Rapid Recall
- Write $\ds \int_{0}^{5}\int_{x}^{5}dydx$ in the form $\ds \int_{?}^{?}\int_{?}^{?}dxdy$.
Solution
The solution is $\ds \int_{0}^{5}\int_{0}^{y}dxdy$. Start by drawing the region.
- Find a vector that is orthogonal to both $(1,2,3)$ and $(4,5,6)$.
Solution
The cross product gives $$(2\cdot 6-3\cdot 5, 3\cdot 4-1\cdot 6, 1\cdot 5-2\cdot 4) = (-3, 6,-3).$$
- Find the area of a parallelogram whose edges are the vectors $(1,2,3)$ and $(4,5,6)$.
Solution
Just compute the magnitude of the previous vector, so $$\sqrt{9+36+9} = \sqrt{54} = 3\sqrt{6}.$$
Group problems
- Consider the region $R$ that is bounded by the lines $x=0$, $y=6$, and $x=y/2$. The density (mass per area) is given by $\delta(x,y)$.
- Set up a double integral to compute the area of $R$.
- Set up a double integral to compute the mass of $R$.
- Set up a double integral formula to compute $\bar x$ and $\bar y$ for the centroid of $R$.
- Set up a double integral formula to compute $\bar x$ and $\bar y$ for the center-of-mass of $R$.
- Consider $\int_{0}^{4}\int_{x}^{4}\cos(y^2)dydx$.
- Draw the region described by the bounds.
- Swap the order of the bounds on the integral (use $dxdy$ instead of $dydx$) and then actually compute the integral.
- Set up an integral formula to compute each of the following:
- The mass of a disc that lies inside the circle $x^2+y^2=9$ and has density function given by $\delta = x+10$
- The $x$-coordinate of the center of mass (so $\bar x$) of the disc above.
- 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 $2x+3y+6z=6$.
- The $y$-coordinate of the center-of-mass (so $\bar y$) of the same object.
- Draw the region described the bounds of each integral. (Use the Mathematica notebook Integration.nb to check your work.)
- $\ds\int_{0}^{3}\int_{0}^{9-x^2}\int_{0}^{3-x}dzdydx$
- $\ds\int_{0}^{1}\int_{0}^{1-z}\int_{0}^{\sqrt{1-x^2}}dydxdz$
- $\ds\int_{0}^{3}\int_{0}^{\pi}\int_{0}^{5}rdzdrd\theta$
- $\ds\int_{-1}^{1}\int_{0}^{1-y^2}\int_{0}^{x}dzdxdy$
- 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 $C$.
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