# Tag Archives: M32B

teaching

## Math 32B Sample Questions

I’ve finished writing up some sample questions for the Math 32B final. The questions are available here and their solutions here. These questions are are my best guess of the sort of material that I feel is likely to appear on a final exam. As usual, feel free to let me know in the comments below if you find any errors.

teaching

## Math 32B Final Review

I have written up a whirlwind review of the material covered in Math 32B this quarter. It is available here. The review is not meant to be entirely comprehensive, but I think it covers the bulk of the computational material from the course. The review is longer than I was hoping (6 pages) but it is certainly more condensed than the text which will hopefully be appealing to some students. As usual, let me know in the comments if you find errors or if anything in the text is unclear. In the next couple days, I will write up some sample problems and solutions for the exam. So stay tuned.

teaching

## Homework 8 Hints

I’ve had a number of requests for hints and/or solutions to several of the homework problems for tomorrow. So here we go.

17.5 #30 This question asks us to compute the gravitational flux through a cylinder $$S$$ from the gravitational field
$\mathbf{F} = – G m \frac{\mathbf{e}_r}{r^2}.$
Here, $$S$$ is the cylinder of radius $$R$$ whose axis of symmetry is the $$z$$-axis with $$a \leq z \leq b$$, and $$r = \sqrt{x^2 + y^2 + z^2}$$ is the distance from $$(x, y, z)$$ to the origin.

Recall that $$\mathbf{e}_r(x, y, z)$$ is the unit vector that points in the direction $$\langle x, y, z \rangle$$. Therefore, we can write
$\mathbf{F}(x, y, z) = – G m \frac{\langle x, y, z \rangle}{(x^2 + y^2 + z^2)^{3/2}}.$
In general, if $$H(u, v)$$ is a parametrization of a surface $$S$$, we compute the flux of $$\mathbf{F}$$ through $$S$$ to be
$\iint_S \mathbf{F} \cdot d\mathbf{S} = \iint_D F(H(u, v)) \cdot \mathbf{n}(u, v)\, du \, dv$
where
$\mathbf{n}(u, v) = \frac{\partial H}{\partial u} \times \frac{\partial H}{\partial v}$
is the normal vector for the surface $$S$$. For this problem, we cannot parametrize the cylinder with a single equation: we must parametrize the top, sides, and bottom separately. Hence, we must compute 3 integrals to determine the flux.

The sides of the cylinder can be parametrized in cylindrical coordinates by
$H(\theta, z) = (R \cos \theta, R \sin \theta, z)$
where $$0 \leq \theta \leq 2 \pi$$ and $$a \leq z \leq b$$. You should check that this parametrization gives a normal vector $$\mathbf{n}(\theta, z) = \langle (R \cos \theta, R \sin \theta, 0 \rangle$$. Therefore, the contribution of the flux from the sides of the cylinder is
$\iint_{\mathrm{sides}} \mathbf{F} \cdot d\mathbf{S} = – G m\int_a^b \int_0^{2 \pi} \frac{R^2}{(R^2 + z^2)^{3/2}}\, d\theta\, dz.$
You can compute the outer integral using the trig substitution $$z = R \tan u$$.

For the top of the cylinder, we must parametrize the disk of radius $$R$$ parallel to the $$xy$$–plane centered at $$(0, 0, b)$$. Again, we can use polar coordinates to parametrize this region as
$H(r, \theta) = (r \cos \theta, r \sin \theta, b)$
for $$0 \leq r \leq R$$ and $$0 \leq \theta \leq 2 \pi$$. This parametrization gives a normal vector of $$\mathbf{n}(r, \theta) = \langle 0, 0, r \rangle$$. Notice that this is the upward pointing normal vector, which is what we want since we need an outward pointing normal for the entire cylinder $$S$$. Therefore, the integral for the flux through the top face of the cylinder is
$\iint_{\mathrm{top}} = – G m \int_0^R \int_0^{2 \pi} \frac{b r}{(b^2 + r^2)^{3/2}}\, d\theta \, dr.$
This integral is easily computed using a $$u$$ substitution.

The flux through the bottom face of the cylinder is almost identical to the top face except that $$z = a$$ on the bottom face, and we must take the downward facing normal vector $$\mathbf{n}(r, \theta) = \langle 0, 0, -r \rangle$$. When computing computing the integral for the bottom face, you will get a term involving $$\sqrt{a^2}$$. Be careful that you simplify $$\sqrt{a^2} = |a|$$, and not just $$a$$, for this will make a difference (in fact the crucial difference) when $$a$$ is negative instead of positive.

18.1 #40 This problem asks you to show that when $$\mathbf{F} = \nabla \phi$$, we have $$\mathrm{curl}_z(\mathbf{F}^*) = \Delta \phi$$. Recall that for $$\mathbf{F} = \langle F_1, F_2 \rangle$$

• $$\mathbf{F}^* = \langle – F_2, F_1 \rangle$$
• $$\mathrm{curl}_z(\mathbf{F}) = \frac{\partial F_2}{\partial x} – \frac{\partial F_1}{\partial y}$$
• $$\Delta \phi = \frac{\partial^2 \phi}{\partial x^2} + \frac{\partial^2 \phi}{\partial y^2}$$.

Using these definitions and $$\mathbf{F} = \nabla \phi = \langle \partial \phi / \partial x, \partial \phi / \partial y\rangle$$, we can compute
$\mathrm{curl}_z(\mathbf{F}^*) = \mathrm{curl}_z( \langle – \partial \phi / \partial y, \partial \phi / \partial x \rangle) = \frac{\partial^2 \phi}{\partial x^2} + \frac{\partial^2 \phi}{\partial y^2},$
which is exactly what we wanted to show.

teaching

## Explanation of 17.5 #48

In this problem, you are asked to compute the gravitational potential from uniform density spherical shell of total mass $$m$$ and radius $$R$$. In general, the gravitational potential of a massive surface $$S$$ is given by
$V(a, b, c) = – G \iint_S \frac{\rho\, dS}{\sqrt{(x – a)^2 + (y – b)^2 + (z – c)^2}}$
where $$\rho$$ is the density (mass per area) of the surface. In this case, $$\rho = m / (4 \pi R^2)$$.

The first part of the question asks you to reduce the problem to the case where $$(a, b, c) = (0, 0, r)$$. We can do this by spherical symmetry: given any $$(a,b,c)$$ we can rotate the entire picture so that the point $$(a, b, c)$$ lies on the positive $$z$$ axis. That is, $$(a, b, c) = (0, 0, r)$$ with $$r = \sqrt{x^2 + y^2 + z^2}$$. Applying this rotation doesn’t change the gravitational potential because the rotated sphere is indistinguishable from the original sphere.

The second part of the question asks you to set up the integral for $$V(0, 0, r)$$ in spherical coordinates. In spherical coordinates (with radius $$R$$) we have
$x = R \cos \theta \sin \varphi, \quad y = R \sin \theta \sin \varphi, \quad z = R \cos \varphi.$
You should verify that the normal vector for this parametrization satisfies $$\|n\| = R^2 \sin \varphi$$. Notice that the denominator of the integral defining the gravitational potential is just the distance from $$(a, b, c)$$ to $$(x, y, z)$$ (where the latter is a point on the surface). In the case where $$(a, b, c) = (0, 0, r)$$ and $$S$$ is the sphere of radius $$R$$ centered at the origin, we can compute this distance using the law of cosines as indicated in the following figure:

In the figure, $$d$$ is the distance from the point $$P = (0,0,r)$$ to a point $$Q = (x, y, z)$$ lying on the sphere — that is, $$d$$ is the denominator of the integral we’re trying to compute. By the law of cosines,
$d^2 = r^2 + R^2 – 2 R r \cos \varphi.$
Therefore, the potential is given by
$V(0, 0, r) = \frac{- G m}{4 \pi} \int_0^{\pi} \int_{0}^{2 \pi} \frac{\sin \varphi \, d\theta\, d\varphi}{\sqrt{R^2 + r^2 – 2 R r \cos \varphi}}.$
The book suggests using the substitution $$u = R^2 + r^2 – 2 R r \cos \varphi$$ to compute this integral.

teaching

## Math 32B/H Midterm 2 Review

I just finished writing up some review materials for the second midterm for Math 32B/H. There are two documents: the problems (here) and solutions (here). If you find any typos or points that need clarification, feel free to let me know in the comments below.

teaching

## Problem 17.2 #31

In this problem, we are asked to compute the line integral
$\int_C \frac{-y\, dx + x\, dy}{x^2 + y^2}$
where $$C$$ is the line segment from $$(1,0)$$ to $$(0,1)$$. It turns out that this integral is particularly easy to compute in polar coordinates (although the work that I did in class probably made it seem hopelessly complicated!).

In polar coorinates, we have $$x = r \cos \theta$$ and $$y = r \sin \theta$$. To get $$dx$$ and $$dy$$ in terms of $$r$$, $$\theta$$ and $$d\theta$$ we have to do a little work differentiating. Since we are integrating over the curve $$C$$, we will eventually have $$r = r(\theta)$$, a function of theta. We compute
$dx = d(r \cos \theta) = \left(\frac{dr}{d\theta} \cos \theta – r \sin \theta \right)\, d\theta$
and similarly
$dy = d(r \sin \theta) = \left(\frac{dr}{d\theta} sin \theta + r \cos \theta \right)\, d\theta.$
Using these formulas, the numerator of the integrand simplifies immensely:
$– y\, dx + x\, dy = r^2 \, d\theta.$
Therefore, the integral becomes (in polar coordinates)
$\int_C \frac{-y\, dx + x\, dy}{x^2 + y^2} = \int_C \, d\theta.$
The curve $$C$$ can be written in polar coordinates as
$r(\theta) = \frac{1}{\cos \theta + \sin \theta}$
where $$0 \leq \theta \leq \pi / 2$$. To see this, notice the line satisfies $$y = 1 – x$$, or equivalently $$x + y = 1$$. In polar coordinates, this equation is $$r \cos \theta + r \sin \theta = 1$$, and solving for $$r$$ gives the equation for $$C$$. Therefore, we can compute the integral
$\int_C\, d\theta = \int_0^{\pi / 2} \, d\theta = \frac{\pi}{2}.$
So despite the imposing setup to the problem, the final integral turns out to be very simple.

teaching

## Explanation of 17.1 #34

This problem asks you to argue that the depicted vector field is not conservative. Recall that a vector field $$\mathbf{F}(x, y)$$ isĀ conservative if we can write
$\mathbf{F}(x, y) = \nabla V(x, y)$
where $$V(x, y)$$ is a scalar function. That is,
$\mathbf{F} = \left\langle \frac{\partial V}{\partial x}, \frac{\partial V}{\partial y} \right\rangle.$
The geometric interpretation of the gradient $$\nabla V$$ is that

1. $$\nabla V$$ is perpendicular to the level curves given by $$V(x, y) = c$$
2. $$\nabla V$$ points in the direction of greatest increase of $$V$$
3. $$\|\nabla V\|$$ is the slope of $$V$$ in the direction of greatest increase.

From the figure, the vector field $$\mathbf{F}$$ always points horizontally to the right. Therefore, by property 1 of the gradient, the level curves of $$V$$ must be vertical lines.

The figure also depicts the lengths of the vectors $$\mathbf{F}(x, y)$$ as decreasing as $$y$$ increases. By properties 2 and 3, this tells us that the supposed potential function $$V$$ is increasing as $$x$$ increases, but that this increase is slower for larger values $$y$$. However, this implies that the level curves are further apart for larger values $$y$$. This contradicts the previous observation that the level curves of $$V$$ must be vertical lines as vertical lines are a constant $$x$$–width apart!

$\mathbf{F} = \langle F_1, F_2 \rangle = \left\langle \frac{\partial V}{\partial x}, \frac{\partial V}{\partial y} \right\rangle,$
then by Clairaut’s theorem, we must have
$\frac{\partial F_1}{\partial y} = \frac{\partial F_2}{\partial x}.$
In the depicted vector field, $$F_1$$ decreases as $$y$$ increases, but $$F_2$$ is constant (in fact, $$0$$). Therefore, $$\frac{\partial F_1}{\partial y}\neq 0$$, but $$\frac{\partial F_2}{\partial x} = 0$$. Thus the vector field cannot possibly be conservative, as these two partial derivatives would be equal for a conservative vector field.

teaching

## Math 32B/H Midterm 1 Review Materials

I have prepared some review materials for the first midterm for Math 32B and Math 32B H. I wrote up some sample problems (here) and solutions (here). As usual, let me know if you find any errors or if an explanation is unclear.

teaching

## Explanation of 16.5 #56

By popular demand, I present here an explanation of problem 16.5 #56 from the homework due on Monday. The problem gives the equation of the wave function for an electron in a hydrogen atom
$\psi_{S1}(\rho) = -\frac{1}{\sqrt{\pi a_0^3}} e^{-\rho / a_0}.$
Here, $$\rho$$ is the distance from the origin, so that the equation for $$\psi_{1S}$$ is in spherical coordinates. The probability that the electron is in some region $$W$$ can be computed by integrating the function
$p(\rho) = \left|\psi_{S1}(\rho)\right|^2 = \frac{1}{\pi a_0^3} e^{-2 \rho / a_0}$
over the region $$W$$. The problem asks you to compute the probability that an electron is found outside of the Bohr radius, $$a_0$$. In this case, the region $$W$$ is everything in $$\mathbf{R}^3$$ outside of the ball of radius $$a_0$$ centered at the origin. In polar coordinates, $$W$$ is easily described by the following inequalities:
\begin{align*} a_0 \leq &\rho < \infty\\ 0 \leq &\varphi \leq \pi\\ 0 \leq &\theta \leq 2 \pi \end{align*} So the integral we must compute in (in spherical coordinates) $\int_0^{2\pi}\int_0^\pi\int_{a_0}^\infty \frac{1}{\pi a_0^3} e^{-2 \rho / a_0} \rho^2 \sin \varphi \, d\rho\, d\varphi \, d\theta.$ The $$\varphi$$ and $$\theta$$ integrals are straightforward to compute, but the $$\rho$$ integral is a little bit tricky. I computed it using integration by parts (applied twice). Using this method, I got a solution of $\frac{5}{e^2}$ which agrees with the answer the book provides.

teaching

## HW1 Problem

I have written up an explanation of problem 51 from section 16.1 in the text. A PDF of the write-up is available here. Feel free to let me know if you have any questions or notice any errors.