Rapid Recall
- Fill in the blanks in $$\ds \int_{0}^{3}\int_{x^2}^{9}dydx= \int_{?}^{?}\int_{?}^{?}dxdy.$$
Solution
We have $$\ds \int_{0}^{3}\int_{x^2}^{9}dydx= \int_{0}^{9}\int_{0}^{\sqrt{y}}dxdy.$$
- Find the area of a parallelogram whose edges are the vectors $(3,4,0)$ and $(0,0,2)$.
Solution
You can first compute the cross product, and then find the magnitude. This gives $$(3,4,0)\times(0,0,2) = (8,-6,0),$$ whose magnitude is $\sqrt{64+36} = \sqrt{100}=10$.
If you first notice that the vectors are orthogonal (their dot product is zero), then you now the object is a rectangle with side lengths 5 and 2, so area is 10.
- Set up an integral formula to find $\bar x$, the $x$ coordinate of the centroid, for the region in the $xy$-plane that lies inside the cardioid $r=2-2\sin\theta$.
Solution
The solution is $$\bar x = \ds \frac{\iint_RxdA}{\iint_RdA} = \ds \frac{\ds \int_{0}^{2\pi}\int_{0}^{2-2\sin\theta}(r\cos\theta)rdrd\theta}{\ds \int_{0}^{2\pi}\int_{0}^{2-2\sin\theta}rdrd\theta}.$$
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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