I'm asked to find the area of the polar region inside $r(\theta)=5(1-\sin(\theta))$ and $r=5$. Below is a plot of the situation. THe black area is the area I want.
I then found the intersection points. So $5=5(1-\sin(\theta))$ or $1=1-\sin(\theta)$ which omplies that $\sin(\theta)=0$ or $\theta=0,\pi,2\pi$.
So the area is $\int_{\pi}^{2\pi} \frac{1}{2} [(5(1-\sin(\theta)))^2-5^2] d\theta=\frac{25\pi}{2}$.
However, the answer should be $\frac{25\pi}{2}+\frac{75\pi}{4}-50$.
Can someone point out my mistake and how I can fix this? Thanks!
Best Answer
I don't quite understand why you though that your integral was the correct answer (as an aside, the integral you were looking at actually is equal to $6.25(8+\pi)$). I feel like you may have a difficulty with the formula we use to find areas bounded by polar curves; if you are having difficulties with it then I will be happy to explain it to you.
The way to find this area is to think of the area split into $2$ regions: the first above the $x$ axis-let's call it $A_1$- and the second below the $x$ axis-let's call it $A_2$.
To find the first area, $A_1$: $$A_1=\frac{1}{2}\int_0^{\pi}25(1-\sin\theta)^2d\theta$$ or note that by symmetry, $$A_1=2\left(\frac{1}{2}\int_0^{\pi/2}25(1-\sin\theta)^2d\theta\right)=\int_0^{\pi/2}25(1-\sin\theta)^2d\theta$$
And the value of the second area, $A_2$ is equal to the area of half a semicircle of radius $5$, which is just $25\pi/2$. If you really wanted, you could also calculate $A_2$ via an integral: $$A_2=\frac{1}{2}\int_{\pi}^{2\pi}5^2d\theta$$ Add $A_1$ and $A_2$ together and you have your answer.
I hope that helps. If you have any questions please don't hesitate to ask :)