Brain Building (30 seconds per problem - just do your best - no worries if wrong)
- Find the area of a parallelogram whose edges are parallel to the vectors $(\cos\theta,\sin\theta)$ and $(-r\sin\theta,r\cos\theta)$.
- When we add up lots of little bits of area, so $\int_R dA$, what do we get?
- Draw the region described by $\pi/6\leq \theta\leq \pi/3$ and $4\leq r\leq 5$.
Brain Building Solutions
- $|r\cos^2\theta +r\sin^2\theta| = |r|$
- Total Area
- See board
- length of arc divided by radius of arc.
Group Work
- Consider the polar curve $r=7$. Compute $\frac{dx}{d\theta}$ and $\frac{dy}{d\theta}$. Then 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 $0\leq \theta\leq \alpha$.
- Draw the region in the $xy$ plane described by $0\leq \theta \leq \pi$ and $0\leq r\leq 2\sin3\theta$.
- Draw the region in the $xy$ plane described by $0\leq \theta \leq \pi$ and $0\leq r\leq 3\sin2\theta$.
- Set up a double integral that gives the area of the region in the $xy$ plane that lies inside one petal of the rose $r=3\cos2\theta$.
- 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 $r=1$. Then draw this region.
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