Welcome to class
This is the right place to be for Math 215 today (starting at 3:15). If I'm not here, then I ran upstairs for a bit. If you are here for office hours, hang tight for a bit and I'll be right back.
Rapid Recall
- Consider the triangular region $R$ in the plane that is bounded by the curves $y=2x$, $y=4$, and $x=0$. Set up a double integral that compute the area of this region.
Solution
There are two possible answers, namely $$ \int_{0}^{2}\int_{2x}^{4}dydx \text{ or } \int_{0}^{4}\int_{0}^{y/2}dxdy. $$ Both are equally correct.
- Using the same region $R$ as above, set up a double integral formula to compute $\bar x$, the $x$-coordinate of the centroid of $R$.
Solution
There are two possible answers, namely $$ \frac{\int_{0}^{2}\int_{2x}^{4}xdydx} {\int_{0}^{2}\int_{2x}^{4}dydx} \text{ or } \frac{\int_{0}^{4}\int_{0}^{y/2}xdxdy} {\int_{0}^{4}\int_{0}^{y/2}dxdy}. $$ Both are equally correct. The bounds are completely independent of the formula $\bar x = \dfrac{\iint_R xdA}{\iint_R dA}$.
- Using the same region $R$ as above but adding a varying density of $\delta(x,y) = x^2y$, set up a double integral formula to compute $\bar y$, the $y$-coordinate of the center-of-mass of $R$.
Solution
There are two possible answers, namely $$ \frac{\int_{0}^{2}\int_{2x}^{4}y(x^2y)dydx} {\int_{0}^{2}\int_{2x}^{4}(x^2y)dydx} \text{ or } \frac{\int_{0}^{4}\int_{0}^{y/2}y(x^2y)dxdy} {\int_{0}^{4}\int_{0}^{y/2}(x^2y)dxdy}. $$ Both are equally correct. The bounds are completely independent of the formula $\bar x = \dfrac{\iint_R xdm}{\iint_R dm}$, where $dm=\delta dA$.
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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