MA 26100 - Spring 2026 Exam 2 Test/Quiz #:...

Accessible transcription generated on 4/29/2026

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MA 26100 - Spring 2026 Exam 2 Test/Quiz #: 21

1. Compute $ \int_{0}^{\pi} \int_{y}^{\pi} \sin(x^2) \, dx \, dy $. Hint: Change the order of integration.

1
$ \frac{1}{2} $
0
$ \frac{1 + \cos \pi^2}{2} $
$ \frac{1 - \cos \pi^2}{2} $ (Circled)

Handwritten notes for the region of integration:

$ y \le x \le \pi, \quad 0 \le y \le \pi $

$Hand-drawn 2D coordinate system showing the region of integration for the double integral. The axes are drawn in black. A diagonal line labeled 'y = x' in blue starts from the origin. A vertical line labeled 'x = \pi' in blue is drawn to the right. The region between the x-axis, the line y = x, and the line x = \pi is shaded with blue diagonal lines. A thick vertical black representative strip is drawn inside the shaded region, extending from the x-axis to the line y = x, with a mark 'x' on the horizontal axis below it. This visualizes the change in limits: when integrating with respect to y first, for a fixed x, y ranges from 0 to x, and the total range for x is from 0 to \pi.$

Visual Description: Hand-drawn 2D coordinate system showing the region of integration for the double integral. The axes are drawn in black. A diagonal line labeled 'y = x' in blue starts from the origin. A vertical line labeled 'x = \pi' in blue is drawn to the right. The region between the x-axis, the line y = x, and the line x = \pi is shaded with blue diagonal lines. A thick vertical black representative strip is drawn inside the shaded region, extending from the x-axis to the line y = x, with a mark 'x' on the horizontal axis below it. This visualizes the change in limits: when integrating with respect to y first, for a fixed x, y ranges from 0 to x, and the total range for x is from 0 to \pi.

By changing the order of integration, the integral becomes:

\[ \int_{0}^{\pi} \int_{0}^{x} \sin(x^2) \, dy \, dx = \int_{0}^{\pi} x \sin(x^2) \, dx = \left[ -\frac{1}{2} \cos(x^2) \right]_{0}^{\pi} \]

Evaluating at the limits:

\[ \left( -\frac{1}{2} \cos(\pi^2) \right) - \left( -\frac{1}{2} \cos(0) \right) = -\frac{1}{2} \cos \pi^2 + \frac{1}{2} = \frac{1 - \cos \pi^2}{2} \]

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MA 26100 - Spring 2026 Exam 2 Test/Quiz #: 21

Question 2

The volume of the region inside $x^2 + y^2 + z^2 = 25$ and outside $x^2 + y^2 + z^2 = 4$ in the region $y \ge 0$ and $z \le 0$ is given as a triple integral using spherical coordinates as

\[ \int_C^D \int_A^B \int_2^5 \rho^2 \sin(\phi) \, d\rho \, d\phi \, d\theta. \]

What are the values of $A, B, C, D$?

$A = \frac{\pi}{2}, B = \pi; C = 0, D = 2\pi$
$A = 0, B = \frac{\pi}{2}; C = 0, D = 2\pi$
$A = 0, B = \pi; C = 0, D = \pi$
$A = \frac{\pi}{2}, B = \pi; C = 0, D = \pi$
$A = 0, B = \frac{\pi}{2}; C = 0, D = \pi$

Step 1: Determine the limits for $\rho$.
From the equations of the bounding spheres: \[ \left. \begin{aligned} x^2 + y^2 + z^2 &= 25 \\ x^2 + y^2 + z^2 &= 4 \end{aligned} \right\} \implies \begin{aligned} \rho &= 5 \\ \rho &= 2 \end{aligned} \] This confirms the inner integral limits are from 2 to 5.

$A 3D coordinate system diagram defining the standard spherical coordinates. The x, y, and z axes are drawn. A vector with length rho ($\rho$) connects the origin to a point (x, y, z). The polar angle phi ($\phi$) is labeled between the positive z-axis and the vector. The projection of the point onto the xy-plane is marked as (x, y, 0), and the azimuthal angle theta ($\theta$) is shown in the xy-plane between the positive x-axis and this projection.$

Visual Description: A 3D coordinate system diagram defining the standard spherical coordinates. The x, y, and z axes are drawn. A vector with length rho ($\rho$) connects the origin to a point (x, y, z). The polar angle phi ($\phi$) is labeled between the positive z-axis and the vector. The projection of the point onto the xy-plane is marked as (x, y, 0), and the azimuthal angle theta ($\theta$) is shown in the xy-plane between the positive x-axis and this projection.

Step 2: Determine the limits for $\phi$ (values $A$ and $B$).
The problem states the region is where $z \le 0$. In spherical coordinates, $\phi$ is measured from the positive $z$-axis ($\phi = 0$). The $xy$-plane corresponds to $\phi = \frac{\pi}{2}$. Since the region is below the $xy$-plane ($z \le 0$), $\phi$ ranges from the horizontal plane to the negative $z$-axis.

$A 2D cross-sectional diagram focusing on the z-axis to find the range of the polar angle phi ($\phi$). The upper vertical axis is labeled z $\ge$ 0 and the lower part is labeled z $\le$ 0. A horizontal dashed line represents the xy-plane. A red arc shows the angle $\phi$ sweeping from the horizontal plane (where $\phi = \pi/2$) down to the negative z-axis (where $\phi = \pi$). Below the diagram, the range is written as $\pi/2 \le \phi \le \pi$.$

Visual Description: A 2D cross-sectional diagram focusing on the z-axis to find the range of the polar angle phi ($\phi$). The upper vertical axis is labeled z $\ge$ 0 and the lower part is labeled z $\le$ 0. A horizontal dashed line represents the xy-plane. A red arc shows the angle $\phi$ sweeping from the horizontal plane (where $\phi = \pi/2$) down to the negative z-axis (where $\phi = \pi$). Below the diagram, the range is written as $\pi/2 \le \phi \le \pi$.

Thus, $A = \frac{\pi}{2}$ and $B = \pi$.

Step 3: Determine the limits for $\theta$ (values $C$ and $D$).
The problem states the region is where $y \ge 0$. We look at the projection of this region onto the $xy$-plane.

$A 2D coordinate diagram of the xy-plane to determine the range of the azimuthal angle theta ($\theta$). The x-axis is horizontal and the y-axis is vertical. The region where y $\ge$ 0 (the upper half-plane) is shaded with blue diagonal lines. A red arc labeled $\theta$ starts at the positive x-axis (0) and sweeps counter-clockwise to the negative x-axis ($\pi$). To the right, the range is mathematically expressed as 0 $\le \theta \le \pi$.$

Visual Description: A 2D coordinate diagram of the xy-plane to determine the range of the azimuthal angle theta ($\theta$). The x-axis is horizontal and the y-axis is vertical. The region where y $\ge$ 0 (the upper half-plane) is shaded with blue diagonal lines. A red arc labeled $\theta$ starts at the positive x-axis (0) and sweeps counter-clockwise to the negative x-axis ($\pi$). To the right, the range is mathematically expressed as 0 $\le \theta \le \pi$.

Thus, $C = 0$ and $D = \pi$.

Conclusion:
Comparing these values to the given options: \[ A = \frac{\pi}{2}, B = \pi; C = 0, D = \pi \] This matches Choice D.

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MA 26100 - Spring 2026 Exam 2 Test/Quiz #: 21

Problem 3

3. Find the surface area of the parametric surface

\[ \vec{r}(u,v) = \langle 6u + 7v, -u + v, 67 \rangle, -7 \le u \le -6, 6 \le v \le 7. \]

A. 7
B. 42
C. 6
D. 13
E. 1

Solution:

\[ S.A. = \iint_{S} 1 \, dA = \iint_{\substack{u, v \\ \text{Bounds}}} |\vec{r}_u \times \vec{r}_v| \, dA \]

Calculate the partial derivatives of $ \vec{r} $ with respect to $ u $ and $ v $: \[ \vec{r}_u = \langle 6, -1, 0 \rangle \] \[ \vec{r}_v = \langle 7, 1, 0 \rangle \]

Find the cross product and its magnitude: \[ \vec{r}_u \times \vec{r}_v = \langle 0, 0, 13 \rangle \] \[ |\vec{r}_u \times \vec{r}_v| = 13 \]

Set up and evaluate the integral using the provided bounds: \[ = \int_{6}^{7} \int_{-7}^{-6} 13 \, du \, dv \] \[ = \underline{13.} \]

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MA 26100 - Spring 2026 Exam 2 Test/Quiz #: 21

Problem 4

4. A solid $ E $ has density $ \rho(x,y,z) = \sin(z) $ and is bounded by $ x^2 + y^2 \le 1, z = 0 $ and $ z = \pi $. Find the mass of the solid.

A. $ 2\pi - 1 $
B. $ 3\pi $
C. $ 2\pi $
D. $ \pi - 1 $
E. $ \pi $

\[ M = \iiint_R \rho(x,y,z) \, dV = \iiint_R \sin z \, dV \]

A hand-drawn 3D sketch depicting the solid region E as a cylinder centered on the vertical z-axis. The base of the cylinder is a circle in the xy-plane, and the cylinder extends upwards along the z-axis. The interior of the cylinder is shaded with diagonal blue hatch marks. The coordinate system includes a vertical line for the z-axis and horizontal lines representing the x and y axes extending from the origin at the center of the cylinder's base. This diagram illustrates the volume bounded by x^2 + y^2 <= 1, z = 0, and z = pi.

Visual Description: A hand-drawn 3D sketch depicting the solid region E as a cylinder centered on the vertical z-axis. The base of the cylinder is a circle in the xy-plane, and the cylinder extends upwards along the z-axis. The interior of the cylinder is shaded with diagonal blue hatch marks. The coordinate system includes a vertical line for the z-axis and horizontal lines representing the x and y axes extending from the origin at the center of the cylinder's base. This diagram illustrates the volume bounded by x^2 + y^2 <= 1, z = 0, and z = pi.

use Cylindrical coordinates:
$ dV = r \, dz \, dr \, d\theta $
$ 0 \le z \le \pi $
$ r, \theta $ is disk of Radius 1
$ 0 \le r \le 1 $
$ 0 \le \theta \le 2\pi $

\[ M = \int_0^{2\pi} \int_0^1 \int_0^\pi \sin z \cdot r \, dz \, dr \, d\theta = \int_0^{2\pi} \int_0^1 2r \, dr \, d\theta = 2\pi \]

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MA 26100 - Spring 2026 Exam 2 Test/Quiz #: 21

Problem 5

5. Volume of the solid in the first octant bounded by the plane $ x + 2y + 3z = 12 $ is represented by

\[ \int_{0}^{A} \int_{0}^{B} \int_{0}^{\frac{12-x-2y}{3}} dz \, dx \, dy. \]

Find the bounds A and B.

(A) $ A = 6, B = 12 - 2y $
B. $ A = 12, B = 12 - 2y $
C. $ A = 12, B = 6 - \frac{x}{2} $
D. $ A = 6, B = 6 - \frac{x}{2} $
E. $ A = 6, B = 12 $

\[ V = \iiint\limits_{R} 1 \, dV \]

3D coordinate system diagram showing the solid in the first octant bounded by the plane x + 2y + 3z = 12. A triangular plane section connects the axes. A vertical red arrow points upward from the xy-plane to the surface of the plane. Handwritten text next to the diagram shows the equation rewritten as z = (12 - x - 2y)/3. Below the diagram, a note says 'fix x, y in shadow'.

Visual Description: 3D coordinate system diagram showing the solid in the first octant bounded by the plane x + 2y + 3z = 12. A triangular plane section connects the axes. A vertical red arrow points upward from the xy-plane to the surface of the plane. Handwritten text next to the diagram shows the equation rewritten as z = (12 - x - 2y)/3. Below the diagram, a note says 'fix x, y in shadow'.

2D coordinate system diagram of the 'shadow' or projection of the solid onto the xy-plane. The region is a triangle bounded by the x-axis, y-axis, and the line x + 2y = 12. The y-intercept is labeled (0, 6) and the x-intercept is labeled (12, 0). The region is shaded with horizontal lines, and a horizontal pink arrow points from the y-axis to the boundary line, representing integration with respect to x first. Below the diagram, handwritten notes state 'fix y' and provide the resulting bounds: 0 ≤ x ≤ 12 - 2y and 0 ≤ y ≤ 6.

Visual Description: 2D coordinate system diagram of the 'shadow' or projection of the solid onto the xy-plane. The region is a triangle bounded by the x-axis, y-axis, and the line x + 2y = 12. The y-intercept is labeled (0, 6) and the x-intercept is labeled (12, 0). The region is shaded with horizontal lines, and a horizontal pink arrow points from the y-axis to the boundary line, representing integration with respect to x first. Below the diagram, handwritten notes state 'fix y' and provide the resulting bounds: 0 ≤ x ≤ 12 - 2y and 0 ≤ y ≤ 6.

Workings:

To find the bounds for the integral set up as $ dy \, dx \, dz $, we first look at the inner integral for $ z $, which goes from $ 0 $ to the plane $ z = \frac{12-x-2y}{3} $.

Next, we project the solid into the $ xy $-plane (the "shadow"). The boundary line is found by setting $ z = 0 $: $ x + 2y = 12 $.

Since the integration order is $ dz \, dx \, dy $, we treat $ y $ as the outermost variable and $ x $ as the middle variable:

For a fixed $ y $, $ x $ varies from the $ y $-axis ($ x = 0 $) to the line $ x = 12 - 2y $. Thus, $ B = 12 - 2y $.
The range of $ y $ values for the entire region is from $ 0 $ to the $ y $-intercept where $ x = 0 $: $ 2y = 12 \Rightarrow y = 6 $. Thus, $ A = 6 $.

Therefore, A = 6 and B = 12 - 2y, which matches option (A).

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6. Let $ C $ be the counterclockwise oriented boundary of the rectangle with vertices $ (0, 0), (1, 0), (1, 2) $, and $ (0, 2) $. Consider the vector field $ \vec{F}(x, y) = \langle y e^x, x^2 \rangle $. Compute $ \oint_C \vec{F} \cdot d\vec{r} $.

A. $ 3e - 1 $
B. $ 4(1 - e) $
C. $ 4 - 2e $ (Circled)
D. $ 3(e - 1) $
E. $ 2(1 - e) $

Can we use Green's Theorem?

Hand-drawn diagram of a rectangle in the first quadrant of an xy-coordinate plane. The rectangle has vertices at (0,0), (1,0), (1,2), and (0,2). The interior of the rectangle is shaded with diagonal lines and labeled 'R'. Arrows along the boundary indicate a counterclockwise orientation, and the boundary is labeled 'C'.

Visual Description: Hand-drawn diagram of a rectangle in the first quadrant of an xy-coordinate plane. The rectangle has vertices at (0,0), (1,0), (1,2), and (0,2). The interior of the rectangle is shaded with diagonal lines and labeled 'R'. Arrows along the boundary indicate a counterclockwise orientation, and the boundary is labeled 'C'.

\[ \oint_C \vec{F} \cdot d\vec{r} = \iint_R 2D \text{ curl } dA \]

Let $ \vec{F} = \langle f, g \rangle $, then $ 2D \text{ curl} = g_x - f_y $

For $ \vec{F} = \langle y e^x, x^2 \rangle \rightsquigarrow 2D \text{ curl} = 2x - e^x $

\[ = \int_0^2 \int_0^1 (2x - e^x) \, dx \, dy \] \[ = 2 \left[ x^2 - e^x \right]_0^1 \] \[ = 2 [ (1 - e) - (0 - 1) ] \] \[ = 2 [ 2 - e ] \] \[ = 4 - 2e \]

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MA 26100 - Spring 2026 Exam 2 Test/Quiz #: 21

7. Rewrite the following double integral using polar coordinates.

\[ \int_{-5}^{0} \int_{0}^{\sqrt{25-x^2}} e^{x^2+y^2} dy \, dx \]

A. $ \int_{\frac{\pi}{2}}^{\pi} \int_{0}^{5} e^{r^2} dr \, d\theta $
B. $ \int_{\frac{\pi}{2}}^{\pi} \int_{0}^{5} e^{r^2} r \, dr \, d\theta $
C. $ \int_{0}^{2\pi} \int_{0}^{5} e^{r^2} r \, dr \, d\theta $
D. $ \int_{0}^{\frac{\pi}{2}} \int_{0}^{5} e^{r^2} r \, dr \, d\theta $
E. $ \int_{0}^{\pi} \int_{0}^{5} e^{r^2} dr \, d\theta $

$ 0 \le y \le \sqrt{25-x^2} $

$ -5 \le x \le 0 $

Region in second quadrant

$A Cartesian coordinate graph illustrating the region of integration. The graph shows a quarter-circle in the second quadrant. The boundary of the arc is labeled with the equation $ y = \sqrt{25-x^2} $. The area between this arc and the axes in the second quadrant is shaded with green diagonal lines. A red ray starts at the origin and extends into the second quadrant, and a red arc indicates the angle $\theta$ measured counter-clockwise from the positive x-axis, showing it covers the range from $\frac{\pi}{2}$ to $\pi$. Points on the x and y axes indicate a radius of 5.$

Visual Description: A Cartesian coordinate graph illustrating the region of integration. The graph shows a quarter-circle in the second quadrant. The boundary of the arc is labeled with the equation $ y = \sqrt{25-x^2} $. The area between this arc and the axes in the second quadrant is shaded with green diagonal lines. A red ray starts at the origin and extends into the second quadrant, and a red arc indicates the angle $\theta$ measured counter-clockwise from the positive x-axis, showing it covers the range from $\frac{\pi}{2}$ to $\pi$. Points on the x and y axes indicate a radius of 5.

$ 0 \le r \le 5 $

$ \frac{\pi}{2} \le \theta \le \pi $

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MA 26100 - Spring 2026 Exam 2 Test/Quiz #: 21

Vector Fields and Divergence

8. Consider the vector field $\vec{F} = \langle x^2 + y^2, 2xy, z^2 \rangle$. Compute the divergence of $\vec{F}$.

A. $2x + 4z$

B. $4x + 4y + 2z$

C. $4x + 2z$

D. $4x - 2z$

E. $2x + 2y + 2z$

Handwritten Solution:

\[ \vec{F} = \langle f, g, h \rangle \] \[ \text{div} \vec{F} = f_x + g_y + h_z \] \[ = \frac{\partial}{\partial x}(x^2 + y^2) + \frac{\partial}{\partial y}(2xy) + \frac{\partial}{\partial z}(z^2) \] \[ = 2x + 2x + 2z \] \[ = 4x + 2z \]

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MA 26100 - Spring 2026 Exam 2 Test/Quiz #: 21

9. Minimize the function $ f(x, y) = -2xy $ subject to the constraint $ x^2 + xy + y^2 = 5 $.

There is no minimum
10
-10
D. $ -\frac{10}{3} $ (Selected answer)
$ \frac{10}{3} $

use Lagrange Multipliers

\[ \vec{\nabla}f = \lambda \vec{\nabla}g \rightsquigarrow \langle -2y, -2x \rangle = \lambda \langle 2x + y, x + 2y \rangle \]

From this vector equation, we set up the following system:

① $ -2y = \lambda(2x + y) $
② $ -2x = \lambda(x + 2y) $
③ $ x^2 + xy + y^2 = 5 $

Divide equation ① by equation ②:

\[ \frac{-2y}{-2x} = \frac{\lambda(2x + y)}{\lambda(x + 2y)} \] \[ \frac{y}{x} = \frac{2x + y}{x + 2y} \] \[ y(x + 2y) = x(2x + y) \] \[ xy + 2y^2 = 2x^2 + xy \] \[ 2y^2 = 2x^2 \implies x^2 = y^2 \]

Case 1: $ y = x $ in ③

\[ x^2 + (x)(x) + x^2 = 5 \] \[ 3x^2 = 5 \] \[ x^2 = \frac{5}{3} \implies x = \pm \sqrt{\frac{5}{3}} \]

Points: $ \left( \sqrt{\frac{5}{3}}, \sqrt{\frac{5}{3}} \right), \left( -\sqrt{\frac{5}{3}}, -\sqrt{\frac{5}{3}} \right) $

Objective function value:
$ f = -2x^2 $
$ f = -\frac{10}{3} $

Case 2: $ y = -x $ in ③

\[ x^2 + (x)(-x) + (-x)^2 = 5 \] \[ x^2 - x^2 + x^2 = 5 \] \[ x^2 = 5 \implies x = \pm \sqrt{5} \]

Points: $ (\sqrt{5}, -\sqrt{5}), (-\sqrt{5}, \sqrt{5}) $

Objective function value:
$ f = 2x^2 $
$ f = 10 $

The minimum value found is $ -\frac{10}{3} $.

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MA 26100 - Spring 2026 Exam 2 Test/Quiz #: 21

Question 10

10. Consider some vector field \[ \vec{F}(x,y) = \langle f(x,y), g(x,y) \rangle. \] If $ f_y = g_x $ and $ f_x \neq g_y $, then which of the following best describes $ \vec{F} $?

(I) $ \vec{F} $ is conservative because $ f_y = g_x $. ✓

(II) $ \vec{F} $ is not conservative because $ f_x \neq g_y $.

(III) There exists a simple closed curve $ C $ such that $ \oint_C \vec{F} \cdot d\vec{r} \neq 0 $.

(IV) For every simple closed curve $ C $, the line integral $ \oint_C \vec{F} \cdot d\vec{r} = 0 $. ✓

A. I and III only
B. II and III only
C. I only
D. I and IV only
E. II only

$ \vec{F} = \langle f, g \rangle = \nabla \phi = \langle \phi_x, \phi_y \rangle $
using mixed partials of $ \phi $
$ f_y = g_x $

\[ \oint_C \nabla \phi \cdot d\vec{r} = \phi(\text{end}) - \phi(\text{start}) = 0 \]

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MA 26100 - Spring 2026 Exam 2 Test/Quiz #: 21

11. Compute the flux of the vector field $\vec{F}(x,y) = \langle x, -y \rangle$ through the piece of the parabolic curve $y = x^2$ that begins at $(0,0)$ and ends at $(2,4)$, using the unit normal vector that is $90^\circ$ counterclockwise from the unit tangent vector.

0
$-8$
$\frac{8}{3}$
8
$-\frac{8}{3}$

A coordinate geometry diagram on a 2D Cartesian plane showing a parabolic curve segment in the first quadrant starting from the origin (0,0). At a point on the curve, two vectors are drawn: a green arrow pointing tangent to the curve in the direction of motion, and a purple arrow pointing perpendicular to the curve, upwards and to the left. A small arc connects the tangent vector to the normal vector, labeled 'ccw 90 degrees', indicating the normal vector is oriented 90 degrees counterclockwise from the tangent vector.

Visual Description: A coordinate geometry diagram on a 2D Cartesian plane showing a parabolic curve segment in the first quadrant starting from the origin (0,0). At a point on the curve, two vectors are drawn: a green arrow pointing tangent to the curve in the direction of motion, and a purple arrow pointing perpendicular to the curve, upwards and to the left. A small arc connects the tangent vector to the normal vector, labeled 'ccw 90 degrees', indicating the normal vector is oriented 90 degrees counterclockwise from the tangent vector.

Parametrization:
\[\vec{r}(t) = \langle t, t^2 \rangle, \quad 0 \le t \le 2\]

Unit Tangent Vector:
\[\vec{T}(t) = \frac{\vec{r}'(t)}{|\vec{r}'(t)|} = \frac{\langle 1, 2t \rangle}{|\vec{r}'(t)|}\]

Unit Normal Vector:
The unit normal vector $\vec{n}$ is $90^\circ$ counterclockwise to the unit tangent. The two candidates are:
\[\vec{n} = \frac{\langle 2t, -1 \rangle}{|\vec{r}'(t)|} \quad \text{or} \quad \vec{n} = \frac{\langle -2t, 1 \rangle}{|\vec{r}'(t)|}\]
Based on the requirement for a $90^\circ$ counterclockwise rotation and the diagram (where $x < 0, y > 0$), we select:
\[\vec{n} = \frac{\langle -2t, 1 \rangle}{|\vec{r}'(t)|}\]

Flux Calculation:
The formula for flux is: \[\text{Flux} = \int_C \vec{F} \cdot \vec{n} \, ds\] Where $ds = |\vec{r}'(t)| \, dt$. Using the definition of the line integral: \[\int_C f \, ds = \int_a^b f(\vec{r}(t)) |\vec{r}'(t)| \, dt\]

\[\text{Flux} = \int_0^2 \vec{F}(\vec{r}(t)) \cdot \frac{\langle -2t, 1 \rangle}{|\vec{r}'(t)|} |\vec{r}'(t)| \, dt\] \[\text{Flux} = \int_0^2 \vec{F}(\vec{r}(t)) \cdot \langle -2t, 1 \rangle \, dt\] \[\text{Flux} = \int_0^2 \langle t, -t^2 \rangle \cdot \langle -2t, 1 \rangle \, dt\] \[\text{Flux} = \int_0^2 (-2t^2 - t^2) \, dt\] \[\text{Flux} = \int_0^2 -3t^2 \, dt\] \[\text{Flux} = \left[ -t^3 \right]_0^2 = -8\]

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MA 26100 - Spring 2026 Exam 2 Test/Quiz #: 21

12. Find the flux of $ \vec{F} = \langle x, y, z \rangle $ through the surface $ z = 4 - x - y $, in the first octant with upward orientation (normal pointing upward).

A. 0
B. 4
C. 16
D. 32
E. 64

Surface integral = $ \iint_{S} \vec{F} \cdot d\vec{S} = \iint_{S} \vec{F} \cdot \vec{n} \, dS $

$ \vec{r}(u,v) $ is your parametrization

Flux = $ \iint_{u,v \text{ Bounds}} \vec{F}(\vec{r}(u,v)) \cdot (\pm (\vec{r}_u \times \vec{r}_v)) \, dA $

Note: chose upward pointing $ \Rightarrow z > 0 $

$ \vec{r}(u,v) = \langle u, v, 4 - u - v \rangle $

\[ \left. \begin{aligned} \vec{r}_u &= \langle 1, 0, -1 \rangle \\ \vec{r}_v &= \langle 0, 1, -1 \rangle \end{aligned} \right\} \vec{r}_u \times \vec{r}_v = \langle 1, 1, 1 \rangle \]

A sketch of the region of integration R in the uv-plane. The vertical axis is labeled 'v'. The region is a right triangle in the first quadrant bounded by the v-axis, the u-axis, and the line u + v = 4. The triangle has vertices at (0,0), (4,0) on the horizontal u-axis, and (0,4) on the vertical v-axis. The interior area of the triangle is shaded with horizontal lines and labeled 'R'.

Visual Description: A sketch of the region of integration R in the uv-plane. The vertical axis is labeled 'v'. The region is a right triangle in the first quadrant bounded by the v-axis, the u-axis, and the line u + v = 4. The triangle has vertices at (0,0), (4,0) on the horizontal u-axis, and (0,4) on the vertical v-axis. The interior area of the triangle is shaded with horizontal lines and labeled 'R'.

\[ \text{Flux} = \iint_{R} \langle u, v, 4 - u - v \rangle \cdot \langle 1, 1, 1 \rangle \, dA \] \[ = \iint_{R} (u + v + 4 - u - v) \, dA \] \[ = 4 \iint_{R} 1 \, dA = 4 \times \text{Area of } R \] \[ = 4 \times \frac{1}{2} (4 \times 4) \] \[ = \underline{32} \]

MA 26100 - Spring 2026 Exam 2 Test/Quiz #:...

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Question 2

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Problem 3

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Problem 4

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Problem 5

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7. Rewrite the following double integral using polar coordinates.

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Vector Fields and Divergence

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9. Minimize the function \( f(x, y) = -2xy \) subject to the constraint \( x^2 + xy + y^2 = 5 \).

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Question 10

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11. Compute the flux of the vector field \(\vec{F}(x,y) = \langle x, -y \rangle\) through the piece of the parabolic curve \(y = x^2\) that begins at \((0,0)\) and ends at \((2,4)\), using the unit normal vector that is \(90^\circ\) counterclockwise from the unit tangent vector.

Page 12

12. Find the flux of \( \vec{F} = \langle x, y, z \rangle \) through the surface \( z = 4 - x - y \), in the first octant with upward orientation (normal pointing upward).