Brain Gains
- Draw the region described by $\pi/6\leq \theta\leq \pi/3$ and $4\leq r\leq 5$.
Solution
The problem didn't specify if you should draw in the $r\theta$ or $xy$ plane, which means (by convention) that we should draw in the $xy$-plane. We'll do this together. This region is the image of a typical polar rectangle in the $xy$-plane. Our goal is to understand the area stretch factor between polar and Cartesian coordinates.
- Find the area of a parallelogram whose edges are parallel to the vectors $(\cos\theta dr,\sin\theta dr)$ and $(-r\sin\theta d\theta,r\cos\theta d\theta)$.
Solution
$|r\cos^2\theta dr d\theta +r\sin^2\theta dr d\theta| = |r dr d\theta|$
- Compute the polar double integral $\ds \int_0^{2\pi}\int_{0}^{7} r\, dr d\theta$.
Solution
We compute $$\ds \int_0^{2\pi}\int_{0}^{7} r\, dr d\theta = \int_0^{2\pi}\frac{r^2}{2}\bigg|_0^{7} d\theta = \int_0^{2\pi}\frac{7^2}{2} d\theta = \frac{7^2}{2}\theta\bigg|_0^{2\pi} = \pi7^2. $$ This shows that that area inside a circle of radius $7$ is $\pi 7^2$. Changing the 7 to an an arbitrary radius $a$, the above work shows that the area inside a circle of radius $a$ is $\pi a^2$.
- For the curve $r=7$, compute $\frac{dx}{d\theta}$ and $\frac{dy}{d\theta}$.
Solutions
Since $x=r\cos\theta = 7\cos\theta$ and $y=r\sin\theta=7\sin\theta$, we have $\frac{dx}{d\theta} = -7\sin\theta$ and $\frac{dy}{d\theta}=7\cos\theta$. The curve is a circle of radius $7$. The integral
Group problems
- Consider the polar curve $r=7$. We already showed that $\frac{dx}{d\theta} = -7\sin\theta$ and $\frac{dy}{d\theta} = 7\cos \theta$. Use the arc length formula $$\int_C\sqrt{\left(\frac{dx}{d\theta}\right)^2+\left(\frac{dy}{d\theta}\right)^2}d\theta$$ to find the arc length for the portion of this curve with $0\leq \theta\leq \alpha$. Your result will show that the arc length on a circle of radius $7$ through an angle $\alpha$ is $s = 7\alpha$. This generalizes to a circle of radius $r$ through an angle $d\theta$ to give $s=rd\theta$.
- Draw the region in the $xy$ plane described by $0\leq \theta \leq \pi$ and $2\leq r\leq 5$.
- Compute the double integral $\ds \int_{0}^{\pi}\left(\int_{2}^{5}rdr\right)d\theta$. [Check: $\frac{25\pi}{2} - \frac{4\pi}{2}$, the difference in areas of two semicircles.]
- Draw the region in the $xy$ plane described by $0\leq \theta \leq \pi/3$ and $0\leq r\leq 2\sin3\theta$.
- Compute the double integral $\ds \int_{0}^{\pi/3}\left(\int_{0}^{2\sin 3\theta}rdr\right)d\theta$. (Use software if you get stuck at some point.)
- Set up a double integral that gives the area of the region in the $xy$ plane that lies inside the cardiod $r=2+2\cos\theta$.
- Draw the region in the $xy$ plane described by $0\leq \theta \leq \pi/4$ and $0\leq r\leq 3\cos2\theta$.
- Set up a double integral that gives the area of the region in the $xy$ plane that lies inside one petal of the rose $r=3\cos2\theta$.
- Draw and shade the region in the $xy$-plane that lies inside the curve $r=3+2\cos\theta$ and outside the curve $r=1$.
- Set up a double integral that gives the area of the region in the $xy$ plane that lies inside the curve $r=3+2\cos\theta$ and outside the curve $r=1$.
- Set up a double integral that gives the area of the region in the $xy$ plane that lies inside the curve $r=2-2\cos\theta$ and inside the curve $r=2\cos\theta$.
- Set up a double integral that gives the area of the region in the $xy$ plane that lies inside the curve $r=2-2\cos\theta$ and outside the curve $r=2\cos\theta$.
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