Rapid Recall
- Compute the eigenvalues of the matrix $\begin{bmatrix}4&3\\2&5\end{bmatrix}$.
Solution
$We solve $(4-\lambda)(5-\lambda)-6=0$. This becomes $0=\lambda^2-9\lambda+14=(\lambda-7)(\lambda-2)$. The eigenvalues are $\lambda=7$ and $\lambda=2$.
- Let $P=(1,2,3)$, $Q=(2,2,2)$, and $R=(4,5,6)$. Find the area of the triangle $\Delta PQR$.
Solution
We form $\vec{PQ} = \left<-1,0,1\right>$ and $\vec{PR} = \left<3,3,3\right>$. Then our area would be $|\vec{PQ} \times \vec{PR}|/2$. However, if we note that $\vec{PQ} \cdot \vec{PR} = 0$, then we know that $A = |\vec{PQ}||\vec{PR}|/2 = \sqrt{2}\sqrt{27}/2=3\sqrt{6}/2$, since the triangle is half a rectangle.
- Choose an order of integration to use in setting up an iterated integral for the volume of the region $D$ bounded by the surfaces $r=\cos\theta$, $r=2\cos\theta$, $z=0$, and $z=3-y$. I'll provide a picture.
Solution
The $z$ surfaces depend on both $r$ and $\theta$, so choose that first. The $r$ surfaces depend only on $\theta$, so do that one next. The last variable should be $\theta$ since its limits will be constants (what are they?).
Group problems
- Draw the region described the bounds of each integral.
- $\ds\int_{0}^{\pi}\int_{0}^{3}\int_{0}^{5}rdzdrd\theta$
- $\ds\int_{-1}^{1}\int_{0}^{1-y^2}\int_{0}^{x}dzdxdy$
- $\ds\int_{0}^{2}\int_{0}^{1-y/2}\int_{6+z}^{6-z}dxdzdy$
- Set up an integral formula to compute the $z$-coordinate of the center-of-mass (so $\bar z$) of the solid object in the first octant (all variables positive) that lies under the plane $2x+3y+6z=6$ (Draw the region as well).
- A wire lies along the curve $C$ parametrized by $\vec r(t) = (t^2+1, 3t, t^3)$ for $-1\leq t\leq 2$.
- Compute $ds$. (Remember - a little distance equals the product of the speed and a little time.)
- Set up an integral to find $\bar x$, then $\bar y$, then $\bar z$, for the centroid of $C$.
- Let $P=(1,2,0)$, $Q=(0,2,-1)$, and $R=(3,0,4)$.
- Find a vector that is orthogonal to both $\vec{PQ}$ and $\vec {PR}$.
- Find the area of triangle $\Delta PQR$.
- Give an equation of the plane PQR. (Let $S=(x,y,z)$ be any point on the plane PQR. Use $\vec {PS}\cdot (\vec {PQ}\times \vec {PR})=0$.)
- Let $T$ be another point in space. The quantity $\left|\text{proj}_{\vec {PQ}\times \vec {PR}}\vec {PT}\right|$ computes the distance between two things. What two things?
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