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 $(x,y)$ and $(p,q)$.
- When we add up lots of little changes in $x$, so $\int_C dx$, what do we get?
- Compute the double integral $\ds \int_{-1}^{2}\left(\int_{x}^{4-x}dy\right)dx$.
- Draw the region whose area is given by the double integral above.
Brain Building Solutions
- $|xq-yp|$
- Total change in $x$. This might represent a width of something.
- We get $$\ds \int_{-1}^{2}\left(\int_{x}^{4-x}dy\right)dx =\int_{-1}^{2}(4-x)-xdx =4x-x^2\bigg|_{-1}^{2} =(8-4)-(4(-1)-(-1)^2) = 9 .$$
- The picture is a triangle with corners at $(-1,-1)$, $(-1,5)$, and $(2,2)$. The height is 6, the width is 3, so the area is 9.
Group Work
- Draw the region in the plane described by $-3\leq y\leq 2$ and $y\leq x\leq 6-y^2$.
- Compute the integral $\ds\int_{y}^{6-y^2}dx$ (assume $y$ is a constant).
- Compute the double integral $\ds \int_{-3}^{2}\left(\int_{y}^{6-y^2}dx\right)dy$.
- Draw the region in the $xy$ plane described by $0\leq \theta \leq \pi$ and $2\leq r\leq 5$.
- Compute the double integral $\ds \int_{0}^{\pi}\left(\int_{2}^{5}rdr\right)d\theta$.
- 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 $\ds \int_{0}^{\pi/3}\left(\int_{0}^{2\sin 3\theta}rdr\right)d\theta$.
- Set up a double integral that gives the area of the region in the $xy$ plane that lies inside the cardiod $r=2+2\cos\theta$.
- Set up a double integral that gives the area of the region in the $xy$ plane that lies inside one petal of the rose $r=3\cos2\theta$.
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