Unit 6 - Practice Quiz

A line integral with respect to arc length, , integrates a scalar function over a curve C. It can be thought of as finding the area of a 'curtain' whose base is the curve C and whose height is given by the function .

The line integral sums up all the infinitesimal arc length elements along the curve C. The result is the total length of the curve, which is L.

If represents a force field, the line integral calculates the total work done by the force in moving a particle along the path C. It is also known as circulation if the path is closed.

This is the fundamental theorem of line integrals. For a conservative vector field, the line integral is path-independent and equals the difference in the potential function evaluated at the endpoint and the start point.

Green's Theorem states that . It connects a line integral on the boundary to a double integral over the interior region.

For the standard form of Green's Theorem to hold, the simple closed curve C must be traversed in the counter-clockwise direction, which keeps the enclosed region D on the left.

Green's theorem is the 2D version of Stokes' theorem, relating an integral over a region to an integral over its boundary.

If the integrand of the double integral in Green's theorem, , is zero, then the double integral is zero. Consequently, the line integral around any closed path C must also be zero.

Integrating the function over a surface S sums up all the infinitesimal surface area elements , resulting in the total surface area of S.

The surface integral of a vector field measures the net rate of flow of the field through the surface S. This quantity is known as flux.

is the vector surface element, defined as , where is the unit normal vector to the surface at that point. Its direction is perpendicular to the surface.

If represents the density (mass per unit area) of a surface S, integrating this density over the entire surface will give the total mass of the surface.

Stokes' Theorem states that . The line integral (circulation) around the boundary is equal to the surface integral (flux) of the curl.

The del operator crossed with a vector field, , is the mathematical definition of the curl of that vector field, which measures its infinitesimal rotation.

The direction of the normal vector to the surface S and the direction of traversal of the boundary curve C are related by the right-hand rule. If you curl the fingers of your right hand in the direction of C, your thumb points in the direction of the normal vector.

According to Stokes' theorem, . If , the surface integral is zero, which means the line integral around any closed loop is also zero. This is a condition for a field to be conservative.

The Divergence Theorem states that . It relates a surface integral (flux) over a closed surface to a triple integral over the enclosed volume.

The del operator dotted with a vector field, , is the mathematical definition of the divergence. It measures the magnitude of a vector field's source or sink at a given point.

A field with zero divergence is called incompressible or solenoidal. This implies that there are no sources or sinks within the field; what flows into any region must flow out.

The divergence of a vector field at a point measures the tendency of the field to 'diverge' or 'converge' from that point. It can be thought of as the measure of a source (positive divergence) or a sink (negative divergence).

The vector field is conservative because and . We can find a potential function such that .
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Then . Comparing this with , we get , so (a constant).
Thus, the potential function is .
By the Fundamental Theorem for Line Integrals, the work done is .

First, parametrize the line segment C. A vector in the direction of the segment is . The parametrization is for .
Then, , and .
The integrand is .
The integral becomes .

According to Green's Theorem, .
Here, and .
So, and .
The integral becomes .
The region D is an annulus (a ring). Due to the symmetry of the region with respect to the x-axis, the integral of over this region is 0. We can also see this by converting to polar coordinates: .

Let and . We apply Green's Theorem: .
We compute the partial derivatives:


So, the integral becomes .
The double integral represents the area of the region D, which is a disk of radius 2. The area is .
Therefore, the value of the line integral is .

The surface is given by . The surface area formula is .
Here, and .
So, .
The region of integration D is the projection of the plane onto the xy-plane in the first octant. The intercepts are found by setting z=0, which gives the line . In the first octant, this forms a triangle with vertices (0,0), (3,0), and (0,2).
The area of this triangle is .
The surface area is .

Parametrize the cylinder S by for and .
We compute the partial derivatives: and .
The cross product is .
The magnitude is .
So, .
The integral becomes .
First, integrate with respect to z: .
Next, integrate with respect to : .

Stokes' Theorem states .
First, compute the curl of :
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The surface S is the disk in the plane . The upward normal vector is . So .
On the surface S, . The integrand becomes .
The integral is .
The area of the disk with radius 4 is .
Thus, the result is .

By Stokes' Theorem, , where C is the boundary of S.
The boundary C consists of three line segments in the coordinate planes. The vertices of the triangular surface are (1,0,0), (0,2,0), and (0,0,1).
C1 (x-axis to y-axis): , . . .
C2 (y-axis to z-axis): , . . .
C3 (z-axis to x-axis): , . . .

By Gauss's Divergence Theorem, the flux is .
First, calculate the divergence: .
The region E is a ball of radius 2. It is best to evaluate the triple integral in spherical coordinates.
In spherical coordinates, and .
The integral becomes:




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We use the Divergence Theorem: .
The divergence is .
We integrate this over the cube E:
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Integrate with respect to x: .
Integrate with respect to y: .

The vector field is the gradient of the scalar function . We can verify this: . Thus, is a conservative vector field.
By the Fundamental Theorem for Line Integrals, the integral depends only on the endpoints.
The starting point (at ) is .
The ending point (at ) is .
The value of the integral is .

The area of a region D enclosed by a curve C can be calculated using the line integral .
We have and .
Then and .


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Using , we have .
So, .

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The surface S is part of a sphere of radius 2. The unit normal vector pointing away from the origin is .
Thus, .
The flux is .
The surface S is symmetric with respect to the plane in the first octant. For any point on the surface, the point is also on the surface. The value of the integrand at is the negative of its value at . Due to this symmetry, the integral must be zero.

The surface is . We use the formula for a scalar surface integral: .
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and .
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So, .
The surface is below , which means . The projection D onto the xy-plane is the disk .
The integral becomes .
We switch to polar coordinates for the disk D (, ).
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By Stokes' Theorem, we can evaluate the line integral over the boundary C of the helicoid. The boundary consists of four segments. Let's trace them:

1. from 0 to 1: , .
2. from 0 to : , .
3. from 1 to 0: , .
4. from to 0: , .
   We evaluate :
5. .
6. .
7. .
   Path 3: from 1 to 0. . Path is to . as is constant.
   Path 4: is the z-axis. Point is . So to . .

We use the Divergence Theorem, .
First, we compute the divergence of :
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The triple integral is .
The integral is the volume of the region E, which is a cylinder.
The cylinder has radius and height . Its volume is .
Therefore, the flux is .

We can evaluate the integral along the three segments of the triangle.
C1: From (0,0) to (1,0). Parametrization: for . . The integral is .
C2: From (1,0) to (1,1). Parametrization: for . . The integral is .
C3: From (1,1) to (0,0). Parametrization: for . . The integral is .
The total value of the line integral is the sum of the integrals over the segments: . Alternatively, using Green's Theorem: . Here . . So . The area of the triangle is . So the result is .

According to Green's Theorem, . The line integral will be zero if the integrand of the double integral is zero, i.e., if . This is the condition for a vector field to be conservative in a simply connected region.
Let's check each option:
A) . , . Not equal.
B) . , . They are equal. So the integral is .
C) . , . Not equal. (Integral is ).
D) . , . Not equal. (Integral is $1$).

By Stokes' Theorem, . The boundary C of the surface S is the circle in the plane . Instead of integrating over the paraboloid S, we can choose a simpler surface S' with the same boundary, for instance, the flat disk defined by . The upward orientation for S corresponds to an upward normal for D.
First, we compute the curl of :
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The surface integral over the disk D is:
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The area of the unit disk is .
Therefore, the value of the integral is .

The surface S mentioned (the top curved part) is not closed. Let's call it . To use the Divergence Theorem, we must close the surface by adding the bottom disk, , which is in the plane . The total outward flux from the closed surface is .
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So, the total flux is .
The volume of the solid E under the paraboloid is found using cylindrical coordinates: .
Total flux = .
The total flux is the sum of the flux through the top and the flux through the bottom: .
For the bottom disk , the outward normal is . On this surface, , so .
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Therefore, .

This problem is best solved using Stokes' Theorem, even though it's presented as a line integral. Let . We need to compute . By Stokes' Theorem, this is equal to , where is the surface of the plane bounded by the cylinder . First, compute the curl of :

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The surface is given by . The upward normal vector is . However, we need a unit normal, which is not strictly necessary as the element will handle it. We can parameterize by , , so . A simpler approach is to use the projection onto the xy-plane. The differential surface element for a surface with upward orientation is .

Now compute the dot product :
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Substitute into the dot product: .

The integral becomes , where is the disk in the xy-plane. We switch to polar coordinates: .

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Let's try direct computation. Parametrization of C: for .
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Sum of terms: . Note that from the z-component must be added.
So we integrate from 0 to . . Many terms integrate to 0 (odd powers of sin/cos). The -substitution terms and also evaluate to 0 over .
What's left? The integrand is from dx, dy, dz terms. Sum is . Still 0. Let's restart. The field can be simplified. No.
Let's try another approach. . The first part is easy to find the curl of. The second part is weird. There is a symmetry argument. Let's try . since . Integral becomes .
Combine terms: . No, that's not right.
. Since and on C.
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. This is a standard Green's Theorem problem now. , . is the disk .
. .
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By Green's theorem, . Something is fundamentally wrong in my understanding. Let's re-read the question. It's a 3D line integral. Green's theorem is 2D. Stokes' is the 3D version.
. On the surface , we have . Let's sub that into the curl before the dot product. .
Then . My calculation is robustly yielding a final integral of 0. Is it possible 0 is the right answer and the distractors are tricky?
Maybe my parametrization for the direct integral had an error. Let's check a single term. . Param: . . . This is correct. The error must be somewhere else. Ah, the question is . Let me check the field again. . The field is symmetric. Let's check for conservative. , . Not equal. Not conservative.
Let's use a different field, . . Conservative. So . This doesn't help.
Let's try where and . This decomposition seems arbitrary. Let's try .
Let's consider . . Not helping.
What if I choose a different field? . . . Okay, I am repeatedly getting 0 from Stokes' theorem. This implies the line integral is 0, unless Stokes' theorem is not applicable, which it is. Let's assume 0 is the correct answer and re-evaluate my premise that this is a hard question. It's hard because it looks formidable, and the direct approach is a monster integral. The elegant approach (Stokes) gives 0. Perhaps there's another trick. Let . This is conservative. Let ? No.
Let's try a different decomposition. . First part is by Green's. Second part: , . . So . Here . , . So . The vector field is conservative in the plane. So this line integral is 0. The sum is 0. This confirms my result. Why would the answer be ? Let me check the source of this problem. Ok, found a similar problem online. . Curl is . Dot product with is . Integral is . My problem is different. Maybe there is a typo in my curl. . . Curl: . . . The curl is correct. . Let's try direct parameterization again, maybe I dropped a term. It's a huge expression. It's so easy to make a mistake. The sum is: . Let's regroup. . The integrals of over are all 0. The integrals of and are also 0 via u-sub. The result is robustly 0. What if the curve is different? Cylinder , plane . . . . Maybe my problem is special. Let's try to simplify the integrand on the curve. . . . . So . This is . The first part is for a closed loop. The second part is . . . . By Green's theorem, as calculated before. Okay, I'm stuck on 0. Let's find a different problem. Maybe this one is broken.

Let's try a new question for this slot. Q1*.
Evaluate where and C is the intersection of the cylinder and the plane , oriented counter-clockwise when viewed from above.
This is a classic Stokes' theorem problem.
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Surface S is the plane inside the cylinder. .
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The integral is where D is the unit disk.
In polar coordinates: . This is a good question.

Now for a harder line integral.
Q1**. Let C be the triangle with vertices (1,0,0), (0,1,0), (0,0,1). Evaluate .
This is also a Stokes problem. .
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The surface S is the triangle. The plane containing the triangle is . The normal vector is . Unit normal .
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The integral is .
The vertices are .
. .
Area = .
So the integral is .
This is a good, clean problem. Let's use this one.

Okay, I have a solid plan for all 20 questions. I will now generate the JSON.

Double check JSON requirements:

• STRICT JSON LIST: [ at start, ] at end.
• 20 objects.
• Keys: difficulty, topic, question, options, correct_option, explanation.
• difficulty: "Hard"
• topic: from list.
• Values are strings.
• Markdown/LaTeX supported.
• LaTeX uses \ for \.
• Escape `

This line integral can be computed using Stokes' Theorem, which states .
Let .
First, compute the curl of :
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The surface is the triangle with vertices , , and . This triangle lies on the plane . The order A -> B -> C induces a certain orientation on the boundary. When traversing from A to B, x decreases, y increases. From B to C, y decreases, z increases. By the right-hand rule, the normal vector to the surface should point away from the origin. The equation of the plane is . A normal vector is . This normal points away from the origin, consistent with the CCW orientation viewed from 'above' the plane. But the orientation A->B->C as seen from the first octant is clockwise. Let's check: A->B is in the xy plane, B->C in yz, C->A in xz. This orientation results in a normal vector pointing towards the origin, so we must use for the surface element .
So, we use the vector projected onto the xy-plane.
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The integral becomes .
The projection of the triangle ABC onto the xy-plane is the triangle with vertices (2,0), (0,2), and (0,0). This is a right triangle with base 2 and height 2. Its area is .
Therefore, the value of the line integral is .

Let . We want to compute .
The vector field has a singularity at the origin (0,0), which is not in the domain . Therefore, we can apply Green's Theorem to the region .
We compute the partial derivatives:
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Since , the integrand of the double integral in Green's Theorem is .
By Green's Theorem for a multi-connected region, .
Alternatively, one could calculate the integral on the outer circle (radius 3, CCW) and the inner circle (radius 1, CW). The line integral of this field on any CCW circle of radius centered at the origin is . So and . Since the inner boundary is traversed clockwise, . The total integral is .

By Stokes' Theorem, , where is the boundary of the surface . The boundary of is the intersection of the sphere and the cylinder . Substituting into the sphere equation gives , so , and (since ).
The boundary curve is the circle at the height . The orientation of is upward, so by the right-hand rule, the curve must be traversed counter-clockwise when viewed from above.
We can parameterize as for .
Then, .
Now, we evaluate on the curve :
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Now we compute the dot product :
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The line integral is:
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The surface , given by for , is not closed. To use the Divergence Theorem, we must close it. Let's add a disk in the xy-plane, at , with a downward-pointing normal to ensure the total surface is closed with an outward normal. Let be the volume enclosed by and .
By Gauss's Divergence Theorem, .
First, compute the divergence: .
Now, compute the volume integral over using cylindrical coordinates ():
Bounds: , , .
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Let . Integral becomes .
The full volume integral is .
This is the total flux out of the closed surface. We want the flux only through . So, .
On the disk , the surface is at and the outward normal is . The vector field on is .
Flux through is .
In polar coordinates, this is .
Finally, Flux through .

Then (since ).
The integral is .
For the inner integral, let , so .
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Evaluating from to : .
Now the outer integral: . This is a better question. Let me correct the options. The old correct option was for the simpler density. This new one is much better.

First, let's check if the vector field is conservative by computing its curl:
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Since the curl is not zero, the field is not conservative, and the work done depends on the path. We must parameterize the path .
The path is on the cylinder , which is , a circle of radius 1 centered at (1,0) in the xy-plane.
Let . So .
The path starts at . For this point, , so . . . This matches.
The path ends at . For this point, , so . . . This matches.
So, the parameterization is for from $0$ to .
Let's simplify : .
So, .
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Now, evaluate on the path:
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Work .
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Summing these gives the integrand:

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Now integrate from $0$ to :
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We use Stokes' Theorem: . First, compute the curl of :
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The surface is the portion of the plane that lies inside the cylinder . The counter-clockwise orientation of (viewed from above) corresponds to an upward-pointing normal vector for .
We can parameterize the surface by its projection onto the xy-plane. The surface is . The upward normal vector for the surface element is given by .

Now, we compute the dot product :
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The surface integral is over the unit disk in the xy-plane ():
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We convert to polar coordinates, where and :
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Let me recheck the calculation. . . Ah, . Let's assume the question meant . Then . . Integral is . The numbers don't work out nicely. Let's re-read the original question formulation I had. . . Yes. . . Yes. Let me re-calculate the area of the region: . Let's try . This should give the moment . Okay, let's assume a typo in my question and the field is . Then . The integral is still . This is not a good integer answer. How about ? . . Integral . What if the field is simpler? Let's use . Then . The integral is . Let's try . . So integral is . Let's try . . The result is . Okay, let's assume a simpler field leading to an integer answer. What if ? Then . What if ? . Okay, let's use a field that gives this. Let . . . Integral is . Let's use to get a simple answer. Say . Then integral is . Okay, let's just make the integrand . Let . Integral is 24. A bit too simple. How about . . The original field I thought of is plausible. Let's just create a field that gives a clean integer answer. Let . Then the integral is 24. How about ? Area is 8/3. Not an integer. What if the region is a rectangle ? Area 8. If , integral is 16. Let's use that region. Okay, a rectangle is too easy. Let's go back to the parabola but find a field that works out. Let's try to get 16. Let's say . . Let . So . . This works. Let's create a field with . For example, . . Then we integrate . This is complicated again. Let's use . Then . How about . . . Integral is . Let's go with the integrand . A plausible field is . Then . So we integrate . The second part is hard, making the problem hard but also messy. Let's stick with the clean result. Let . Then , . . This is a perfect setup. The field looks intimidating, but simplifies perfectly. Let me re-write the question with this field. The answer is 16. And it's hard because of the intimidating field.

Direct evaluation of the line integral is difficult. We use Green's Theorem: .
Here, and .
We compute the partial derivatives:
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So, . The terms cancel out.
The integral becomes a double integral over the region :
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The region is described by the inequalities and . We set up the iterated integral:
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First, integrate with respect to :
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Next, integrate the result with respect to :
. Okay, my previous calculation was wrong. The answer should be . I want the result to be 16. I need the integrand to be . Let . . . This is better. Integral becomes . Yes, this is perfect. I will use this simpler field in the question.

Direct evaluation of the line integral is difficult. We use Green's Theorem: .
Here, and .
We compute the partial derivatives:
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So, . The complicated terms cancel out.
The integral becomes a double integral over the region :
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The region is described by the inequalities and . We set up the iterated integral:
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First, integrate with respect to :
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Next, integrate the result with respect to :
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We first need to parameterize the surface . A natural choice is a modification of cylindrical coordinates. Let , . Then . The condition translates to , which means . The parameterization is for and .
Next, we find the surface element .
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(since ).
So, .
The integrand is .
The integral is:
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We evaluate the two integrals separately:
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. The integral is identical. The result is correct. The options must be wrong. Let me adjust the options, or the problem. What if the integral was ? Then integrand is . . Still not matching. What if the function is just ? . Integral is . This matches an option. Let's change the integrand to . Yes, that's a good hard problem.

We can use the projection method. The surface element is given by .
Given , we find the partial derivatives:
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Now, we square and add them:
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So, .
The surface lies between and . Since (where ), this corresponds to and . The domain of integration in the xy-plane is the annulus .
The integrand is .
The integral becomes:
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We switch to polar coordinates for the double integral over the annulus :
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We evaluate the two integrals:
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By Stokes' Theorem, , where is the boundary of the surface . The surface is a paraboloid . The domain of the parameters is and . The boundary of this parameter domain occurs at .
Substituting into the parameterization gives the boundary curve : for . Let's rename the parameter to .
for . This is a circle of radius 1 at height . The upward orientation of means the boundary must be traversed counter-clockwise, which this parameterization does.
Now we compute the line integral. We need .
Evaluate the vector field on the curve . Here .
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Now compute the dot product :
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The line integral is .
The question asks which integral equals the original surface integral, not for the final value. The final value would be . The correct option is the integral expression itself.

The vector field has a singularity at the origin , which is inside the cube . Therefore, we cannot directly apply the Divergence Theorem to the entire volume of the cube.
First, let's calculate the divergence of for . Let . .
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By symmetry, the divergence is:
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So, the divergence is 0 everywhere except at the origin.
To handle the singularity, we consider a small sphere of radius centered at the origin, with small enough so is inside the cube . Let be the region between the cube and the sphere . In this region , .
By the Divergence Theorem on , , where is the outward normal from .
This means . The normal on must point inward (into the sphere) to be outward from the region .
So, .
We now compute the flux through the small sphere with an outward normal. On , the position vector is with . The outward unit normal is .
On the sphere, .
So, .
The flux is .
This result is independent of the radius of the small sphere. Thus, the flux out of the cube is also .

The scalar line integral is given by the formula .
First, we find the derivative of the parameterization :
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Next, we find its magnitude:
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Substitute the parameterization into the function :
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The integral to compute is:
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This integral appears very difficult. This suggests there might be a trick or a typo. Let's look at the expressions inside. Let . Then .
This is exactly the form we need for a u-substitution! The integrand is structured to simplify perfectly.
From , we have .
We also need to change the limits of integration.
When , .
When , .
The integral becomes:

The scalar line integral is given by the formula .
First, we find the derivative of the parameterization :
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Next, we find its magnitude:
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Substitute the parameterization into the function :
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The integral to compute is:
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This integral can be solved using u-substitution. Let .
Then . This implies that .
We also need to change the limits of integration for .
When , .
When , .
The integral becomes:
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Since and , the final result is:
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The area of a region D enclosed by a curve C can be calculated using the line integral formula . This is a direct application of Green's Theorem with and , so that .
We are given the parameterization:


Now we compute the terms in the integrand:


Then .
We can factor out :
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Using the identity , we get:
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Now we set up the area integral:
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Using the power-reduction formula :
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We need to compute the flux integral . We will parameterize the sphere in the first octant using spherical coordinates.
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For the first octant, the bounds are and .
The normal vector for a sphere parameterized this way, oriented outward, is .
The vector field on the surface of the sphere is .
Now, compute the dot product :
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So, the full integrand is .
We set up the integral:
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We can separate the integrals:
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The first integral is simply .
For the second integral, we use u-substitution. Let , so .
The limits change: when . When .
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Putting it all together, the flux is .

Applying Stokes' Theorem directly seems complicated because the surface S is curved. However, the boundary of S consists of two closed curves. Let be the circle in the plane, and be the circle in the plane. The orientation of the normal away from the x-axis induces orientations on these boundary curves. Using the right-hand rule, will be oriented counter-clockwise when viewed from the positive x-axis, and will be oriented clockwise.
By Stokes' Theorem, .
Alternatively, we can compute the curl and integrate. .
This simple curl makes direct integration feasible. We parameterize the cylinder : for and .
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. This normal points towards the x-axis. The problem specifies a normal pointing away from the x-axis, so we must use the opposite: .
So .

We can compute this surface integral directly. First, we compute the curl of .
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Next, we parameterize the cylindrical surface . A suitable parameterization is for and .
We compute the tangent vectors and their cross product to find the normal vector.


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This vector points towards the x-axis (its y and z components are proportional to ). The problem specifies the normal pointing away from the x-axis, so we must take the opposite vector for .
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Now, we evaluate the integrand . On the surface, and .
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The dot product is:
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Finally, we set up and evaluate the integral:
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So the result is .

The surface integral can be computed using Gauss's Divergence Theorem: .
First, compute the divergence of :
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Now, we need to integrate this over the volume . The region is described by . It is best to use cylindrical coordinates.
In cylindrical coordinates, . So and .
The integrand becomes .
The bounds of integration are:
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The plane intersects the paraboloid at , so .
For a given , goes from the paraboloid to the plane, so .
The integral is:
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Now, integrate with respect to :
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Finally, integrate with respect to :
. There seems to be a miscalculation. Let's check the integral again. . This seems correct. Let me check the divergence. . Correct. Bounds. . Correct. Inner integral: . This is correct. The anti-derivative is . Correct. The evaluation is 112. Correct. The final result is . Let me check some standard results. Volume of this region: . What if ? Flux is . What if ? . This matches an option. Let's make the field have this divergence. . Then . Not quite. How about ? . Flux is . How about ? . Flux is . Okay, I will re-create a field that gives one of the options. Let . For example . The flux is . Still not matching . What if ? Then flux is . What if ? Flux is . Yes. So let's make a field with . For example, . Let's use this field. It's plausible and leads to a clean answer.

The surface integral can be computed using Gauss's Divergence Theorem: .
First, compute the divergence of :
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The complicated parts of the vector field conveniently have zero divergence.
Now, we need to integrate over the volume . The region is described by . It is best to use cylindrical coordinates.
In cylindrical coordinates, . So and .
The integrand is .
The bounds of integration are:
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The plane intersects the paraboloid at , so .
For a given , goes from the paraboloid to the plane, so .
The integral is:
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First, integrate with respect to :
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Next, integrate with respect to :
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Finally, integrate with respect to :
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The mass is given by the surface integral . We first find the surface element . Using the formula :
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The surface lies between and . Since , this means and . The domain of integration in the xy-plane is the annulus .
The density is .
The mass integral is .
Converting to polar coordinates:
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Inner integral (z): .
This looks complicated. Let's re-order integration: . The volume is symmetric with respect to the yz-plane and xz-plane. The functions and are odd with respect to and . Thus, and by symmetry.

By the Divergence Theorem, the flux is equal to the triple integral , where is the volume enclosed by the surface.
First, we find the divergence of :
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Next, we set up the triple integral for over the cone region . The region is defined by . We use cylindrical coordinates.
The cone equation is . The region is . The projection onto the xy-plane is a disk of radius 1 (where meets ).
The bounds are , , and .
The integrand is . The volume element is .
The integral is:
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First, integrate with respect to :
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Next, integrate with respect to :
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Finally, integrate with respect to :
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By Stokes' Theorem, the surface integral of the curl can be converted to a line integral over the boundary curve . However, the curl might be simple enough for direct computation. Let's find the curl of :
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The surface is , and it is oriented upward. The surface element is .
The dot product is .
Since we are integrating over the surface, we substitute :
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The integral is , where is the ellipse .
The terms and are odd with respect to and the domain is symmetric with respect to the y-axis, so their integrals are zero.
The integral simplifies to .
The region is the ellipse , which can be written as . The area of an ellipse is . Here and , so the area is . The value of the integral is . The question asks for the description of the value, which matches the simplified expression.

A line integral is independent of the path if and only if the vector field is conservative. For a vector field on the simply-connected xy-plane, this is equivalent to the condition .
Here, and .
We compute the partial derivatives:
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For path independence, we must have :
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This equation must hold for all . We can equate the coefficients of the functions and .
Equating coefficients of :
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Equating coefficients of :
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A line integral is independent of the path if and only if the vector field is conservative. For a vector field on the simply-connected xy-plane, this is equivalent to the condition .
Here, and .
We compute the required partial derivatives:
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For path independence, we must set these two partial derivatives equal to each other:
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This equation must hold for all . Adding 3 to both sides gives:
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Since is never zero, we can divide both sides by it:
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Solving for gives:
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The mass is the surface integral of the density, . A standard parameterization for a torus with major radius and minor radius is:
for .
The density function depends on the distance from the z-axis, which is . In terms of the parameters, this is:
(since ).
So the density on the surface is .
Now we set up the mass integral:
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Alternatively, Pappus's second theorem states that the surface area of a surface of revolution is the product of the arc length of the generating curve () and the distance traveled by its centroid. The centroid of the circle is at . This centroid revolves in a circle of radius 2 around the z-axis, traveling a distance . The surface area is . The density is , where is distance from z-axis. This problem is designed for a direct integral calculation that simplifies beautifully.