# 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.