Task 28.1

We have seen how to compute the center of mass of a rod (a 1 dimensional object) and triangle (a 2 dimensional object). This task will do so with a circular region (2 dimensional object) and 3D solid.

  1. Consider the semicircular disc $R$ that lies above the $x$-axis and below the circle of radius $a$. If you would rather work with numbers instead of variables, feel free to let $a=5$ for this problem.
    1. We know the area of $R$ is $\frac{1}{2}\pi a^2$. Set up a double integral using polar coordinates to compute this area. Then compute the integral by hand and simplify your work to obtain the correct area.
    2. Let's assume the density for this problem is $\delta = 1$, so that $dm=dA$. When the density is constant, we use the word "centroid" instead of "center-of-mass" to talk about the geometric center of an object. The centroid in the $x$-direction is given by the formula $$\bar x = \frac{\ds\iint_R xdA}{\ds\iint_R dA}= \frac{\ds\int_0^\pi\int_0^{a} \overbrace{(r\cos\theta)}^{x} \overbrace{r dr d\theta}^{dA}}{\ds\int_0^\pi\int_0^{a} \underbrace{r dr d\theta}_{dA}}.$$ Compute the integrals above, by hand, to show that $\bar x=0$.
    3. Set up an integral formula, like the one above, to compute $\bar y$. Show the integral formula you used, and then compute it (feel free to use software) to obtain $\bar y$. You can check your answer is correct by referring to a list of centroid of regions (such as this Wikipedia list).
  2. The triple integral $\ds\int_{0}^{5}\int_0^7\int_{0}^{10-2x}dzdydx$ gives the volume of a solid domain $D$ in space.
    1. Draw the solid domain $D$ described by the bounds of the integral above. This is the solid satisfying the inequalities $0\leq x\leq 5$, $0\leq y\leq 7$, and $0\leq z\leq 10-2x$.
    2. Let $\delta =1$ so that $dm=\delta dV = 1dV$. The centroid of $D$ has three coordinates $(\bar x, \bar y, \bar z)$. The $x$-coordinate is given by the integral formula $$\bar x = \frac{\ds\iiint_R xdV}{\ds\iiint_R dV}= \frac{\ds\int_{0}^{5}\int_0^7\int_{0}^{10-2x}(x)dzdydx}{\ds\int_{0}^{5}\int_0^7\int_{0}^{10-2x}1dzdydx}.$$ Compute this triple integral and simplify to show that $\bar x = \frac{5}{3}$.
    3. Modify the above formula to obtain integral formulas for both $\bar y$ and $\bar z$. Then state the values of $\bar y$ and $\bar z$, either by using facts we've already proven or by computing the integrals directly (use software).

Task 28.2

For each region $R$ below, draw the region in the $xy$-plane. Set up an iterated integral in polar coordinates ($x=r\cos\theta$, $y=r\sin\theta$) that gives the area of the region and then use the given density to set up an iterated double integral that gives the mass of a metal plate that occupies the region and has the given variable density. Use software to compute each integral.

For example, consider the region that is inside the circle $x^2+y^2=9$, along with the density $\delta(x,y)=y^2$. We can describe the region using the polar inequalities $0\leq \theta \leq 2\pi$ and $0\leq r\leq 3$, which gives us the bounds needed for our integral.

  • The area is $\ds A=\iint_R \delta dA = \int_0^{2\pi}\int_0^3\underbrace{rdrd\theta}_{dA} = 9\pi.$
  • The mass is $\ds m=\iint_R \delta dA = \int_0^{2\pi}\int_0^3\underbrace{(r\sin\theta)^2}_{\delta=y^2}\underbrace{rdrd\theta}_{dA} = \frac{81\pi}{4}.$

The Mathematica code below was used to compute the integrals above (along with a graphical check that the region is the correct region - the last line is there for backwards compatibility, and can be ignored if you have Mathematica 13.0 or greater). 

OuterBounds = {theta, 0, 2 Pi};
InnerBounds = {r, 0, 3};

Integrate[r, OuterBounds, InnerBounds]
Integrate[(r Sin[theta])^2 r, OuterBounds, InnerBounds]

CoordinateSystem = {r Cos[theta], r Sin[theta]};
ParametricPlot[CoordinateSystem, OuterBounds, InnerBounds, Mesh -> {10, 0}]

ParametricPlot[Evaluate[CoordinateSystem, OuterBounds, InnerBounds], Mesh -> {10, 0}]
  1. The region $R$ is the quarter disc in the first quadrant that lies inside the circle $x^2+y^2=25$. The density is $\delta(x,y)=x$.
  2. The region $R$ is bounded above by $y=\sqrt{9-x^2}$, bounded below by $y=x$, and bounded on the left by the $y$-axis. The density is $\delta(x,y)=xy^2$.
  3. The region $R$ is the inside of the cardioid $r=3+3\cos\theta$. The density is $\delta(x,y)=7$.
  4. The region $R$ is the triangular region below $y=\sqrt 3 x$, above the $x$-axis, and to the left of $x=1$. The density is $\delta(x,y)=7$.

Task 28.3

This task provides you with a couple integrals that cannot be done, without first making some change. The first requires a change of order of integration. The second requires a complete change of coordinates.

  1. Compute by hand the iterated integral $$\ds \int_0^{2\sqrt{\pi}}\int_{y/2}^{\sqrt{\pi}} \sin(x^2)dxdy.$$ (Hint, you will need to swap the order of integration first.)
  2. The double integral $\ds\int_0^{\sqrt{2}}\int_{y}^{\sqrt{4-y^2}} e^{x^2+y^2}dxdy$ computes the mass of a region in the plane with density $\delta = e^{x^2+y^2}$ that is bounded by the curves $y=0$, $y=\sqrt{2}$, $x=y$, and $x=\sqrt{4-y^2}$.
    1. Draw the region described by these bounds. (Did you get a sector of a circle, something like a 1/8th of a pizza?)
    2. Give bounds of the form $?\leq \theta\leq ?$ and $?\leq r\leq ?$ that describe the region using polar coordinates. (The new bounds are all constants.)
    3. Convert the Cartesian integral to an integral in polar coordinates (don't forget the $r$ that appears as $dxdy = dA = rdrd\theta$).
    4. Compute the integral by hand. Show your steps.

Task 28.4

Pick some problems related to the topics we are discussing from the Text Book Practice page.