens4 said:

Homework Statement



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


Homework Equations





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


r=5cos(theta) so radius is 5/2
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)
 
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.
 
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.
 
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?
 
How did you go from
[tex]\int_0^{\pi/3}\left((5cos(\theta))^2 - (4 - 3cos(\theta))^2\right)d\theta[/tex]

to 2sin(2*theta)?

You can see my LaTeX code by clicking the integral expression above.
 
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.
 
OK, I see what your mistake was. You misinterpreted the integral to be
[tex]\int_0^{\pi/3}\left(5cos^2(\theta) - (4 - 3cos^2(\theta))\right)d\theta[/tex]

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
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
RKOwens4 said:
Hmmm, so is 8cos^2(theta)-4 the integral?
No, it is not. Try to read what I write more carefully. The right integral is in post #7.
RKOwens4 said:
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.
12 and -12 are entirely different numbers.
 
  • #12
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
RKOwens4 said:
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.
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.
RKOwens4 said:
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.

If you get a negative value for an area, then you have set up the integral incorrectly. Area is always nonnegative.
 
  • #14
Ah, I gotcha. Thanks.
 

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