Presentations
5.? - 5.?Pacing 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
- Consider the region $R$ in the plane that lies below $y = f(x)$ and above $y=g(x)$, for $a\leq x\leq b$. Set up formulas involving iterated double integrals to compute both $\bar x$ and $\bar y$.
Solution
The centroid is $$\bar x = \frac{\int_a^b\int_{g(x)}^{f(x)}x dy dx}{\int_a^b\int_{g(x)}^{f(x)} dy dx} \quad\text{and}\quad \bar y = \frac{\int_a^b\int_{g(x)}^{f(x)}y dy dx}{\int_a^b\int_{g(x)}^{f(x)} dy dx}.$$ Note that if we had a density function $\delta (x,y)$, then the center-of-mass would be $$\bar x = \frac{\int_a^b\int_{g(x)}^{f(x)}x \delta(x,y) dy dx}{\int_a^b\int_{g(x)}^{f(x)} \delta(x,y)dy dx} \quad\text{and}\quad \bar y = \frac{\int_a^b\int_{g(x)}^{f(x)}y \delta(x,y) dy dx}{\int_a^b\int_{g(x)}^{f(x)} \delta(x,y) dy dx}.$$
Let's look at 39 now.
- Set up an integral to find the volume of the region in space above the $xy$-plane that is bounded above by the plane $z=8$ (so $\rho \cos\phi = 8$) and below by the cone $z^2=x^2+y^2$ (so $\phi = \pi/4$).
Solution
$$\int_{0}^{2\pi}\int_{0}^{\pi/4}\int_{0}^{8\sec\phi}\rho^2\sin\phi d\rho d\phi d\theta$$
- Set up an integral to find $\bar z$ for the center-of-mass of the region above if the density is $\delta = x^2+y^2+z^2$. If you forgot, recall $z=\rho\cos\phi$ and the Jacobian is $\rho^2\sin\phi$.
Solution
The bounds don't change at all, rather we just have to add the correct pieces from the center-of-mass formulas. This gives $$\bar z = \frac{\iiint_D z\delta dV}{\iiint_D\delta dV}=\frac{\ds\int_{0}^{2\pi}\int_{0}^{\pi/4}\int_{0}^{8\sec\phi}\overbrace{(\rho\cos\phi)}^{z}(\rho^2) (\rho^2\sin\phi) d\rho d\phi d\theta}{\ds\int_{0}^{2\pi}\int_{0}^{\pi/4}\int_{0}^{8\sec\phi}\underbrace{(\rho^2)}_{\delta} \underbrace{(\rho^2\sin\phi) d\rho d\phi d\theta}_{dV}}$$
- The curves $y=8-x^2$ and $y=x+2$ intersect at $x=2$ and $x=-3$. The area of the region in space bounded by these two curves is $\ds \int_{-3}^{2}\int_{x+2}^{8-x^2}dydx$. Set up an integral to compute the average temperature of a metal plate in the $xy$-plane that lies in this region, provided the temperature at points on the plate is given by $f(x,y)=x+y^2$.
Solution
The average temperature is $$\bar f = \frac{\int_{-3}^{2}\int_{x+2}^{8-x^2}(x+y^2)dydx}{ \int_{-3}^{2}\int_{x+2}^{8-x^2}dydx}.$$
- The length of a wire that lies along the helix $\vec r(t) = (3\cos t,3\sin t, 4t)$ for $0\leq t\leq 4\pi$ is $$\int_{0}^{4\pi} \sqrt{(-3\sin t)^2+(3\cos t)^2+(4)^2}dt.$$ Set up an integral to compute the average charge density, provided the charge density at each point on the wire is given by $\sigma(x,y,z) = x^2+y^2+z$.
Solution
The average charge density is $$\bar \sigma = \frac{ \int_{0}^{4\pi} [(3\cos t)^2+(3\sin t)^2 +4t]\sqrt{(-3\sin t)^2+(3\cos t)^2+(4)^2}dt}{ \int_{0}^{4\pi} \sqrt{(-3\sin t)^2+(3\cos t)^2+(4)^2}dt}.$$
- Set up an integral to compute the average pressure in a solid region in space inside the sphere $x^2+y^2+z^2=9$, provided the pressure at each point in the sphere is given by $P(x,y,z) = 10+x$. In case you need them, remember that in spherical coordinates, we have $x=\rho\sin\phi\cos\theta$ and the Jacobian is $|\rho^2\sin\phi|$.
Solution
The AVERAGE pressure is $$\bar P = \frac{\int_{0}^{2\pi}\int_{0}^{\pi}\int_{0}^{3}(10+\rho\sin\phi\cos\theta)\rho^2\sin\phi d\rho d\phi d\theta}{ \int_{0}^{2\pi}\int_{0}^{\pi}\int_{0}^{3}\rho^2\sin\phi d\rho d\phi d\theta}.$$
Group problems
- Draw the solid whose volume is given by the integral $\ds\int_{0}^{\pi}\int_{\pi/3}^{\pi/2}\int_{0}^{3}\rho^2\sin\phi \,d\rho \,d\phi \,d\theta$. Check with Integration.nb.
- Set up an integral formula to compute the $z$-coordinate of the center of mass of the solid above, provided the density is given by $\delta = x^2+y^2+z^2$.
- Let $f = xy^2+3x$.
- Compute $\vec F = \vec \nabla f$. (We'll use capital $F$ for vector fields, and lower case $f$ for real-valued functions, i.e. the output is a single real number, rather than a vector.)
- Compute $D^2f$ (so $D\vec F$). It's a square matrix.
- Compute $\int_{ (2,1) }^{ (1,3) } df$ (total change in $f$ from $(2,1)$ to $(1,3)$). [Check: Plugging the points into $f$ yields $12-8 = 4$.]
- Let $\vec F = (2xy+4, x^2+2y)$.
- Compute $D\vec F$. It's a square matrix.
- Find a real-valued function $f$ so that $\vec \nabla f = \vec F $. In particular, this means $df = \vec F \cdot d\vec r$. [Check: $f = x^2y+4x+y^2$.]
- Find the work done by $\vec F$ (so $\int_C \vec F\cdot d\vec r$) to get from $(2,0)$ to $(0,3)$. (Or simpler, just compute $\int_C df$, the total change in $f$ from $(2,0)$ to $(0,3)$, which you can do because $df = \vec F \cdot d\vec r$. ) [Check: $9-4=5$.]
- Let $\vec F = (2x+3y, 4x+5y)$.
- Compute $D\vec F$.
- Why it is impossible to find a function $f$ so that $\vec F = \vec \nabla f$.
- Given a vector field $\vec F$, what condition must be true about $D\vec F$ for there to be a function $f$ such that $\vec\nabla f = \vec F$? We call such function $f$ a potential for $\vec F$. When $\vec F$ has a potential, we say that $\vec F$ is a gradient field.
- Compute the derivative of each vector field $\vec F$ below (obtaining a square matrix). Then find a potential for $\vec F$ or explain why the vector field has no potential.
- $\vec F = (2x,3y)$ [Check: $D\vec F = \begin{bmatrix}2&0\\0&3\end{bmatrix}$ and $f = x^2+\frac{3}{2}y^2$ yields $\vec \nabla f = (2x,3y)$. We can quickly verify that $\vec\nabla f = \vec F$ by a direct computation. ]
- $\vec F = (2y,3x)$
- $\vec F = (3y,3x)$
- $\vec F = (4x,5y,6z)$
- $\vec F = (4x,5z,6y)$
- $\vec F = (4x,5z,5y)$
- $\vec F = (2x-y,-x+4y)$
- $\vec F = (y^2+2x,2xy)$
- $\vec F = (x+yz,xz+4yz,xy+2y^2)$
- $\vec F = (x+yz,4yz,xy+2y^2)$
- $\vec F = (x+yz,xz+4yz,xy)$
- $\vec F = (yz,xz+4yz,xy+2y^2)$
- (Cylindrical Coordinates - Disc and Shell method) This sequence of problems develops both the shell and disc method as by-products of cylindrical coordinates. The only difference is the order of integration. We'll use the solid region in space that is bounded above by $z=9-x^2-y^2$ (so $z=9-r^2$) and below by the $xy$-plane. In Cartesian coordinates, the volume of this region is given by $$\int_{-3}^{3}\int_{-\sqrt{9-x^2}}^{\sqrt{9-x^2}}\int_{0}^{9-x^2-y^2}dzdydx.$$ We now work with the region using cylindrical coordinates.
- Set up a triple integral in cylindrical coordinates to compute the volume of this solid using the order $d\theta dzdr$. [Check: $\int_{0}^{3}\int_{0}^{9-r^2}\int_{0}^{2\pi}rd\theta dzdr.$ ]
- Compute the two inside integrals and simplify to show that $V = \int_{0}^{3} 2\pi r (9-r^2) dr$. Recall the shell-method $$V = \int dV = \int_a^b \underbrace{(2\pi r)(\text{height of shell at $r$})}_{\text{shell surface area = (circumference)(height)}} \underbrace{dr}_{\text{shell thickness}}.$$
- Set up a triple integral in cylindrical coordinates to compute the volume of this solid using the order $d\theta drdz$. You will end up with $r=\sqrt{9-z}$ as one of the bounds. You can use Integration.nb to check.
- Compute the two inside integrals and simplify to show that $V = \int_{0}^{9} \pi (\sqrt{9-z})^2 dz$. Recall the disc-method $$V = \int dV =\int_a^b \underbrace{\pi (\text{radius of disc at height $z$})^2}_{\text{area of disc at height $z$}} \underbrace{dz}_{\text{little height}}.$$
Textbook practice
If you want more integrals to work with. These all come from Thomas's calculus, the 14th edition. You can also get lots of practice with integration from chapter 5 of OpenStax's text. If you want to complete an SDL project that involves working through many problems from these sections, and then sharing a video or slide show of how you solve a few that helped you learn the most, I'll happily approve it. Here are the sections in Thomas's Calculus.
- Double integrals - Swapping order - 15.2: 33-54
- Polar integrals - 15.4: 1-8, 9-22 (swap to polar and then use software to check)
- Triple integrals - 15.5: 21-36
- Cylindrical and Spherical- 15.7: 37-42 (cyl), 55-60 (sph), 65-84 (BEST ones, you have to pick the system, draw the region, set things up - use the Mathematica notebook Integration.nb to check if your bounds are right).
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