# Math Help - Question on Area between Polar Curves

1. ## Question on Area between Polar Curves

Hey everyone,
I have two questions regarding the area of polar curves.

1. Find the area of the region lying the polar curve r=1 + cos(theta), and outside the polar curve r= 2cos(theta)

2. Find the area of the shaded region inside the graph of r= 1+2cos(theta); (the graph shows the top half of the cardioid shaded)

Basically, I know how to solve these problem and how to draw the graphs. What I need help on is finding the limits of integration. For the 1st problem, I picked my limits of integration to be 0 to pi, and then I multiplied the integral by 2 to find the total area. (Is this method correct)
For the 2nd, I solved for r, and got cos(theta)= -1/2. So would my limits of integration be 2pi/3 to 4pi/3?

Any help and feedback appreciated.

Thanks

2. ## Re: Question on Area between Polar Curves

Originally Posted by Beevo
Hey everyone,
I have two questions regarding the area of polar curves.

1. Find the area of the region lying the polar curve r=1 + cos(theta), and outside the polar curve r= 2cos(theta)

2. Find the area of the shaded region inside the graph of r= 1+2cos(theta); (the graph shows the top half of the cardioid shaded)

Basically, I know how to solve these problem and how to draw the graphs. What I need help on is finding the limits of integration. For the 1st problem, I picked my limits of integration to be 0 to pi, and then I multiplied the integral by 2 to find the total area. (Is this method correct)
For the 2nd, I solved for r, and got cos(theta)= -1/2. So would my limits of integration be 2pi/3 to 4pi/3?

Any help and feedback appreciated.

Thanks
In the first one, when you set up your double integral, for the top half, the radii are bounded above by \displaystyle \begin{align*} r = 1 + \cos{\theta} \end{align*} and the radii are bounded below by \displaystyle \begin{align*} r = 2\cos{\theta} \end{align*}. I agree with your bounds for \displaystyle \begin{align*} \theta \end{align*}. So to find the area, your double integral is \displaystyle \begin{align*} A = 2\int_0^{\pi}{\int_{2\cos{\theta}}^{1 + \cos{\theta}}{r\,dr}\,d\theta} \end{align*}.

For part 2, you follow a similar process with the same \displaystyle \begin{align*} \theta \end{align*} limits, and your radii are bounded above by \displaystyle \begin{align*} r = 1 + 2\cos{\theta} \end{align*} and below by \displaystyle \begin{align*} r = 0 \end{align*}, giving your double integral for the area as \displaystyle \begin{align*} A = \int_0^{\pi}{ \int_0^{1 + 2\cos{\theta}}{r\,dr} \,d\theta} \end{align*}.

3. ## Re: Question on Area between Polar Curves

That makes sense, appreciate your help, thanks.

4. ## Re: Question on Area between Polar Curves

Hello, Beevo!

1. Find the area of the region lying the polar curve $r\:=\:1 + \cos\theta$
and outside the polar curve $r\:=\: 2\cos\theta$

Basically, I know how to solve these problem and how to draw the graphs.
What I need help on is finding the limits of integration.
For the 1st problem, I picked my limits of integration to be 0 to pi,
and then I multiplied the integral by 2 to find the total area.
Is this method correct?

Yes . . . good work!

2. Find the area of the shaded region inside the graph of $r \:=\:1 + 2\cos\theta$
(The graph shows the top half of the cardioid shaded.)

For the 2nd, I solved for r, and got cos(theta)= -1/2.
So would my limits of integration be 2pi/3 to 4pi/3? . . No

You solved $r \,=\,0$

You found the angles at which the curve passes through the pole (origin).
These are not the limits of integration. .(Well, probably not.)

There is yet another problem.
This is not a cardioid.

It is a limacon with an internal "loop".

. . The entire upper half?
. . The upper half minus the loop?
. . Just the loop?

5. ## Re: Question on Area between Polar Curves

Originally Posted by Soroban
Hello, Beevo!

Yes . . . good work!

You solved $r \,=\,0$

You found the angles at which the curve passes through the pole (origin).
These are not the limits of integration. .(Well, probably not.)

There is yet another problem.
This is not a cardioid.

It is a limacon with an internal "loop".