Tutorial note 5

Problem 1. Done

Problem 2. Done

Problem 3. The correct matches are B, E, D, F respectively. It is hard to plot graphs here, you can plot the remaining four curves as a good practice.

Problem 4. The area formula is given by $\frac{1}{2}\int_0^{\theta_0}r^2(\theta)\,d\theta$, we need to determine the smallest value $\theta_0$ such that the graph of $r=r(\theta)$ completes in polar coordinate on $[0,\theta_0]$.

For example, the graph of $r=2+\cos\theta$ completes at $\theta_0=2\pi$, while the graph of $r=\cos3\theta$ completes at $\theta_0=\pi$. The determination is usually not easy, you may need to plot on your own.

In the formula $\frac{1}{2}\int_0^{\theta_0}r^2(\theta)\,d\theta$ if your choice of $\theta_0$ is too large, then this formula will count repeated area, this is of course not the exact area bounded by your graph!

Problem 5. Forget about this.

Problem 6. We can choose the coordinate in this way:

Recall that the blued circle can be parametrized by $r=2\cos\theta$, and the larger circle is given by $r=\sqrt{3}$, the area indicated need to be computed by two integrals. We may just find the area in the first quadrant and then multiply it by $2$. The intersection occurs when \[2\cos\theta=\sqrt{3}\implies \theta=\pi/6,
\]hence desired area is \[
2\times \brac{\frac{1}{2}\int_{0}^{\pi/6} (\sqrt{3})^2\,d\theta + \frac{1}{2}\int_{\pi/6}^{\pi/2}(2\cos\theta)^2\,d\theta} = \frac{7\pi}{6}-\frac{\sqrt{3}}{2}.
\]