Brain Building (30 seconds per problem - just do your best - no worries if wrong)
- Draw a bear in polar coordinates, and then Cartesian coordinates.....
- If we know $x=3u+2v$ and $y=-u+4v$, then areas in the $uv$-plane are multiplied by how much to obtain an area in the $xy$-plane.
- Let $d\theta$ be a small angle, and let $dr$ be a small distance. Draw a region described by $\pi/3\leq \theta\leq \pi/3+d\theta$ and $4\leq r\leq 4+dr$.
- Draw the region whose area is given by the integral $\ds\int_{0}^{\pi}\int_{1}^{5+3\sin\theta}r dr d\theta$.
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
- Draw the region in the $xy$ plane whose area is given by $\ds\int_{\pi}^{2\pi}\int_{0}^{3+3\cos\theta}r dr d\theta$.
- If we know $x=r\cos\theta$ and $y=r\sin\theta$, then small areas in the $r\theta$-plane are multiplied by how much to obtain the area of the transformed region in the $xy$-plane? (Compute differentials and then use the parallelogram area rule.)
- Draw the region in the $xy$ plane described by $0\leq \theta \leq \pi/4$ 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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