# Math Help - cardiods

1. ## cardiods

Can anyone help me with this problem?

Find the area shaded by the cardiods r=2+2cos(x) and r=2-2cos(x).

Thanks!

2. Originally Posted by runner21
Can anyone help me with this problem?

Find the area shaded by the cardiods r=2+2cos(x) and r=2-2cos(x).

Thanks!
Is this the intersection of the cardiods or the area individually? Do you know the area formula in polar coordinates $\text{Area}=\frac{1}{2}\int_{\alpha}^{\beta}f^2(\t heta)~d\theta$?

3. Originally Posted by Mathstud28
Is this the intersection of the cardiods or the area individually? Do you know the area formula in polar coordinates $\text{Area}=\frac{1}{2}\int_{\alpha}^{\beta}f^2(\t heta)~d\theta$?
Its the area that they overlap. I know of the $\text{Area}=\frac{1}{2}\int_{\alpha}^{\beta}f^2(\t heta)~d\theta$ formula but how would you set it up so that you were only finding the shared region?

4. Originally Posted by runner21
Its the area that they overlap. I know of the $\text{Area}=\frac{1}{2}\int_{\alpha}^{\beta}f^2(\t heta)~d\theta$ formula but how would you set it up so that you were only finding the shared region?
Start by drawing sketch graphs of each cardioid. Calculate the polar coordinates of where they intersect. Then read this: Pauls Online Notes : Calculus II - Area with Polar Coordinates

5. We have symmetry in all four of the little intersection 'loopies'.

Find the area of one and multiply by 4. That is one way to go about it.

$2\int_{0}^{\frac{\pi}{2}}\left[2+2cos({\theta})\right]^{2}d{\theta}$

6. Originally Posted by galactus
We have symmetry in all four of the little intersection 'loopies'.

Find the area of one and multiply by 4. That is one way to go about it.

$2\int_{0}^{\frac{\pi}{2}}\left[2+2cos({\theta})\right]^{2}d{\theta}$
I hope that you don't mind me using that haha

7. Originally Posted by galactus
We have symmetry in all four of the little intersection 'loopies'.

Find the area of one and multiply by 4. That is one way to go about it.

$2\int_{0}^{\frac{\pi}{2}}\left[2+2cos({\theta})\right]^{2}d{\theta}$
I finally figured it out! Galactus, you were close, and I actually didn't quite get it until I saw your integral, but the equation should read:

$\int_{0}^{\frac{\pi}{2}}\left[2+2cos({\theta})\right]^{2}-\left[2-2cos({\theta})\right]^{2}d{\theta}=16$

Thanks for all your help, everyone!

8. I started out with that, but then noticed 4 way symmetry. If I am wrong, then Soroban is wrong as well. I am not so sure we are wrong.

I also checked this with tech and got the same as Soroban and I. Therefore, I am going with our solution. Even thought the solutions are close.

S.O.S. Mathematics CyberBoard :: View topic - cardiods

9. Originally Posted by runner21
I finally figured it out! Galactus, you were close, and I actually didn't quite get it until I saw your integral, but the equation should read:

$\int_{0}^{\frac{\pi}{2}}\left[2+2cos({\theta})\right]^{2}-\left[2-2cos({\theta})\right]^{2}d{\theta}=16$

Thanks for all your help, everyone!
I've attached a sketch of the area in question. By inspection I would guess that the area is around 2.5.

So I assume that there is a typo or a minor arithmetic mistake ...