Rapid Recall
- Fill in the blanks below, using the substitution $u=x^2$: $$\int_{3}^{7}xe^{x^2}dx = \int_{\_}^{\_}xe^{\_}(\_)du .$$
Solution
$$\int_{3}^{7}xe^{x^2}dx = \int_{9}^{49}xe^{u}(\frac{1}{2x})du .$$
- Fill in the missing pieces below: $$\int_{0}^{3}\int_{0}^{\sqrt{9-x^2}}x dydx=\int_{?}^{?}\int_{?}^{?}(?)(?)drd\theta.$$
Solution
$$\int_{0}^{3}\int_{0}^{\sqrt{9-x^2}}x dydx=\int_{0}^{\pi/2}\int_{0}^{3}(r\cos\theta)(r)drd\theta.$$
- Find a nonzero vector that is orthogonal to both $(a,b,c)$ and $(d,e,f)$.
Solution
The cross product gives $(bf-ce, cd-af,ae-bd)$. Ask me in class how to use the information below to rapidly remember this. $$\begin{matrix}a&b&c&a&b&c\\d&e&f&d&e&f\end{matrix}$$
Group problems
- For each region below, (1) set up a double integral that gives the area of the region. Feel free to use any coordinate system you want. Then set up a formula to compute $\bar x$, the $x$ coordinate of the centroid.
- The region in the first quadrant of the $xy$-plane that lies above the line $y=x$ and below the semicircle $y=\sqrt{16-x^2}$.
- The region in the first quadrant that lies below the line that passes through the two points $(4,0)$ and $(0,6)$.
- For each solid region below, (1) set up a triple integral that gives the volume of the region. Then set up a formula to compute $\bar z$, the $z$-coordinate of the centroid of the region.
- The solid hemiball of radius 3, above the $xy$-plane.
- The solid region in the first octant that lies under the paraboloid $z=9-x^2-y^2$.
- Compute $\ds \int_{0}^{4}\int_{8}^{2x}e^{y^2}dydx$.
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