PAGE 1

Vector Field Analysis

Let \( \vec{F}(x, y, z) = 2y\vec{i} - x\vec{j} + 3\vec{k} \). Which of the following statements is true?

(i) \( \text{Curl } \vec{F} = \vec{0} \).
(ii) \( \vec{F} \) is a gradient of some function \( f \).
(iii) Line integrals, \( \int_C \vec{F} \cdot d\vec{r} \), over a curve \( C \) from a point \( P \) to a point \( Q \), are path independent.
(iv) \( \int_C \vec{F} \cdot d\vec{r} = 0 \) for all smooth closed curves \( C \).

Multiple Choice Options

A. (i) only
B. (ii) only
C. (ii) and (iii) only
D. (ii), (iii) and (iv) only
E. None of the above

Handwritten Solution

\[ \vec{F} = \langle 2y, -x, 3 \rangle \]

\[ \text{curl } \vec{F} = \begin{vmatrix} \vec{i} & \vec{j} & \vec{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ 2y & -x & 3 \end{vmatrix} \]

\[ = \langle 0-0, 0-0, -1-2 \rangle = \langle 0, 0, -3 \rangle \]

If \( \vec{F} \) is conservative, then \( \vec{F} = \nabla f \), and \( \text{curl } \vec{F} = \vec{0} \).

Here, \( \vec{F} \) is NOT conservative, so is NOT \( \nabla f \).

PAGE 2

Analysis of Statements (iii) and (iv)

iii)

\( \int_C \vec{F} \cdot d\vec{r} \) is path independent only if \( \vec{F} = \nabla \phi \)

iv)

no, for the same reason as iii)

PAGE 3

Line Integrals and Green's Theorem

Evaluate \(\int_C y dx + \cos y dy\) where \(C\) is the polygonal path from \((0,0)\) to \((1,0)\) to \((1,2)\) to \((0,2)\) to \((0,0)\).

The path \(C\) is a closed rectangle in the \(xy\)-plane, bounded by the vertices mentioned. We can identify the vector field as:

\[ \vec{F} = \langle y, \cos y \rangle = \langle f, g \rangle \]\[ d\vec{r} = \langle dx, dy \rangle \]

Figure: Coordinate graph showing a rectangular path C with segments C1 through C4 enclosing region R.

Multiple Choice Options

A. \(-1\)
B. \(0\)
C. \(-2\) (Correct)
D. \(-\frac{3}{2}\)
E. \(\frac{2}{3}\)

Solution Methods

There are two ways to solve this:

Parametrize paths \(C_1, C_2, C_3, C_4\) then do \(\int_C \vec{F} \cdot d\vec{r}\).
Use Green's Theorem (since the path is closed).

Green's Theorem:

\[ \oint_C \vec{F} \cdot d\vec{r} = \iint_R (g_x - f_y) dA \]

Here, \(g_x - f_y = 0 - 1 = -1\).

\[ \iint_R (-1) dA = (-1) \iint_R dA = (-1) \times (\text{area of } R) \]\[ = (-1)(1)(2) = -2 \]

Alternatively, using iterated integrals:

\[ \text{or } = (-1) \int_0^1 \int_0^2 dy dx \]

PAGE 4

Identifying Quadric Surfaces

Identify the surface \(2x^2 + 3z^2 = 4x + 2y^2\) through completing the square.

Multiple Choice Options

A. Cone
B. Ellipsoid (signs)
C. Parabolic hyperboloid
D. Hyperboloid of one sheet (Correct)
E. Hyperboloid of two sheets

Para. hyp: saddle / pringle shape

Algebraic Derivation

\[ 2x^2 - 4x - 2y^2 + 3z^2 = 0 \]\[ 2(x^2 - 2x) - 2y^2 + 3z^2 = 0 \]\[ 2(x^2 - 2x + 1) - 2y^2 + 3z^2 = 2 \]\[ 2(x-1)^2 - 2y^2 + 3z^2 = 2 \]\[ (x-1)^2 - y^2 + \frac{3}{2}z^2 = 1 \]

Center: \((1, 0, 0)\). Let \(w = x - 1\).

\[ w^2 - y^2 + \frac{3}{2}z^2 = 1 \]

Figure: Sketch of a hyperboloid of two sheets, consisting of two separate bowl-like surfaces.

Trace Analysis

\(wy\)-trace (\(z=0\)): \(w^2 - y^2 = 1\) (hyperbola), \(w\)-ints: \(\pm 1\)
\(wz\)-trace (\(y=0\)): \(w^2 + \frac{3}{2}z^2 = 1\) (ellipse)
\(yz\)-trace (\(w=0\)): \(\frac{3}{2}z^2 - y^2 = 1\) (hyperbola)

Figure: Sketch of a hyperboloid of one sheet, a continuous surface with a narrow waist.

Figure: 3D coordinate axes w, y, z showing hyperbolic and elliptical traces of the surface.

PAGE 5

If \(\vec{F}(x, y, z) = x\vec{i} + y\vec{j} + z\vec{k}\), \(\Sigma\) is the unit sphere \(x^2 + y^2 + z^2 = 1\) and \(\vec{n}\) is the outward unit normal on \(\Sigma\), then \(\iint_{\Sigma} \vec{F} \cdot \vec{n} dS =\)

A. \(-4\pi\)

B. \(\frac{2\pi}{3}\)

C. 0

D. \(\frac{4\pi}{3}\)

E. \(4\pi\)

\[ \vec{F} = \langle x, y, z \rangle \quad \Sigma : x^2 + y^2 + z^2 = 1 \quad \vec{n} : \text{outward} \]

calculate flux \(\iint_{\Sigma} \vec{F} \cdot \vec{n} dS\)

Figure: A sketch of a unit sphere in a 3D coordinate system with axes x, y, and z, labeled with radius 1.

choices: parametrize \(\Sigma\), find \(\vec{r}_u \times \vec{r}_v\) or \(\vec{r}_v \times \vec{r}_u\) (whichever is out)

then do \(\iint_{R} \vec{F} \cdot (\vec{r}_u \times \vec{r}_v) dA\)

\(\rightarrow\) Divergence Theorem: closed surface

\(\vec{F}\) defined in the volume

\[ \text{Div. Theorem: } \iint_{S} \vec{F} \cdot \vec{n} dS = \iiint_{E} \text{div} \vec{F} dV \]

\(\text{div} \vec{F} = 3\)

volume of \(E\) (sphere)

\[ \iiint_{E} \text{div} \vec{F} dV = 3 \iiint_{E} dV = 3 \cdot \frac{4}{3} \pi (1)^3 = 4\pi \]

PAGE 6

1. Find an equation of the plane that contains the point \((1, 2, -3)\) and the line with symmetric equations \(x - 2 = y - 1 = \frac{z + 2}{2}\).

A. \(5x + y + z = 4\)

B. \(2x - y + z = -3\)

C. \(3x + y - 2z = 11\)

D. \(4x - 2y - 3z = 9\)

E. \(x + y - 2z = 9\)

sym. eq of line:

\[ t = x - 2 = y - 1 = \frac{z + 2}{2} \] \[ x = 2 + t, \] \[ y = 1 + t \] \[ z = 2t - 2 \]

Figure: A geometric sketch of a plane in 3D space containing a point (1, 2, -3) and a line with direction vector <1, 1, 2>.

\[ \vec{r}(t) = \langle 2 + t, 1 + t, 2t - 2 \rangle \] \[ = \langle 2, 1, -2 \rangle + t \langle 1, 1, 2 \rangle \]

\[ \vec{u} = \langle -1, 1, -1 \rangle \] \[ \vec{n} = \vec{u} \times \vec{v} \quad \text{or} \quad \vec{v} \times \vec{u} \] \[ = \begin{vmatrix} \vec{i} & \vec{j} & \vec{k} \\ -1 & 1 & -1 \\ 1 & 1 & 2 \end{vmatrix} = \langle 3, 1, -2 \rangle \]

plane: \(a(x - x_0) + b(y - y_0) + c(z - z_0) = 0\)

\[ 3(x - 1) + (y - 2) - 2(z + 3) = 0 \]

PAGE 7

Triple Integrals in Rectangular and Spherical Coordinates

An object occupies the region bounded above by the sphere radius \(\sqrt{32}\) \(x^2 + y^2 + z^2 = 32\) and below by the upper nappe of the cone \(z^2 = x^2 + y^2\). The mass density at any point of the object is equal to its distance from the \(xy\) plane. Set up a triple integral in rectangular coordinates for the total mass \(m\) of the object.

A. \(\int_{-4}^{4} \int_{-\sqrt{16-x^2}}^{\sqrt{16-x^2}} \int_{-\sqrt{x^2+y^2}}^{\sqrt{32-x^2-y^2}} z \, dz \, dy \, dx\)

B. \(\int_{-4}^{4} \int_{-\sqrt{16-x^2}}^{\sqrt{16-x^2}} \int_{\sqrt{x^2+y^2}}^{\sqrt{32-x^2-y^2}} z \, dz \, dy \, dx\)

C. \(\int_{-2}^{2} \int_{-\sqrt{4-x^2}}^{\sqrt{4-x^2}} \int_{\sqrt{x^2+y^2}}^{\sqrt{32-x^2-y^2}} z \, dz \, dy \, dx\)

D. \(\int_{0}^{4} \int_{0}^{\sqrt{16-x^2}} \int_{\sqrt{x^2+y^2}}^{\sqrt{32-x^2-y^2}} z \, dz \, dy \, dx\)

E. \(\int_{-4}^{4} \int_{-\sqrt{16-x^2}}^{\sqrt{16-x^2}} \int_{\sqrt{x^2+y^2}}^{\sqrt{32-x^2-y^2}} xy \, dz \, dy \, dx\)

Figure: 3D coordinate system showing a cone z = sqrt(x^2+y^2) capped by a sphere x^2+y^2+z^2=32.

Now do the same in spherical

A. \(\int_{0}^{2\pi} \int_{0}^{\frac{\pi}{4}} \int_{0}^{\sqrt{32}} \rho^3 \cos \phi \sin \phi \, d\rho \, d\phi \, d\theta\)

B. \(\int_{0}^{2\pi} \int_{0}^{\frac{\pi}{4}} \int_{0}^{\sqrt{32}} \rho \cos \phi \sin \phi \, d\rho \, d\phi \, d\theta\)

C. \(\int_{0}^{2\pi} \int_{0}^{\frac{\pi}{4}} \int_{0}^{\sqrt{32}} \rho^3 \sin^2 \phi \, d\rho \, d\phi \, d\theta\)

D. \(\int_{0}^{2\pi} \int_{0}^{\frac{\pi}{2}} \int_{0}^{\sqrt{32}} \rho^3 \cos \phi \sin \phi \, d\rho \, d\phi \, d\theta\)

E. \(\int_{0}^{2\pi} \int_{0}^{\frac{\pi}{4}} \int_{0}^{\sqrt{32}} \rho \cos \phi \, d\rho \, d\phi \, d\theta\)

PAGE 8

density: \(\rho(x,y,z) = z\)

In Cartesian: intersection of surfaces projected on xy-plane

\[ \begin{aligned} x^2 + y^2 + z^2 = 32 &\implies z = \sqrt{32 - x^2 - y^2} \\ z = \sqrt{x^2 + y^2} & \\ \sqrt{x^2 + y^2} &= \sqrt{32 - x^2 - y^2} \\ x^2 + y^2 &= 32 - x^2 - y^2 \\ x^2 + y^2 &= 16 \end{aligned} \]

Figure: A circle of radius 4 centered at the origin in the xy-plane, with y = sqrt(16-x^2) and y = -sqrt(16-x^2) labeled.

\(-4 \le x \le 4\)

\(-\sqrt{16-x^2} \le y \le \sqrt{16-x^2}\)

\(\sqrt{x^2+y^2} \le z \le \sqrt{32-x^2-y^2}\) (Sphere)

\[ \int_{-4}^{4} \int_{-\sqrt{16-x^2}}^{\sqrt{16-x^2}} \int_{\sqrt{x^2+y^2}}^{\sqrt{32-x^2-y^2}} z \, dz \, dy \, dx \]

Spherical:

\(0 \le \rho \le \sqrt{32}\)

\(0 \le \phi \le \pi/4\)

\(0 \le \theta \le 2\pi\)

density: \(\rho = z = \rho \cos \phi\)

\(dV = \rho^2 \sin \phi \, d\rho \, d\phi \, d\theta\)

\[ \int_{0}^{2\pi} \int_{0}^{\pi/4} \int_{0}^{\sqrt{32}} \rho^3 \cos \phi \sin \phi \, d\rho \, d\phi \, d\theta \]