Day 35 - Prep

Task 35.1

1. Use the area of a parallelogram formula we developed to explain why the Jacobian can be computed using the formula $$\frac{\partial (x,y)}{\partial (u,v)} = \left|\frac{\partial x}{\partial u}\frac{\partial y}{\partial v}-\frac{\partial x}{\partial v}\frac{\partial y}{\partial u}\right|.$$
2. For the polar coordinate transformation given by the change-of-coordinates $x=r\cos\theta$ and $y=r\sin\theta$, compute the differential $(dx,dy)$, and then show that the Jacobian is $\ds\frac{\partial (x,y)}{\partial (r,\theta)} = |r|$. This is precisely the reason we use the notation $dA = |r|drd\theta$ when setting up double integrals using polar coordinates.
3. For the change-of-coordinates $x=2u-v$ and $y=u+2v$, compute the differential $(dx,dy)$, and then obtain the Jacobian $\ds \frac{\partial (x,y)}{\partial (u,v)}$.
4. For the change-of-coordinates $x=au$ and $y=bv$, show that the Jacobian is $\ds \frac{\partial (x,y)}{\partial (u,v)}=|ab|$.
5. For the change-of-coordinates $u=x+y$ and $v=x-y$, show that $\ds \frac{\partial (u,v)}{\partial (x,y)} = 2$.
6. For the change-of-coordinates $u=x+y$ and $v=x-y$, first solve for $x$ and $y$ in terms of $u$ and $v$ (so obtain $x = ....$ and $y= ...$). Then show that $\ds \frac{\partial (x,y)}{\partial (u,v)} = \frac{1}{2}$.

Did you notice that $\ds \frac{\partial (u,v)}{\partial (x,y)} = \left(\ds \frac{\partial (x,y)}{\partial (u,v)}\right)^{-1}$?

Task 35.2

In three dimensions, some common coordinate systems are cylindrical and spherical coordinates. In this task, for each of these coordinates systems, we'll (1) develop the change-of-coordinates formula, (2) compute the Jacobian using the triple product, and then (3) use the Jacobian to compute the volume of an object in this coordinate system.

Let's first tackle cylindrical coordinates. Let $P=(x,y,z)$ be a point in space. This point lies on a cylinder of radius $r$, where the cylinder has the $z$ axis as its axis of symmetry. The height of the point is $z$ units up from the $xy$ plane. The point casts a shadow in the $xy$ plane at $Q=(x,y,0)$. The angle between the ray $\vec{Q}$ and the $x$-axis is $\theta$. See the image below.

1. Explain why the equations for cylindrical coordinates are $$x=r\cos\theta, \quad y=r\sin\theta,\quad z=z.$$ [Hint: Compute the sine and cosine of $\theta$ in terms of $x$, $y$, and $r$, and then solve for $x$ and $y$.]
2. Compute the Jacobian $\dfrac{\partial(x,y,z)}{\partial(r,\theta,z)}$ for cylindrical coordinates. The steps below are a guide, if needed.
   • For cylindrical coordinates we have $x=r\cos\theta$, $y=r\sin\theta$, and $z=z$. Write the differential $d(x,y,z)$ as the linear combination of partial derivatives $$\begin{pmatrix}dx\\dy\\dz\end{pmatrix} = \begin{pmatrix}\cos\theta\\\sin\theta\\0\end{pmatrix}dr+\begin{pmatrix}?\\?\\?\end{pmatrix}d\theta+\begin{pmatrix}?\\?\\?\end{pmatrix}dz.$$
   • Compute the volume of the parallelepiped formed by the three vectors (partial derivatives) above, using the triple product. Software can make quick work of this. Simplify your result to show that $\dfrac{\partial(x,y,z)}{\partial(r,\theta,z)} = |r|$.
3. Consider the solid domain $D$ in space that lies inside the right circular cylinder $x^2+y^2=a^2$ (or $r=a$) for $0\leq z\leq h$. Start by drawing the domain $D$. The volume in cylindrical coordinates is given by the iterated integral $$V = \iiint_D dV = \ds\int_{0}^{2\pi}\int_{0}^{a}\int_{0}^{h}rdzdrd\theta.$$ Compute this integral to obtain a familiar formula $V = \pi a^2 h$.

Now let's tackle spherical coordinates. Let $P=(x,y,z)$ be a point in space. This point lies on a sphere of radius $\rho$ ("rho"), where the sphere's center is at the origin $O=(0,0,0)$. The point casts a shadow in the $xy$ plane at $Q=(x,y,0)$. The angle between the ray $\vec{Q}$ and the $x$-axis is $\theta$, which some call the azimuth angle. The angle between the ray $\vec{P}$ and the $z$-axis is $\phi$ ("phi"), which some call the inclination angle, polar angle, or zenith angle. See the image below.

4. Explain why the equations for spherical coordinates are $$x=\rho\sin\phi\cos\theta,\quad y=\rho\sin\phi\sin\theta,\quad z=\rho\cos\phi.$$ [Hint: Compute the sine and cosine of $\phi$ in terms of $\rho$, $r$, and $z$, and then solve for $r$ and $z$. Substitution into $x=r\cos\theta$ and $y=r\sin \theta$ will get the rest.]
5. Compute the Jacobian $\dfrac{\partial(x,y,z)}{\partial(\rho,\phi,\theta)}$ for spherical coordinates. The steps below are a guide, if needed.
   • For spherical coordinates we have $$x=\rho\sin\phi\cos\theta,\quad y=\rho\sin\phi\sin\theta,\quad z=\rho\cos\phi.$$ Write $d(x,y,z)$ as a linear combination of partial derivatives, so $$\begin{pmatrix}dx\\dy\\dz\end{pmatrix} = \begin{pmatrix}\sin\phi\cos\theta\\\sin\phi\sin\theta\\\cos\phi\end{pmatrix}d\rho+\begin{pmatrix}?\\?\\?\end{pmatrix}d\phi+\begin{pmatrix}?\\?\\?\end{pmatrix}d\theta.$$
   • Compute the volume of the parallelepiped formed by the three vectors (partial derivatives) above. Software can make quick work of this. You can do it all by hand, but you'll have to use a Pythagorean identity several times to complete the simplification. Simplify your result to show that $\frac{\partial(x,y,z)}{\partial(\rho,\phi,\theta)} = |\rho^2\sin\phi|$.
6. Consider the solid domain $D$ in space that lies inside the sphere $x^2+y^2+z^2=a^2$ (or $\rho=a$). Start by drawing the domain $D$. The volume in spherical coordinates is given by the iterated integral $$V = \iiint_D dV = \ds\int_{0}^{2\pi}\int_{0}^{\pi}\int_{0}^{a}\rho^2\sin\phi d\rho d\phi d\theta.$$ Compute this integral to obtain a familiar formula $V = \frac{4}{3}\pi a^3$.

Task 35.3

For a differentiable function $f(x)$, the second part of the fundamental theorem of calculus states that $\ds\int_a^b \frac{df}{dx} dx =f(b)-f(a)$. To find the total change in a function $f$ (so $f(b)-f(a)$), we just sum the little changes $\int_C df = \int_C \frac{df}{dx}dx$. We'll use this fact to greatly simplify work computations for vector fields that have a potential.

Let $\vec F$ be a vector field with $f$ being a potential for $\vec F$, meaning $\vec F = \vec \nabla f$. Let $\vec r(t)$ for $a\leq t\leq b$ be a differentiable parametrization of a curve $C$. Let $A = \vec r(a)$ and $B= \vec r (b)$ be the end points of the curve. The composite function $g(t) = f(\vec r(t)) = (f\circ \vec r)(t)$ gives the potential at points along the curve. In particular $f(A)$ and $f(B)$ give the potential at the end points of the curve. The difference in potential is $f(B)-f(A)$.

1. Pause. Reread the above. Do you understand what each of $\vec F$, $f$, $\vec r$, $a$, $b$, $A$, $B$, and $g$ represent? If not, what parts do you have questions about? Then continue reading.

From the chain rule, the composite function $g(t) = f(\vec r(t)) = (f\circ \vec r)(t)$ has the derivative $\frac{dg}{dt}=\vec \nabla f(\vec r(t))\cdot\frac{d\vec r}{dt}$. We now compute $$\begin{align*} f(B)-f(A) &= f(\vec r(b))-f(\vec r(a))&\text{($A$ and $B$ are the end points of the curve)}\\ &= g(b)-g(a)&\text{(recall $g(t) = f(\vec r(t))$}\\ &= \int_a^b \frac{dg}{dt}dt&\text{(the fundamental theorem of calculus)}\\ &= \int_a^b \vec \nabla f(\vec r(t))\cdot \frac{d\vec r}{dt} dt&\text{(using the chain rule to compute $\frac{dg}{dt}$)}\\ &= \int_C \vec F\cdot d\vec r&\text{(recall that $\vec \nabla f=\vec F$)}. \end{align*}$$ This shows that the work done by $\vec F$ along $C$ is the difference in the potential $f$.

Let's try using this theorem in a few situations.

2. Let $\vec F(x,y) = (2x+y,x+4y)$ and $C$ be the parabolic path $\vec r(t) = (t,9-t^2)$ for $-3\leq t\leq 2$.
   1. Use the work formula we developed earlier in the semester, so $\int_C Mdx+Ndy$ or $\int_C \vec F\cdot d\vec r$, to set up and compute the work done by $\vec F$ along $\vec r$. This is a review of a previous learning target. Feel free to use software to complete the integral.
   2. Find a potential $f$ for $\vec F$, state $A$ and $B$, and then compute $f(B)-f(A)$.
   3. Identify the function $g(t) = f(\vec r(t))$, compute $\frac{dg}{dt}$, state $\vec F(\vec r(t))$ and $\frac{d\vec r}{dt}$, and then verify that $\frac{dg}{dt} = \vec F(\vec r(t))\cdot \frac{d\vec r}{dt}$ (these were the terms that appeared in the proof of the fundamental theorem of line integrals).
3. Let $\vec F(x,y,z) = (2x+yz,2z+xz,2y+xy)$ and $C$ be the straight segment from $(2,-5,0)$ to $(1,2,3)$. Compute the work done by $\vec F$ along $C$ by first finding a potential for $\vec F$.
4. Let $\vec F = (x,2yz,y^2)$. Let $C$ be the curve which starts at $(1,0,0)$ and follows a helical path $(\cos t, \sin t, t)$ to $(1,0,2\pi)$ and then follows a straight line path to $(2,4,3)$. Find the work done by $\vec F$ to get from $(1,0,0)$ to $(2,4,3)$ along this path.
5. Suppose a vector field $\vec F$ has a potential. Compute $\int_C \vec F\cdot d\vec r$ where $C$ is the path parametrized by $\vec r(t) = (3\cos t, 3\sin t)$ for $0\leq t\leq 2\pi$.

Task 35.4

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

Day 36 - Prep

Task 36.1

This task has you practice using spherical and cylindrical coordinates. Mathematica's ParametricRegion[] command allows us to plot a region, using any coordinate system, which is extremely useful for visualizing regions defined by bounds of an integral. As an example, the code below visualizes the region whose volume is given by the cylindrical coordinate iterated triple integral $$\int_{1}^{3}\int_{\pi/2}^{2\pi}\int_{0}^{r}rdz d\theta dr,$$ and then computes the triple integral.

coordinates = {r Cos[theta], r Sin[theta], z}
R = ParametricRegion[coordinates, {{r, 1, 3}, {theta, Pi/2, 2 Pi}, {z, 0, r}}];
Region[R, Axes -> True, AxesLabel -> {x, y, z}, AxesOrigin -> {0, 0, 0}]
Integrate[r, {r, 1, 3}, {theta, Pi/2, 2 Pi}, {z, 0, r}]

To compute integrals in spherical coordinates, we just update the change-of-coordinates and bounds. Here's code to plot the region whose volume is given by the spherical coordinate iterated triple integral $$\ds \int_{0}^{\pi}\int_{\pi/6}^{\pi/3}\int_{1}^{3}\rho^2\sin\phi d\rho d\phi d\theta.$$

coordinates = {rho Sin[phi] Cos[theta], rho Sin[phi] Sin[theta], rho Cos[phi]}
R = ParametricRegion[coordinates, {{theta, 0, Pi/2}, {phi, Pi/6, Pi/3}, {rho, 1, 3}}];
Region[R, Axes -> True, AxesLabel -> {x, y, z}, AxesOrigin -> {0, 0, 0}]
Integrate[rho^2 Sin[phi], {theta, 0, Pi/2}, {phi, Pi/6, Pi/3}, {rho, 1, 3}]

1. Consider the solid domain $D$ in space which is above the cone $z=\sqrt{x^2+y^2}$ and below the paraboloid $z=6-x^2-y^2$.
   1. Sketch the region by hand.
   2. Explain why an equation of the cone in cylindrical coordinates is $z=r$. Then obtain an equation of the paraboloid in cylindrical coordinates.
   3. Use cylindrical coordinates to set up an iterated triple integral that would give the volume of the region. You'll need to find where the surfaces intersect, as their intersection will help you determine the appropriate bounds. Use the ParametricRegion[] command to verify that the bounds you gave do indeed produce the correct region.
   4. By symmetry, it should be clear that for the centroid of this region, we have $\bar x = \bar y = 0$. Set up a formula involving iterated triple integrals that would give $\bar z$ for this solid, and then use software to compute $\bar z$.
2. Consider the solid domain $D$ in space that lies below the cone $z=\sqrt{x^2+y^2}$, above the $xy$-plane, and inside the sphere $x^2+y^2+z^2=25$.
   1. Provide a sketch of the domain $D$.
   2. Explain why an equation of the cone in spherical coordinates is $\phi = \pi/4$. Then given an equation of the sphere in spherical coordinates.
   3. Set up an integral in spherical coordinates that gives the volume of $D$. Use the ParametricRegion[] command to verify that the bounds you gave do indeed produce the correct region.
   4. Set up an integral in spherical coordinates that would give the $z$-coordinate of the centroid of $D$.

Task 36.2

Two important vector fields show up over and over again when studying gravity and electrostatics. In this task we will develop a common formula for these fields, show that these fields have a potential, and then practice using the fundamental theorem of line integrals to perform work computations, using the potential.

1. We need a formula for a vector field where at each point in space, the vector points towards the origin with a magnitude that proportional to 1 over the square of the distance to the origin. To obtain this field, complete the following steps.
   1. Let $P=(x,y,z)$ be a point in space. At the point $P$, let $\vec F_1(x,y,z)$ be the vector which points from $P$ to the origin. Give a formula for $\vec F_1(x,y,z)$.
   2. Give an equation of the vector field where at each point $P$ in space, the vector $\vec F_2(P)$ is a unit vector that points towards the origin.
   3. Give an equation of the vector field where at each point $P$ in space, the vector $\vec F_3(P)$ is a vector of length 7 that points towards the origin.
   4. Give an equation of the vector field where at each point $P$ in space, the vector $\vec F(P)$ points towards the origin, and has a magnitude equal to $G/d^2$ where $d = \sqrt{x^2+y^2+z^2}$ is the distance to the origin, and $G$ is a constant.
2. The gravitational vector field is directly related to the radial field $\ds\vec F(x,y,z) = \frac{\left(-x,-y,-z\right)}{(x^2+y^2+z^2)^{3/2}}$.
   1. We say that a vector field is conservative when the work done by the field is independent of the path traveled. Show that this vector field is conservative, by finding a potential for $\vec F$.
   2. Compute the work done by $\vec F$ to move an object from $(1,2,-2)$ to $(0,-3,4)$ along ANY path that avoids the origin.

Task 36.3

We need to gain some familiarity with the notation related to gradients, divergence, and curl. As you work on the tasks below, you are welcome to use subscript notation (such as $f_x$ and $M_y$) to simplify writing.

1. Suppose $f(x,y,z)$ is twice continuously differentiable.
   1. Compute the curl of the gradient of $f$, so compute $\vec \nabla \times \vec \nabla f$. Simplify the result as much as possible.
   2. If a vector field $\vec F = (M,N,P)$ has a potential, then what is the curl of $\vec F$?
2. Suppose $\vec F(x,y,z) = (M,N,P)$ is a vector field and $f(x,y,z)$ is a function, both of which are twice continuously differentiable.
   1. Compute the divergence of the curl of $\vec F$, so compute $\vec \nabla \cdot \left(\vec \nabla \times \vec F\right)$, and simplify the result as much as possible.
   2. Compute the divergence of the gradient of $f$, so compute $\vec \nabla \cdot \vec \nabla f$, and simplify the result as much as possible.

Task 36.4

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

The vector field has a potential, namely $f(x,y) = x^2-3xy +\frac{4}{3}y^3$. The start and end points of the curve are the same (the curve is a closed curve). The work done is zero, as the potential at the start and end points are the same.

The code below graphs the curve along with the vector field, and then computes the work directly from the formula (without finding a potential).

F[x_, y_] := {2 x - 3 y, -3 x + 4 y^2}
r[t_] := {3 Cos[t], 3 Sin[t]}
Show[
 VectorPlot[F[x, y], {x, -4, 4}, {y, -4, 4}], 
 ParametricPlot[r[t], {t, 0, 2 Pi}]
 ]
Integrate[Dot[F @@ r@t, D[r[t], t]], {t, 0, 2 Pi}]

The volume is given by $\ds V= \int_{0}^{\pi/2}\int_{0}^{\pi/4}\int_{0}^{7}\rho^2\sin\phi d\rho d\phi d\theta$, which means $$\bar x = \frac{\iiint_D x dV}{\iiint_D dV}= \frac{\int_{0}^{\pi/2}\int_{0}^{\pi/4}\int_{0}^{7}(\rho\sin\phi\cos\theta)\rho^2\sin\phi d\rho d\phi d\theta}{\int_{0}^{\pi/2}\int_{0}^{\pi/4}\int_{0}^{7}\rho^2\sin\phi d\rho d\phi d\theta}.$$ The following Mathematica code will visualize the region, and compute the centroid.

coordinates = {rho Sin[phi] Cos[theta], rho Sin[phi] Sin[theta], rho Cos[phi]}
R = ParametricRegion[coordinates, {{theta, 0, Pi/2}, {phi, 0, Pi/4}, {rho, 0, 7}}];
Region[R, Axes -> True, AxesLabel -> {x, y, z}, AxesOrigin -> {0, 0, 0}]
Integrate[rho Sin[phi] Cos[theta] rho^2 Sin[phi], {theta, 0, Pi/2}, {phi, 0, Pi/4}, {rho, 0, 7}]/
 Integrate[rho^2 Sin[phi], {theta, 0, Pi/2}, {phi, 0, Pi/4}, {rho, 0, 7}]

With Mathematica, we have the following solutions.

F[x_, y_, z_] := {a x + b y + c z, d x + e y + f z, g x + h y + i z }
D[F[x, y, z], {{x, y, z}}] // MatrixForm
Curl[F[x, y, z], {x, y, z}]
Div[F[x, y, z], {x, y, z}]

Here is the region for part d.

coordinates = {rho Sin[phi] Cos[theta], rho Sin[phi] Sin[theta], rho Cos[phi]}
R = ParametricRegion[coordinates, {{theta, 0, 2 Pi}, {phi, 0, Pi/4}, {rho, 0, 3}}];
Region[R, Axes -> True, AxesLabel -> {x, y, z}, AxesOrigin -> {0, 0, 0}]

The corresponding integral is $$V=\int_{0}^{2\pi}\int_{0}^{\pi/4}\int_{0}^{3}\rho^2\sin\phi d\rho d\phi d\theta.$$

Integrate[rho^2 Sin[phi], {theta, 0, 2 Pi}, {phi, 0, Pi/4}, {rho, 0, 3}]

Here is the region for part f.

coordinates = {rho Sin[phi] Cos[theta], rho Sin[phi] Sin[theta], rho Cos[phi]}
R = ParametricRegion[coordinates, {{theta, 0, 2 Pi}, {phi, 0, Pi/4}, {rho, 0, 8/Cos[phi]}}];
Region[R, Axes -> True, AxesLabel -> {x, y, z}, AxesOrigin -> {0, 0, 0}]

The corresponding integral is $$V=\int_{0}^{2\pi}\int_{0}^{\pi/4}\int_{0}^{8/ \cos\phi}\rho^2\sin\phi d\rho d\phi d\theta.$$

Integrate[rho^2 Sin[phi], {theta, 0, 2 Pi}, {phi, 0, Pi/4}, {rho, 0, 8/Cos[phi]}]