We show how to do the area calculation using polar coordinates.
There is a potential minor complication. There are two different conventions about the meaning of a polar equation when $r<0$. Some say the curve is then not defined. Some say it is defined, but one must reflect across the origin. For our particular curve, the two conventions give the same region, so we don't need to worry. But for other curves, you may need to know what convention your course uses.
The curve $r=6\sin \theta$ and the circle $r=1$ meet where $\sin\theta=1/6$. Let $\theta_0=\arcsin(1/6)$. From $\theta=\theta_0$ to $\theta=\pi-\theta_0$, the curve $r=6\sin\theta$ is "outside" the circle $r=1$. By symmetry, we can just look at the part from $\theta=\theta_0$ to $\theta=\pi/2$, and double the resulting area.
The area from $\theta=\theta_0$ to $\theta=\pi/2$, bounded by our curve $r=6\sin\theta$, is
$$\int_{\theta_0}^{\pi/2} (1/2)r^2 d\theta.$$
From this we must subtract the area inside the circle $r=1$, which is
$$\int_{\theta_0}^{\pi/2} (1/2)\,1^2 \,d\theta.$$
Subtract, and multiply by $2$ to take care of the other half of the region.
The desired area is therefore
$$\int_{\theta_0}^{\pi/2} (36\sin^2\theta-1)\, d\theta.$$
Now integrate. Using the identity $\cos 2\theta=1-2\sin^2\theta$, we find that $36\sin^2\theta=18-18\cos 2\theta$. Thus our area is
$$\int_{\theta_0}^{\pi/2} (17-18\cos 2\theta)\, d\theta.$$
The rest is routine. We will need $\sin 2\theta_0$. When you calculate this, you will bump into the $35$ that you already met in your rectangular coordinates analysis. The ultimate answer is not "nice," because $\arcsin(1/6)$ is not a pleasant number. Too bad that the polar curve was not $r=2\sin\theta\:$!
Comment: We can also obtain the result by switching to rectangular coordinates, as you did. The switching in this case was straightforward, and showed we were dealing with a circle. However, in many cases the rectangular coordinates version of a curve given in polar coordinates can be very unpleasant. It is therefore useful to learn to work directly in polar coordinates.
In situations where there is strong circular symmetry, polar coordinates come up very naturally. For example, the force of gravity exerted by the Earth on a small object is directed towards the center of the Earth. In studying motions of satellites, polar coordinates are the way to go.
A rough picture is very useful. There is curve for $\theta=0$ to $\frac{\pi}{4}$. Then since $r^2$ cannot be negative, there is no curve from $\theta=\frac{\pi}{4}$ to $\frac{\pi}{2}$.
Then there is curve from $\theta=\frac{3\pi}{4}$ to $\theta=\pi$,and there is still curve from $\theta=\pi$ to $\theta=\frac{5\pi}{4}$, then no curve for a while, then curve from $\theta=\frac{7\pi}{4}$ to $2\pi$.
That gives a flower with two petals. The easiest way to compute the area is probably to think of the flower as made up of $4$ parts identical to the part between $\theta=0$ and $\theta=\frac{\pi}{4}$. The total area is
$$4\int_0^{\pi/4} \frac{1}{2}(50) \cos(2\theta)\,d\theta.$$
Best Answer
Draw a picture. We use the standard formula for area in polar coordinates. In principle there could be a problem with the interval of integration, but here there is no problem, because $3+\sin\theta$ is always positive. So our area is $$\int_0^{2\pi}\frac{1}{2}(3+\sin\theta)^2\,d\theta.$$ For the details, expand. The integral of $\frac{9}{2}$ is easy. For the "middle" term, the integral is easily $0$. Finally, we need to deal with $\int_0^{2\pi}\frac{1}{2}\sin^2\theta\,d\theta$. There are many ways to do this. One of them is to use $\cos2\theta=1-2\sin^2\theta$. Another way is to use symmetry, and note that the integral over our interval of $\cos^2\theta$ is the same as the integral of $\sin^2\theta$. But the sum of these integrals is the integral of $1$.