Rapid Recall
1. 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.
Solution
Note that $dx = 3du+2dv$ and $dy=-1du+4dv$. We can write this as $$(dx,dy)=(3,-1)du+(2,4)dv.$$ We'll finish in class.
2. 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 $4\leq r\leq 4+dr$.
Solution
Drawn on the board.
3. The radian measure of an angle is defined as the quotient of two distances (which is why the measure is unitless). What are those two distances?
Solution
The arc length divided by the radius. This is why 360 degrees equals $2\pi$ radians, and $2\pi = \frac{2\pi r}{r}$. Radian measure is defined as the directed arc length traversed (so going clockwise counts as negative), divided by the radius of the circle.
4. Draw the region whose area is given by the integral $\ds\int_{0}^{\pi}\int_{1}^{5+3\sin\theta}r dr d\theta$.
Solution
Group problems
- Draw the region in the $xy$ plane described by $0\leq \theta \leq \pi/3$ and $0\leq r\leq 2\sin3\theta$.
- 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 $r=3\cos2\theta$.
- 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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