What is the area of the region inside the limacon with equation $r=3+2 \sin (\theta)$ that lies below the line $y=x$?

Question

What is the area of the region inside the limacon with equation $r=3+2 \sin (\theta)$ that lies below the line $y=x$?

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I know I should use Riemann sum but how to write the equation? And what about the $y=x$?

Answer 1

Notice that, as mentioned in the comments as well, the line is a combination of the rays corresponding to $\displaystyle \theta = \frac{π}{4}$ and $\displaystyle \theta = -\frac{3π}{4}$.

Just reminding you, polar curves are traced out by angle, and so are their integrals.

We have that the area enclosed by the two rays $\theta=a$ and $\theta=b$, the origin, and $r=f(\theta)$ is $\displaystyle \frac{1}{2}\int_a^b f(\theta)^2 \,d\theta$.

We have the rays $\displaystyle \theta = \frac{π}{4}$ and $\displaystyle \theta = -\frac{3π}{4}$ and $r=f(\theta)=3+2\sin(\theta)$.

We have $\displaystyle \frac{1}{2}\int_{-\frac{3π}{4}}^{\frac{π}{4}}(3+2\sin(\theta))^2\,d\theta=\boxed{\frac{11π}{2}-6\sqrt{2}}$ units$^2$.