18.02 Practice Exam 4 -- for April 25, 1997

Directions: Suggested time: 50 minutes.

1.

(20 points) The solid $D$ is represented as the region of space between the graph of $z=1-x^2-y^2$ and the $xy$-plane. Find its moment of inertia about the $z$-axis, if the density is $\ell$.

Solution: The moment of inertia around the z-axis is the integral

$$\begin{multline*} \iiint_D\ell (x^2+y^2)dV=\iint_R\int_0^{1-x^2-y^2}\ell(x^2+y^2) dz\,dx\,dy\\ =\iint_R \ell(x^2+y^2)(1-x^2-y^2)dx\,dy \end{multline*}$$

where the shadow region $R$ is $x^2+y^2\le1.$ Introducing polar coordinates this becomes

$$=\int_0^{2\pi}\ell\int_0^1(1-r^2)r^2\,rdr\,d\theta =2\pi\ell\big[\frac14r^4-\frac16r^6\big]_0^1=\frac\pi6\ell.$$

2.

(20 points) A solid $D$ has the shape of a filled ice cream cone: a right circular cone, surmounted by a slice from a sphere of radius $a$.

The cone has slant height $a$ and a vertex angle $60$ degrees as pictured.

Take the density to be $1$, and find the gravitational attraction of $D$ on a unit mass placed at the vertex. (Set up and evaluate a triple integral in spherical coordinates.)

Solution: By symmetry, the gravitational attraction on a unit mass at the origin is in the direction of the z-axis and hence equal to its ${\bf k}$ component. This is

$$\begin{multline*} G\iiint_D \frac{\cos\varphi}{\rho ^2}dV =\int_0^{2\pi}\int_0^{... ...\theta\\ =G2\pi a\big[\frac12\sin^2\varphi ]^{\pi/6}_0=Ga\pi/4. \end{multline*}$$

3.

(5 points) Modify the iterated triple integral of the preceding problem so it now calculates the gravitational attraction at the vertex due to the part of $D$ lying above the top of the cone (i.e., to the spherical slice). Just set up the integral; don't evaluate it.

Solution: Instead the integral becomes

$$\int_0^{2\pi}\int_0^{\pi/3}\int_{\frac{a\cos\pi/6}{\cos\varphi}}^a \cos\varphi \sin\varphi d\rho \,d\varphi \,d\theta.$$

4.

(10 points) Suppose the vector field $F$ in the $xy$-plane satisfies $F= \nabla f$, where

$$f_{xx} + f_{yy} =0 \quad \hbox{for all } x,y \, .$$

Show that $F$ has no net flux through any simple closed curve $C$.

This should not really have been on this practice exam. Better do one for flux through a surface.

Solution: ${\bf F}=f_x{\bf i}+f_y{\bf j}.$ The flux form of Green's theorem for a simple closed curve $C$ is

$$\oint_C{\bf F}\cdot{\bf\hat n}ds=\iint_R \mathop{div}{\bf F}dA.$$

Since $\mathop{div}{\bf F}=\frac{\partial}{\partial x}(f_x)+\frac{\partial}{\partial y}(f_y)=f_{xx}+f_{yy}=0$ by assumption. Thus the flux integral must vanish.

5.

(45 points; 5, 15, 25)

Let $S$ be the hemisphere of radius $1$, with base in the $xy$-plane and center at the origin, and $T$ be its base (the disc of radius $1$ and center at the origin in the $xy$-plane). Both surfaces are oriented so their unit normal vectors point upwards, i.e., this is the direction of positive flux. Let ${\bf F}=(3z+1){\bf k}$.

(a)

Find the flux of $F$ over $T$, by inspection.

(b)

Using your answer to part (a), find the flux of $F$ over $S$ by using the divergence theorem.

(c)

Find the flux of $F$ over $S$ directly, by calculating a surface integral using spherical coordinates.

Solution:

(a)

On $T$ the normal upward normal is ${\bf k}$ so ${\bf F}\cdot{\bf j}=1$ and hence the flux upward through $T$ is the area $\pi.$

(b)

If ${\bf\hat n}$ is the upward unit normal to both $S$ and $T$ then by the divergence theorem

$$\iint_S{\bf F}\cdot{\bf\hat n}dS-\iint_T{\bf F}\cdot{\bf\hat n} dS=\iint_D{\nabla}\cdot{\bf F}dV.$$

The divergence is ${\nabla}\cdot{\bf F}=3$ so

$$\iint_S{\bf F}\cdot{\bf\hat n}dS=3 (D)+\pi=3\pi.$$

(c)

The flux integral written explicitly is

$$\iint_S {\bf F}\cdot{\bf\hat n}dS=\int_0^{2\pi}\int_0^{\pi/2}(3z+1)z \sin\varphi d\varphi \,d\theta .$$

Here $z=\cos\varphi$ on $S$ so this flux integral is

$$\begin{multline*} 2\pi\int_0^{\pi/2}(3\cos^2\varphi\sin\varphi+\cos\varphi \sin\... ...\big[-\cos^3\varphi -\frac12\cos^2\varphi \big]_0^{\pi/2} =3\pi. \end{multline*}$$