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18.02 Exam 4 -- April 30, 1999 2:05-2:55

Attempt all questions. Each problem is worth 20 points, although some are harder than others!

1.

Using the Divergence Theorem, evaluate the flux integral

$${\text{\LARGE O}}\kern-14pt\int\kern-9pt\int_S {\bf F}\cdot {\hat n}\,dS$$

where ${\bf F}=x^2{\bf i}+(y+z^3){\bf j}+y^2{\bf k},$ $S$ is the surface of the sphere $x^2+y^2+z^2=4$ and $\hat n$ is the unit outward normal.

Solution: By the divergence theorem this flux integral is equal to the volume integral

$$\iiint_{x^2+y^2+z^2\le4}{\text{div}{\bf F}}\,dV= \iiint_{x^2+y^2+z^2\le4}(2x+1)\,dV=\frac{32\pi}3$$

by symmetry and the formula for the volume of a sphere.

2.

Compute the surface area of the part of the surface $z+x^2+y^2=1$ in $z\ge0.$

Solution: The surface area is given by the integral $\iint_S dS.$ We may use the variables $x$ and $y$ on the surface, the `shadow region' is $x^2+y^2\le 1$ and the surface measure is

$$dS=(1+f_x^2+f_y^2)^{\frac12}\,dx\,dy=(1+4x^2+4y^2)^{\frac12}\,dx\,dy,\ f(x,y)=1-x^2-y^2.$$

Thus the surface area is

$$\begin{multline*} \iint_{\{x^2+y^2\le1\}}(1+4x^2+4y^2)^{\frac12}\,dx\,dy =\int_0... ...\frac1{12}(1+4r^2)^{\frac32}\big]_0^1=\frac16\pi(5^{\frac32}-1). \end{multline*}$$

3.

Let $D$ be the three dimensional region between the two surfaces

$$z = (x^2 + y^2)^2$$

$$z = 8 + 2(x^2 + y^2) \, .$$

(a)

Sketch the region $D$.

(b)

Using cylindrical (polar) coordinates compute the moment of inertia of $D$, assumed to have unit density, around the $z$-axis.

Solution: a) Anything reasonable! It lies over $x^2+y^2\le 4$ with $(x^2+y^2)^2\le z\le 8+2(x^2+y^2).$

b)

$$\hbox{MoI}=\int\int\int_D (x^2+y^2)dV=\int_0^{2\pi}\int_0^2\int_{r^4}^{8+2r^2} r^3dzdrd\theta$$

$$=2\pi\int_0^2 r^3(8+2r^2-r^4)dr=2\pi(2\cdot2^4+2^6/3-2^8/8)=128\pi/3.$$

4.

Let $D$ be the region in three dimensional space consisting of the points $(x,y,z)$ with $x \geq 0$, $y \geq 0$, $z \geq 0$ and $x^2 + y^2 + z^2 \leq 4$.

Assuming where needed that $D$ has unit density, write down integral formulae in terms of spherical coordinates (DO NOT EVALUATE) for

(a)

The average over $D$ of the distance from the $z$-axis.

(b)

The moment of inertia of $D$, with unit density, around the $z$-axis.

(c)

The $\hat{k}$ component of the gravitational force exerted by $D$ on a unit mass at the origin.

Solution: The region in spherical coordinates is $0\le\phi\le \pi/2,$ $0\le\theta\le\pi/2$ and $0\le\rho\le2.$ Thus the three integrals are

(a)

$\int_0^{\pi/2}\int_0^{\pi/2}\int_0^2 \rho^3\sin^2\phi d\rho d\phi d\theta/(4\pi/3)$ (or an integral for the volume can be written out).

(b)

$\int_0^{\pi/2}\int_0^{\pi/2}\int_0^2 \rho^4\sin^3\phi d\rho d\phi d\theta.$

(c)

$G\int_0^{\pi/2}\int_0^{\pi/2}\int_0^2\sin\phi\cos\phi d\rho d\phi d\theta.$

5.

Let $D$ be `ice-cream cone' consisting of the points in the unit ball (=solid unit sphere) where the `azimuth' (angle with the positive direction of the z-axis) $\varphi<\pi/4.$ Assuming that it has unit density, compute the gravitational force $D$ exerts on a unit mass at the origin.

Solution: Place $D$ so that its axis is the z-axis (as indicated). Then in spherical polar coordinates the ${\bf k}$ component (and hence the actual) gravitational force on a unit mass at the origin is

$$G\int_0^{2\pi}\int_0^{\pi/4}\int_0^1 \frac{\cos\varphi }{\rho ^2}... ...phi d\rho\, d\varphi \,d\theta =2\pi G[\frac12\sin^2\varphi]_0^{\pi/4}=\pi G/2.$$