Finding the area, two polar curves given

In summary, the conversation discusses finding the area of a polar curve and the process of setting up the integral. The initial attempt did not take into account the symmetry of the curve, leading to an incorrect answer. The correct integral involves expanding the integrand and using a trig identity. The final answer should be a positive number representing the area of the region.
  • #1
RKOwens4
33
0

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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  • #2
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.
 
  • #3
? 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)
 
  • #4
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
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
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
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.
 
  • #8
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
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.
 

FAQ: Finding the area, two polar curves given

What is the formula for finding the area between two polar curves?

The formula for finding the area between two polar curves is A = 1/2 ∫θ1θ2 [r2(θ)]2 - [r1(θ)]2 dθ, where r1(θ) and r2(θ) are the two polar curves and θ1 and θ2 are the angles at which the curves intersect.

How do you determine the limits of integration for finding the area between two polar curves?

The limits of integration for finding the area between two polar curves are determined by finding the θ values at which the curves intersect. These values will be the limits of integration for the integral in the area formula.

Can you use the same method for finding the area between two polar curves as you would for finding the area between two Cartesian curves?

No, the method for finding the area between two polar curves is different from the method used for finding the area between two Cartesian curves. The formula and limits of integration are different for polar curves.

What is the significance of the 1/2 coefficient in the area formula for polar curves?

The 1/2 coefficient in the area formula for polar curves is necessary because the area between two polar curves is not a simple rectangle or triangle, but a sector of a circle. The 1/2 coefficient accounts for the fact that we are finding the area of a sector, not a full circle.

Can you use a calculator or graphing software to find the area between two polar curves?

Yes, you can use a calculator or graphing software to find the area between two polar curves. Many graphing calculators have a built-in function for finding the area between curves, and graphing software like Desmos also has this capability. However, it is still important to understand the formula and process for finding the area by hand.

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