PAGE 1

The tangent plane to the graph of the surface \( z = e^{2x} \ln y \) at the point \( (1, 1, 0) \) is

A \( e^2 y - z = e^2 \)
B. \( 2e^2 y - z = 1 \)
C. \( x - e^2 y + z = 1 \)
D. \( 2e^2 x - e^2 y = e^2 \)
E. \( 2e^2 x - e^2 y + z = e^2 \)

normal vector

point ✓

let \( F = e^{2x} \ln y - z \)

\( \vec{\nabla} F \) is normal to the surface \( z = e^{2x} \ln y \)

\[ \vec{\nabla} F = \left\langle \frac{\partial F}{\partial x}, \frac{\partial F}{\partial y}, \frac{\partial F}{\partial z} \right\rangle = \left\langle 2e^{2x} \ln y, \frac{e^{2x}}{y}, -1 \right\rangle \]

at \( (1, 1, 0) \) \( \vec{\nabla} F = \langle 0, e^2, -1 \rangle \)

plane: \( 0(x - 1) + e^2(y - 1) - 1(z - 0) = 0 \)

\[ e^2 y - e^2 - z = 0 \]

PAGE 2

Find the maximum value of the function \( f(x, y) = 8x - 6y \) subject to the constraint \( (x - 1)^2 + y^2 = 1 \)

A 18
B. 19
C. -2
D. 10
E. 11/5

Lagrange multipliers

\( g(x, y) = (x - 1)^2 + y^2 - 1 = 0 \)

\( f(x, y) = 8x - 6y \)

Solve \( \vec{\nabla} f = \lambda \vec{\nabla} g \) for \( x, y \) then find max of \( f(x, y) \)

\( \vec{\nabla} f = \langle 8, -6 \rangle \quad \vec{\nabla} g = \langle 2(x - 1), 2y \rangle \)

\( \langle 8, -6 \rangle = \lambda \langle 2(x - 1), 2y \rangle \)

\( 8 = \lambda \cdot 2(x - 1) \rightarrow \lambda = \frac{8}{2(x - 1)} \rightarrow \lambda = \frac{4}{x - 1} \)

\( -6 = \lambda \cdot 2y \rightarrow \lambda = \frac{-3}{y} \)

\( \frac{4}{x - 1} = \frac{-3}{y} \)

\( x - 1 = -\frac{4}{3} y \)

\( x = 1 - \frac{4}{3} y \)

\( (x - 1)^2 + y^2 = 1 \)

\( \frac{16}{9} y^2 + y^2 = 1 \)

\( \frac{25}{9} y^2 = 1 \)

PAGE 3

Calculations for Critical Points

\[ y^2 = \frac{9}{25} \quad \implies \quad y = \frac{3}{5}, \, -\frac{3}{5} \]\[ x = 1 - \frac{4}{3}y \quad \implies \quad x = \frac{1}{5}, \, \frac{9}{5} \]

Points to check: \( \left( \frac{9}{5}, -\frac{3}{5} \right) \), \( \left( \frac{1}{5}, \frac{3}{5} \right) \)

\[ f\left( \frac{9}{5}, -\frac{3}{5} \right) = 18 \quad \leftarrow \text{max} \]\[ f\left( \frac{1}{5}, \frac{3}{5} \right) = \dots \]

PAGE 4

The function \( f(x,y) = x^3 - y^3 - 3xy + 6 \) has local extrema consisting of:

A. One local maximum and one local minimum.
B. One local maximum and one saddle point.
C. One local minimum and one saddle point.
D. One local maximum, one local minimum, and one saddle point.
E. One local minimum and two saddle points.

Finding Critical Points

Critical points: \( f_x = 0 \) and \( f_y = 0 \)

\[ f_x = 3x^2 - 3y = 0 \implies x^2 = y \]\[ f_y = -3y^2 - 3x = 0 \implies x = -y^2 \]

Substituting \( x = -y^2 \) into the first equation:

\[ (-y^2)^2 = y \]\[ y^4 = y \implies y^4 - y = 0 \implies y(y^3 - 1) = 0 \]\[ y = 0, \quad y = 1 \]\[ x = 0, \quad x = -1 \]

Critical pts: \( (0,0), (-1,1) \)

Second Derivative Test

\[ f_{xx} = 6x, \quad f_{yy} = -6y, \quad f_{xy} = -3 \]\[ D = f_{xx}f_{yy} - (f_{xy})^2 \]\[ D = -36xy - 9 \]

\( D(0,0) < 0 \implies \text{saddle pt} \)

\( D(-1,1) > 0, \, f_{xx}(-1,1) < 0 \implies \text{max} \)

PAGE 5

Change the Order of Integration and Evaluate

Change the order of integration and evaluate:

\[ \int_{0}^{1} \int_{\sqrt{x}}^{1} e^{y^3} dy \, dx \]

The original limits of integration are:

\( 0 \le x \le 1 \)
\( \sqrt{x} \le y \le 1 \)

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

Figure: Graph of the region in the xy-plane bounded by y=1 and y=sqrt(x) from x=0 to x=1.

To change the order, we look at the region horizontally:

\( 0 \le y \le 1 \)
\( 0 \le x \le y^2 \)

The new integral is:

\[ \int_{0}^{1} \int_{0}^{y^2} e^{y^3} dx \, dy \]

Figure: Graph of the same region showing a horizontal strip for integration with respect to x first.

Evaluation

\[ = \int_{0}^{1} x \Big|_0^{y^2} e^{y^3} dy \]\[ = \int_{0}^{1} y^2 e^{y^3} dy \]

Substitution:

\( u = y^3 \)

\( du = 3y^2 dy \)

\[ \int_{0}^{1} \frac{1}{3} e^u du = \frac{1}{3} e^u \Big|_0^1 = \frac{1}{3} e^1 - \frac{1}{3} e^0 = \frac{1}{3} e - \frac{1}{3} \]\[ = \frac{1}{3}(e - 1) \]

PAGE 6

Triple Integral in Cylindrical Coordinates

Do NOT evaluate. Rewrite the integral in cylindrical coordinates.

\[ \int_{0}^{2} \int_{0}^{\sqrt{4-y^2}} \int_{\sqrt{x^2+y^2}}^{\sqrt{8-x^2-y^2}} \sqrt{x^2+y^2} dz \, dx \, dy \]

A. \( \int_{0}^{\pi/2} \int_{0}^{2} \int_{r}^{\sqrt{8-r^2}} r^2 dz \, dr \, d\theta \)
B. \( \int_{0}^{\pi/2} \int_{0}^{2} \int_{r}^{\sqrt{8-r^2}} r dz \, dr \, d\theta \)
C. \( \int_{0}^{\pi} \int_{0}^{2} \int_{r}^{\sqrt{8-r^2}} r^2 dz \, dr \, d\theta \)
D. \( \int_{0}^{\pi} \int_{0}^{2} \int_{r}^{\sqrt{8-r^2}} r dz \, dr \, d\theta \)
E. None of the above.

Region Analysis

Floor: xy-plane

\( 0 \le y \le 2 \)
\( 0 \le x \le \sqrt{4-y^2} \)

This describes a quarter circle in the first quadrant.

Figure: Graph of a quarter circle in the first quadrant of the xy-plane, bounded by x=sqrt(4-y^2).

Coordinate Conversion

Polar equivalent for the base:

\( 0 \le \theta \le \frac{\pi}{2} \)
\( 0 \le r \le 2 \)

For the z-limits:

\[ \sqrt{x^2+y^2} \le z \le \sqrt{8-x^2-y^2} \]\[ \sqrt{r^2} \le z \le \sqrt{8-r^2} \]\[ r \le z \le \sqrt{8-r^2} \]

Integrand and differential volume element:

Integrand: \( \sqrt{x^2+y^2} = r \)
\( dV = r dz \, dr \, d\theta \)

New Integral

\[ \int_{0}^{\pi/2} \int_{0}^{2} \int_{r}^{\sqrt{8-r^2}} r \cdot r dz \, dr \, d\theta = \int_{0}^{\pi/2} \int_{0}^{2} \int_{r}^{\sqrt{8-r^2}} r^2 dz \, dr \, d\theta \]

PAGE 7

Let \( E \) be the solid region enclosed by the cylinder \( x^2 + y^2 = 1 \), and the planes \( z = 0 \) and \( y + z = 2 \). Which of the following triple integrals is equal to the volume of \( E \)?

Figure: 3D graph of a cylinder x^2+y^2=1 bounded by z=0 and a slanted plane y+z=2.

Figure: Sketch of the solid E: a cylinder with a flat base and a slanted top face.

A. \( \int_{0}^{2\pi} \int_{0}^{1} \int_{0}^{2-r \sin \theta} r \, dz \, dr \, d\theta \) (Correct)
B. \( \int_{0}^{2\pi} \int_{0}^{1} \int_{0}^{2-\sin \theta} r \, dz \, dr \, d\theta \)
C. \( \int_{0}^{\pi} \int_{0}^{1} \int_{0}^{2-r \sin \theta} r \, dz \, dr \, d\theta \)
D. \( \int_{0}^{\pi} \int_{0}^{1} \int_{0}^{2-\sin \theta} r \, dz \, dr \, d\theta \)
E. \( \int_{0}^{2\pi} \int_{0}^{\sin \theta} \int_{0}^{2} r \, dz \, dr \, d\theta \)

Projection onto xy-plane

Figure: 2D unit circle in the xy-plane representing the projection of the solid.

\( 0 \le r \le 1 \)

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

From plane \( y + z = 2 \):

\( 0 \le z \le 2 - y \)

Since \( y = r \sin \theta \):

\( 0 \le z \le 2 - r \sin \theta \)

\[ \int_{0}^{2\pi} \int_{0}^{1} \int_{0}^{2-r \sin \theta} r \, dz \, dr \, d\theta \]

PAGE 8

Find the volume of the solid that is enclosed by \( x^2 + y^2 + z^2 = 1 \), \( x^2 + y^2 + z^2 = 4 \), and \( z = \sqrt{x^2 + y^2} \).

A. \( \frac{14\pi}{3} \left( 1 - \frac{\sqrt{2}}{2} \right) \)
B. \( \frac{28\pi}{3} \)
C. \( \frac{14\pi}{3} \left( 1 + \frac{\sqrt{2}}{2} \right) \)
D. \( 3\pi \left( 1 - \frac{\sqrt{2}}{2} \right) \)
E. \( 3\pi \)

\( x^2 + y^2 + z^2 = 4 \) : Sphere radius 2 center origin

\( x^2 + y^2 + z^2 = 1 \) : Sphere radius 1 center origin

\( z = \sqrt{x^2 + y^2} \) : cone

Figure: 3D sketch of the region between two spheres of radius 1 and 2, bounded by a cone.

Spherical is good

\( 1 \le \rho \le 2 \)
\( 0 \le \theta \le 2\pi \) (all around z-axis)
\( 0 \le \phi \le \frac{\pi}{4} \)

y-z plane

Cone \( z = \sqrt{x^2 + y^2} \)

On yz-plane \( x = 0 \)

\( z = \sqrt{y^2} = y \)

Slope is 1, bisects QI

So angle \( \phi \) is \( \frac{\pi}{4} \)

PAGE 9

volume:

\[ \int_{0}^{2\pi} \int_{0}^{\pi/4} \int_{1}^{2} \rho^2 \sin \phi \, d\rho \, d\phi \, d\theta = \frac{14\pi}{3} \left( 1 - \frac{\sqrt{2}}{2} \right) \]

Note: \( \rho^2 \sin \phi \, d\rho \, d\phi \, d\theta \) corresponds to the volume element \( dV \).

What if slope of cone is not 1?

Let's use \( \sqrt{3} z = \sqrt{x^2 + y^2} \) instead.

On yz-plane:

\[ \sqrt{3} z = \sqrt{y^2} = y \]\[ \frac{z}{y} = \frac{1}{\sqrt{3}}, \quad \frac{y}{z} = \sqrt{3} \]

\[ \tan \phi = \frac{y}{z} = \sqrt{3} \]\[ \phi = \tan^{-1}(\sqrt{3}) = \tan^{-1}\left( \frac{\sqrt{3}/2}{1/2} \right) = \frac{\pi}{3} \]\[ 0 \le \phi \le \frac{\pi}{3} \]

Figure: Right triangle diagram with angle phi, vertical side y, and horizontal side z.

\[ \tan \phi = \sqrt{3} \]\[ \phi = \tan^{-1}(\sqrt{3}) \]

PAGE 10

Review

Mass (moments, center of mass)

Average value of functions