Presentations
14, 16-22Pacing Tracker
- The quizzes have included questions for 22 objectives. How many have you passed? What are you plans to master those that you haven't mastered yet?
- We've finished units 1 through 4. Have you started your self-directed learning project for each unit?
- The 6th project can be over any topic from the entire semester. Feel free to get started on this one as soon as you have an idea.
- Remember you can submit only one SDL project per week. Plan ahead and don't let yourself get behind.
Brain Gains
- Set up an integral formula to compute the $x$ coordinate of the center-of-mass of a disc that lies inside the circle $x^2+y^2=9$ and has density function given by $\delta = x+5$.
Solution
Recall that $dm = \delta dA$ and $dA = r dr d\theta$ in polar coordinates. Bounds for the region in polar coordinates are $0\leq \theta\leq 2\pi$ and $0\leq r\leq 2$. We can use Mathematica to verify these bounds are correct.
outerB = {theta, 0 , 2 Pi}
innerB = {r, 0, 3}
coordinates = {r Cos[theta], r Sin[theta]};
ParametricPlot[coordinates, outerB, innerB, Mesh -> {20, 0}]
The bounds above give the integral $$\begin{align} \bar x &= \frac{\int\int_R x dm}{\int\int_Rdm}\\ &= \frac{\int\int_R x (x+5)dA}{\int\int_R(x+5) dA}\\ &= \frac{\int_0^{2\pi}\int_{0}^{3} (r\cos\theta) (r\cos\theta+5)rdrd\theta} {\int_0^{2\pi}\int_{0}^{3} (r\cos\theta+5)rdrd\theta} . \end{align}$$ We can quickly compute this integral using Mathematica.
density = r Cos[theta] + 5;
mass = Integrate[density*r, outerB, innerB]
xbar = Integrate[r Cos[theta]*density*r, outerB, innerB]/mass
ybar = Integrate[r Sin[theta]*density*r, outerB, innerB]/mass
{xbar, ybar} // N
We can instead use Cartesian coordinates to set up the integral, which gives $$\begin{align} \bar x &= \frac{\int_{-3}^{3}\int_{-\sqrt{9-x^2}}^{\sqrt{9-x^2}} x(x+5)dydx} {\int_{-3}^{3}\int_{-\sqrt{9-x^2}}^{\sqrt{9-x^2}} (x+5)dydx}\\ &= \frac{\int_{-3}^{3}\int_{-\sqrt{9-y^2}}^{\sqrt{9-y^2}} x(x+5)dxdy} {\int_{-3}^{3}\int_{-\sqrt{9-y^2}}^{\sqrt{9-y^2}} (x+5)dxdy} . \end{align}$$ The code below draws the region, and computes the integrals, using Cartesian coordinates.
outerB = {x, -3, 3}
innerB = {y, -Sqrt[9 - x^2], Sqrt[9 - x^2]}
coordinates = {x, y};
ParametricPlot[coordinates, outerB, innerB, Mesh -> {10, 0}]
density = x + 5;
mass = Integrate[density, outerB, innerB]
xbar = Integrate[x*density, outerB, innerB]/mass
ybar = Integrate[y*density, outerB, innerB]/mass
{xbar, ybar} // N
- A wire lies along the curve $C$ parametrized by $\vec r(t) = (3t, t^3)$ for $-1\leq t\leq 2$. Set up an integral formula to find $\bar y$ for the centroid of $C$.
Solution
Recall that a little length is given by $ds = v(t) dt$ (speed multiplied by a little time). Recall that the velocity $\vec v(t) = (3,3t^2)$ gives the speed $v(t) = \sqrt{9+9t^4}$. We then have $$\begin{align} \bar y &= \frac{\int_C y ds}{\int_C ds}\\ &= \frac{\int_{-1}^{2} t^3 \sqrt{9+9t^4} dt}{\int_{-1}^{2}\sqrt{9+9t^4} dt} . \end{align}$$ Here are the computations done with Mathematica.
ClearAll[r]
r[t_] = {3 t, t^3}
ds = Norm[r'[t]]
y = r[t][[2]]
Integrate[y ds, {t, -1, 2}]/Integrate[ ds, {t, -1, 2}]
% // N
ParametricPlot[r[t], {t, -1, 2}]
- Draw the region described the bounds of the integral $\ds\int_{0}^{3}\int_{0}^{9-x^2}\int_{0}^{5}dzdydx$.
Solution
We'll draw this together, and use Integration.nb to check our work. We'll also discuss how changing the inner bounds will affect the integral with a few examples, such as $z=7-x$ instead of $z=5$, and $z=x$ or $z=-x^2$ instead of $z=0$.
There is a command to plot regions in Mathematica. Sometimes the corner's don't quite come together nicely, but the code is very short.
R = ParametricRegion[{x, y, z}, {{x, 0, 3}, {y, 0, 9 - x^2}, {z, 0, 5}}];
Region[R, Axes -> True, AxesLabel -> {x, y, z}, AxesOrigin -> {0, 0, 0}]
- Draw the region described the bounds of the integral $\ds\int_{0}^{1}\int_{0}^{1-z}\int_{0}^{\sqrt{1-x^2}}dydxdz$.
Solution
We'll draw this together, and use Integration.nb to check our work.
R = ParametricRegion[{x, y, z}, {{z, 0, 1}, {x, 0, 1 - z}, {y, 0, Sqrt[1 - x^2]}}];
Region[R, Axes -> True, AxesLabel -> {x, y, z}, AxesOrigin -> {0, 0, 0}]
Group problems
- Draw the region described the bounds of each integral. (Either use the Mathematica notebook Integration.nb to check your work, or adapt the code below to use the built in region plotter.)
R = ParametricRegion[{x, y, z}, {{z, 0, 1}, {x, 0, 1 - z}, {y, 0, 1 - x^2}}]; Region[R, Axes -> True, AxesLabel -> {x, y, z}, AxesOrigin -> {0, 0, 0}]- $\ds\int_{0}^{3}\int_{0}^{9-x^2}\int_{0}^{3-x}dzdydx$
- $\ds\int_{0}^{\pi}\int_{0}^{3}\int_{0}^{5}rdzdrd\theta$
- $\ds\int_{-1}^{1}\int_{0}^{1-y^2}\int_{0}^{x}dzdxdy$
- 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 volume of the solid object in the first octant (all variables positive) that lies under the plane $2x+3y+6z=6$ (so $\frac{x}{3}+\frac{y}{2}+\frac{z}{1}=1$).
- The $y$-coordinate of the center-of-mass (so $\bar y$) of previous object.
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