• I-Learn, Class Pictures, Learning Targets, Text Book Practice
• Prep Tasks: Unit 1 - Motion, Unit 2 - Derivatives, Unit 3 - Integration, Unit 4 - Vector Calculus

Day 28 - Prep

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.

Day 28 - In class

Brain Gains (Rapid Recall, Jivin' Generation)

• Consider the region in the first quadrant that lies left of the line $y=x/2$ and below the line $y=1$. A metal plate occupies this region, and has density given by $\delta = \cos(y^2)$. Set up an iterated integral of the form $\int_?^?\int_?^? ? dydx$ that would give the mass of the plate.
• For the same region, set up an iterated integral of the form $\int_?^?\int_?^? ? dxdy$ that would give the mass of the plate.
• Examine the two integrals above. One of them has a problem an cannot be computed. Which one, and why?
• Set up the integrals to compute $\bar x$, the $x$-coordinate of the center of mass of the metal plate.

Integrate[Cos[y^2], {y, x/2, 1}]
Integrate[Cos[y^2], {x, 0, 2}, {y, x/2, 1}]
NIntegrate[Cos[y^2], {x, 0, 2}, {y, x/2, 1}]

Integrate[Cos[y^2], {x, 0, 2 y}]
Integrate[Cos[y^2], {y, 0, 1}, {x, 0, 2 y}]

Integrate[x Cos[y^2], {y, 0, 1}, {x, 0, 2 y}]/Integrate[Cos[y^2], {y, 0, 1}, {x, 0, 2 y}]
NIntegrate[x Cos[y^2], {y, 0, 1}, {x, 0, 2 y}]/NIntegrate[Cos[y^2], {y, 0, 1}, {x, 0, 2 y}]

Group Problems

1. Consider the integral $\ds\int_{0}^{3}\int_{0}^{x}dydx$.
   • Shade the region whose area is given by this integral.
   • Compute the integral.
   • Now compute $\ds\int_{0}^{x}\int_{0}^{3}dxdy$. Do you get a single number or an expression involving a variable? What's wrong with this integral? How can you recognize there will be a problem without even computing a single integral?
   • Adjust the bounds on the previous integral (use the order $dxdy$) so that the bounds describe the same region as original integral, but make sure the bounds on $y$ are between two constants. In other words, fill in the question marks below so that the integral's bounds describe the same region as the first part of this problem. $$\ds\int_{?}^{?}\int_{?}^{?}dxdy.$$
2. 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 $2x+3y+6z=6$ (so $\frac{x}{3}+\frac{y}{2}+\frac{z}{1}=1$).
   • The $y$-coordinate of the center-of-mass (so $\bar y$) of the same object.
3. Consider $\int_{0}^{4}\int_{x}^{4}e^{y^2}dydx$.
   • Draw the region described by the bounds.
   • Swap the order of the bounds on the integral (use $dxdy$ instead of $dydx$).
   • Compute the integral.

Day 29 - Prep

Task 29.1

This task has you practice setting up triple integrals.

1. The iterated triple integral $\ds\int_{-1}^1\int_0^4\int_0^{y^2}dzdxdy$ gives the volume of the solid $D$ that lies under the surface $z=y^2$, above the $xy$-plane, and bounded by the planes $y=-1$, $y=1$, $x=0$, and $x=4$. Sketch this region.
2. Set up an iterated triple integral that gives the volume of the solid in the first octant that is bounded by the coordinate planes ($x=0$, $y=0$, $z=0$), the plane $y+z=2$, and the surface $x=4-y^2$, using the order of integration $dxdzdy$. Make sure you sketch the region.
3. Set up an integral to give the volume of the pyramid in the first octant that is below the planes $\ds\frac{x}{3}+\frac{z}{2}=1$ and $\ds\frac{y}{5}+\frac{z}{2}=1$. [Hint, don't let $z$ be the inside bound. Try an order such as $dydxdz$.]
4. (Optional Challenge) Set up an iterated triple integral that gives the volume of the region $D$ that is inside both right circular cylinders $x^2+z^2=1$ and $y^2+z^2=1$. Don't forget to draw the region.

Task 29.2

We've now found the mass and center-of-mass for straight wires, thin flat metal plates, and solid regions in space. Earlier in the semester we used $$s = \int_C ds = \int_C \left|\frac{d\vec r}{dt}\right|dt$$ to obtain the length of a thin wire lying on the curve $C$ with parametrization $\vec r(t)$. For such a wire, we use the differential $$\underbrace{ds}_{\text{little distance}} = \underbrace{ \left| \frac{d\vec r}{dt}\right| }_{ \text{speed} }\underbrace{dt}_{\text{little time}}$$ instead of $dx$ (little length in a straight rod), $dA$ (little area in a thin metal plate), or $dV$ (little volume in a solid). The differential $ds$ can replace $dx$, $dA$, or $dV$ in any of our previous formulas to help us determine, for a curved wire, the length, mass, center-of-mass, and more. This task has you set up several integrals that do this.

Consider a wire that lies along the curve $C$ with parametrization $\vec r(t) = (5\cos t,5\sin t)$ for $0\leq t\leq\pi$.

1. Draw the curve, compute $\frac{d\vec r}{dt}$, and show that $ds = 5 dt$.
2. Evaluate $\int_C ds$ to obtain the length of the wire. Since the curve is half a circle, the length you obtain from integration should be half the circumference of the circle.
3. Assuming the density is constant, why do we know $\bar x=0$?
4. Set up an integral formula for $\bar y$ and compute the integrals involved to obtain $\bar y$, showing your integration steps.
5. If instead, the density is $\delta = xy^2+7$, then set up an integral formula to find $\bar x$. Use software to compute the integral.

Task 29.3

A sphere of radius $a$ centered at the origin is described by the equation $x^2+y^2+z^2 = a^2$. A right circular cone whose tip is at the origin is given by $z=\sqrt{x^2+y^2}$ or $z^2=x^2+y^2$.

1. Draw the surface $x^2+y^2+z^2 = a^2$ and then set up an iterated triple integral using Cartesian coordinates to compute the volume inside the sphere $x^2+y^2+z^2=a^2$.
2. Draw the surface $z^2=x^2+y^2$ and then set up an iterated triple integral using Cartesian coordinates to compute the volume of the solid cone that lies above the cone $z^2=x^2+y^2$ and below the plane $z=h$.
3. Use software to compute both integrals above. If software can't compute one of these integrals (the program hangs, never finishes, etc.), it's OK. These integrals are brutal, and we'll soon learn that different coordinate systems will make quick work of these integrals.

Task 29.4

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

Today

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