# Thread: Area between Polar Curves

1. ## Area between Polar Curves

Hello, I have the problem: Find the area of the region outside , but inside .
So I need to figure out the limits, so I set the r's equal and got $\sin \theta =\frac{1}{9}$ So I have a lower limit of .11134 and the upper limit would be pi/2 right?
$\frac{1}{2}\int_{.11134}^{\frac{\pi}{2}}(27\sin\th eta)^2-(9+9\sin\theta)^2$
So I got the answer and then I multiplied it by two but I'm getting the wrong answer. Did I set it up right?
Thanks,
Matt

2. You solved incorrectly ...

\begin{aligned} 9 + 9\sin \theta & = 27 \sin \theta \\ 18 \sin \theta & = 9 \\ \sin \theta & = \frac{1}{2} \\ \theta & = \frac{\pi}{6}, \ \frac{5\pi}{6} \end{aligned}

Note for these questions, draw a sketch if possible. You have to make sure that that you don't miss any intersection points and that you are sweeping across the desired enclosed area.

So using formula for finding the area enclosed by the two curves :
\begin{aligned} A & = \frac{1}{2}\int_a^b \left( r_1^2 - r_2^2\right) \ d \theta \\ & = \int_{\frac{\pi}{6}}^{\frac{5\pi}{6}} \left(27 \sin \theta \right)^2 - \left(9 + 9\sin \theta\right)^2 \ d\theta \\ & \ \ \vdots \end{aligned}

3. Thanks! At first I was thinking of 1/2 but I put 1/9 down instead for some reason.