Area between 2 Polar Curves

[uberathlete]

Hi everyone. I've come up with a solution for a problem but I'm not very sure if it makes sense. If someone could check it, it would be greatly appreciated.

Please click the link for the problem and my solution:

http://www.geocities.com/uberathlete/index.htm

Thank you!

 

[stapel]

FYI for those who may not be able (or care) to access the remote page and image:

Question: Find the area of the region of the polar grid which lies inside both the circle $\displaystyle r\,=\,\cos{(\theta)}$ and the curve $\displaystyle r\,=\,2\,-\,\cos{(\theta)}$.

My solution:

. . .$\displaystyle \large{A\,= \,2\,\left[{\,\int_0^{\frac{\pi}{3}}{\,\frac{1}{2} \,\left(\,2\,-\,\cos{(\theta)}\,\right)\,}d\theta\,+ \int_{\frac{\pi}{3}}^{\frac{\pi}{2}}{\,\frac{1}{2}\,\left(\,3\cos{(\theta)}\,\right)\,}d\theta} \right]\,}$

 

[uberathlete]

FYI for those who may not be able (or care) to access the remote page and image:

Question: Find the area of the region of the polar grid which lies inside both the circle $\displaystyle r\,=\,\cos{(\theta)}$ and the curve $\displaystyle r\,=\,2\,-\,\cos{(\theta)}$.

My solution:

. . .$\displaystyle \large{A\,= \,2\,\left[{\,\int_0^{\frac{\pi}{3}}{\,\frac{1}{2} \,\left(\,2\,-\,\cos{(\theta)}\,\right)\,}d\theta\,+ \int_{\frac{\pi}{3}}^{\frac{\pi}{2}}{\,\frac{1}{2}\,\left(\,3\cos{(\theta)}\,\right)\,}d\theta} \right]\,}$

Thanks Stapel.

 

[galactus]

deleted.

 

[Unco]

Hi everyone. I've come up with a solution for a problem but I'm not very sure if it makes sense. If someone could check it, it would be greatly appreciated.

Please click the link for the problem and my solution:

http://www.geocities.com/uberathlete/index.htm

Thank you!

Provided you remember that r is squared and to subtract, I agree with you; nice work.

I did it the long way . . .

\(\displaystyle \L \mbox{ A = \frac{1}{2}\int^{\frac{\pi}{3}}_{-\frac{\pi}{3}} \left(2 - \cos{(\theta)}\right)^2 d\theta - \frac{1}{2}\int^{\frac{2\pi}{3}}_{\frac{\pi}{3}} \left(3\cos{(\theta)}\right)^2 d\theta}\)

 

[soroban]

Hello, uberathlete!

Did anyone notice the typo?

Find the area of the region which lies inside both the circle $\displaystyle r\,=\,\cos{(\theta)}$ and the curve $\displaystyle r\,=\,2\,-\,\cos{(\theta)}$

I made a quick sketch ... and chuckled . . . The circle is inside the limacon!

Then I saw a glaring error in your answer . . . you had $\displaystyle 3\cdot\cos\theta$

Stretching my imagination, I conjectured that we're supposed to use $\displaystyle r\,=\,3\cdot\cos\theta$
$\displaystyle \;\;$That would make the problem less silly.

But why are people using two integrals and two sets of limits?

The curves intersect at: $\displaystyle \,\theta\,=\,\pm\frac{\pi}{3}$

Isn't this the area? \(\displaystyle \L\;\;\frac{1}{2}\int^{\;\;\;\frac{\pi}{3}}_{-\frac{\pi}{3}} \left[(2\,-\,\cos\theta)^2\,-\,(3\cdot\cos\theta)^2\right]\,d\theta\)

 

[Unco]

Did anyone notice the typo?

Did you even bother to look at my graph?

The curves intersect at: $\displaystyle \,\theta\,=\,\pm\frac{\pi}{3}$

Correct.

Isn't this the area? \(\displaystyle \L\;\;\frac{1}{2}\int^{\;\;\;\frac{\pi}{3}}_{-\frac{\pi}{3}} \left[(2\,-\,\cos\theta)^2\,-\,(3\cdot\cos\theta)^2\right]\,d\theta\)

Incorrect. From $\displaystyle \theta = -\frac{\pi}{3}$ to $\displaystyle \theta = \frac{\pi}{3}$ the circle goes everywhere except where we want it. That is, outside the limacon.

 

[galactus]

Please pardon my drawing, the larger one is supposed to be the 'limacon'.

If we integrate the 'limacon' from 0 to $\displaystyle \frac{\pi}{3}$ we have the

integral, \(\displaystyle 2(\frac{1}{2}\int_{0}^{\frac{\pi}{3}}(2-cos({\theta}))^{2})d

{\theta}\)...the area in blue.

If we integrate the remaining circular segment from $\displaystyle \frac{\pi}{3}$ to

$\displaystyle \frac{\pi}{2}$, we have the integral: \(\displaystyle 2(\frac{1}{2}\int_{\frac{\pi}{3}}

^{\frac{\pi}{2}}(3cos({\theta}))^{2})d{\theta}\)...the area in red.

Put them together and we have:

$\displaystyle \int_{0}^{\frac{\pi}{3}}(2-cos({\theta}))^{2}d{\theta}+\int_{\frac{\pi}{3}}^{\frac{\pi}{2}}(3cos({\theta}))^{2}d{\theta}$

=$\displaystyle \frac{9{\pi}}{4}-3sqrt{3}=1.87243......$

This seems OK to me. Is my reasoning incorrect?.

 

[Unco]

Yeah, I should have added.