# Area under polar curve inside cardioid and circle

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!

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 :)

• Sorry! I just saw this. Yes this was perfect. Thank you :) . Commented Mar 25, 2021 at 20:25
• @FutureMathperson My pleasure! :) Commented Mar 25, 2021 at 20:28