Fall 1998 Exam Solutions

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1.

The answers are: T, F, F, C, F, F, T, F, T, F, T, F.

2.

(a)

1

(b)

$ye^{xy}\vec{i} + xe^{xy}\vec{j}$

(c)

9

(d)

$\vec{0}$

3.

(a)

$$\begin{eqnarray*}{\frac{\partial E}{\partial v}} & = &mc^2 {\frac{\partial }{\pa... ...} \left(\frac{-v^2}{c^2}\right) = \frac{mv}{(1-v^2/c^2)^{3/2}}. \end{eqnarray*}$$

(b)

Since the kinetic energy of a particle increases with increasing speed ( $K = \frac{1}{2}mv^2$) one expects the sign of the derivative of E with respect to v to be positive. This is definitely true, since (1) m>0, (2) v is speed, which is the magnitude of velocity, so v > 0, and (3) v < c implies that 1-v²/c² > 0.

4.

(a)

Taking the volume to be above the cone and below the sphere, we find that the volume is given by (NOTE: There is a typographic error in the following formula. The limits of integration on z should start at r and end at SQUARE_ROOT(4-r²).)

$$\int_W dV = \int_0^{2\pi} \int_0^{\sqrt{2}} \int_r^{4-r^2} r dz dr d\theta.$$

(b)

In spherical coordinates, the cone is just $\phi = \pi/4$ so

$$\int_W dV = \int_0^{2\pi} \int_0^{\pi/4} \int_0^2 \rho^2 \sin \varphi d\rho d\varphi d\theta.$$

There are other possible ways to compute the volume. These are just the simplest expressions in these two coordinate systems.

5.

(a)

The gradient vector should be drawn perpendicular to the contour, from P toward the direction of increasing f, which is toward the f=25 contour.

(b)

The gradient vector at P is shorter than the gradient vector at Q because the contours at Q are closer together, which means the function is steeper so that the rate of change of f is larger.

(c)

By the Fundamental Theorem of Calculus for Line Integrals, we have $\int_C \nabla f \cdot d\vec{r} = f(Q) - f(P) = 23 -21 = -2$.

6.

(a)

No, the vector field shown is not conservative. If C is a square path centered at the origin and oriented counter-clockwise, then the circulation of $\vec{F}$ around C is clearly positive, since the vector field always points in the same direction as the curve, except for the two segments where it is perpendicular to the curve. If $\vec{F}$ were a conservative vector field, the circulation around any closed path would be zero.

(b)

Using the same path C as in (a), we see that the circulation of $\vec{F}$ is non-zero. In fact, the right-hand rule shows that the direction of curl $\vec{F}$ is in the $+\vec{k}$ direction, since curl $\vec{F} \cdot \vec{k}$ is just the circulation density in the xy plane.

(c)

We know the following about $\vec{F}$: (1) there is no y or z dependence in the vector field, (2) there is no $\vec{i}$ or $\vec{k}$ component to $\vec{F}$, (3) the magnitude of $\vec{F}$ increases as you move farther from the y axis, and (4) $\vec{F}$ points toward $+\vec{j}$ when x > 0 and $-\vec{j}$ when x < 0. So, $\vec{F}(x,y,z) = c x \vec{j}$ (where c is a positive constant) is one posible formula.

7.

(a)

The curve C₁ should be a straight line from the point (0,-1) along the y-axis to the point (0,1) and the curve C₂ should be a semicircle of radius 1 oriented counter-clockwise from (0,1) through the point (-1,0) and down to the point (0,-1).

(b)

Since the curve C = C₁ + C₂ is a closed curve and is oriented counter-clockwise, and the vector field $(x+3y)\vec{i} + y\vec{j}$ is smooth, we can use Green's theorem:

$$\begin{eqnarray*}\int_C \vec{F} \cdot d\vec{r} & = & \int_R \left( {\frac{\parti... ...x{area of R}) = (-3) \frac{1}{2} \pi (1)^2 = -\frac{3\pi}{2}. \end{eqnarray*}$$

This problem can also be done by using the two parameterizations given. However, this is much longer.

$$\begin{eqnarray*}\int_C \vec{F} \cdot d\vec{r} & = & \int_{C_1} \vec{F} \cdot d\... ...\frac{\sin(2t)}{4} \right]_{\pi/2}^{3\pi/2} = -\frac{3\pi}{2}. \end{eqnarray*}$$

8.

(a)

Since the vector field $\vec{F}$ points radially outward from the origin, it is normal to the curve C₁ at all points. Thus, $\int_{C_1} \vec{F} \cdot d\vec{r} = 0$.

(b)

The vector field $\vec{G}$ is constant and the path has a constant tangent vector $d\vec{r} = \vec{j}dt$ so the line integral is just $\int_{C_2} \vec{G} \cdot d\vec{r} = \int_0^{10} 2 dt = 20.$

(c)

The unit normal vector for the surface is simply $\frac{x\vec{i} + y\vec{j}}{\sqrt{x^2 + y^2}}$ so the $\vec{k}$ component of $\vec{F}$ is irrelevant. Computing $\vec{F} \cdot \hat{n}$ we get $\frac{x^2 + y^2}{\sqrt{x^2 + y^2}} = \sqrt{x^2 + y^2}$ which is just 1 on the surface of the cylinder. So $\int_{S_1} \vec{F} \cdot \hat{n} dS = \int_{S_1} (1) dS$ which is just (1) times the surface area of the cylinder, which is $2\pi (1)^2 (1) = 2\pi$.

(d)

Since S₂ is a closed surface oriented outward and the vector field $\vec{G}$ is defined on all of S₂ and the volume W inside of S₂, we can use the divergence theorem. Further, since $\vec{G}$ is a constant vector field, we have div $\vec{G} = 0$ so that $\int_{S_2} \vec{G} \cdot \hat{n} dS = \int_W \mbox{div} \vec{G} dV = 0$.

9.

(a)

The unit normal vector to the sphere is $\hat{n} = \frac{\vec{r}}{\vert\vert\vec{r}\vert\vert}$ so that $\vec{F} \cdot \hat{n} = \frac{\vec{r} \cdot \vec{r}}{\vert\vert\vec{r}\vert\vert^4} = \frac{1}{\vert\vert\vec{r}\vert\vert^2}$ since $\vec{r} \cdot \vec{r} = \vert\vert\vec{r}\vert\vert^2$. On the surface of the sphere, $\vert\vert\vec{r}\vert\vert = 1$ so the flux is simply $\int_S \vec{F} \cdot \hat{n} dS = \int_S (1) dS$ which is just the surface are of the sphere, $4\pi (1)^2 = 4\pi$.

(b)

To calculate the divergence of $\vec{F} = \vec{r}/r^3$ we must first write the vector field in Cartesian coordinates. We find

$$\vec{F} = \frac{\vec{r}}{\vert\vert\vec{r}\vert\vert^3} = \frac{x\vec{i} + y\vec{j} + z\vec{k}}{(x^2 + y^2 + z^2)^{3/2}}.$$

Then,

$$\begin{eqnarray*}\mbox{div} \, \vec{F} & = & {\frac{\partial }{\partial x}} \lef... ...{r}\vert\vert^3} - \frac{3}{\vert\vert\vec{r}\vert\vert^3} = 0. \end{eqnarray*}$$

(c)

We know that we cannot directly use the divergence theorem, since the vector field is undefined at the origin, which is inside the region of integration. However, the vector field is radially symmetric, so one would rather compute the flux of this vector field through a sphere centered at the origin, rather than through a cube. Let S be the surface of the cube and let S₂ be the surface of a sphere of radius 1, centered at the origin and oriented outward. Let W be the volume of space inside the cube and outside the sphere. We then have that

$$\begin{eqnarray*}\int_S \vec{F} \cdot \hat{n} dS & = & \int_S \vec{F} \cdot \hat... ...vec{F} \cdot \hat{n} dS + \int_{S_2} \vec{F} \cdot \hat{n} dS. \end{eqnarray*}$$

The first of these integrals involves a closed surface (and is oriented outward, due to the -S₂) and does not include the origin, so that we can use the divergence theorem. Thus, using part (b) and letting W represent the volume between the sphere and the cube, we have

$$\int_{S - S_2} \vec{F} \cdot \hat{n} dS = \int_W \mbox{div} \, \vec{F} dV = 0.$$

The second integral is simply the integral from part (a) which is just $4\pi$. Thus, the divergence theorem can be used to evaluate the integral, if one manipulates the problem correctly.