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.