# A particle moves along line segments from the origin to the points (3, 0, 0), (3, 3, 1), (0, 3, 1), and back to the origin under the influence of the force field

F(x, y, z) = z2i + 5xyj + 4y2k.

I have tried multiple times to solve this problem but the answer is still not right.

## Well, it seems like you really had a "forceful" encounter with this problem! Don't worry, I'll try to give you a hand (or a laugh)!

To solve this problem, we need to find the work done by the force field along the given line segments. The work done by a force field can be calculated using the line integral formula:

โซFโdr

Where F is the force field and dr is an infinitesimal displacement vector along the path.

Let's break it down step by step:

1. From the origin to (3, 0, 0):
The displacement vector dr = dx*i. Substituting the values into the force field F(x, y, z), we get F = (0^2, 5(0)(x), 4(0)^2) = (0, 0, 0). Therefore, the work done along this segment is 0. It's like the force field is taking a break too!

2. From (3, 0, 0) to (3, 3, 1):
Here, dr = dy*j + dz*k. Substituting the values into the force field, we get F = (1^2, 5(3)(y), 4(y)^2) = (1, 15y, 4y^2). Integrating this along the segment should give you the work done over this interval.

3. From (3, 3, 1) to (0, 3, 1):
Again, dr = -dx*i. Substituting the values into the force field, we get F = (1^2, -5(x)(3), 4(3)^2) = (1, -15x, 36). Integrating this along the segment should give you the work done over this interval.

4. From (0, 3, 1) back to the origin:
One last time, dr = -dy*j - dz*k. Substituting the values into the force field, we get F = (1^2, 5(0)(y), 4(y)^2) = (1, 0, 4y^2). Integrating this along the segment should give you the work done over this interval.

Now, it's time to put it all together and calculate the total work done along the entire path. Don't give up, you're almost there!

## To find the work done by the force field on the particle as it moves along the given path, you can use the line integral formula:

W = โซ F ยท dr,

where F is the force field and dr is an infinitesimal displacement vector along the path. In this case, we need to find the line integral for each segment of the path and then sum them up.

Let's go step by step through each segment of the path and calculate the line integral.

1) The particle moves from the origin to the point (3, 0, 0). The displacement vector for this segment is dr = dx i. Substituting the force field F and the displacement dr into the line integral formula, we have:

W1 = โซ F ยท dr = โซ(z^2 dx) = โซ 0 dx = 0.

2) The particle moves from (3, 0, 0) to (3, 3, 1). The displacement vector for this segment is dr = dy j + dz k. Substituting into the line integral formula, we have:

W2 = โซ F ยท dr = โซ(5xy dy + 4y^2 dz).

To evaluate this integral, we need to express y and z in terms of a parameter. Let's use t as the parameter and parametrize the path from (3, 0, 0) to (3, 3, 1) as:
y = 3t, z = t, x = 3.

dy = 3 dt, dz = dt, dx = 0.

Substituting these values into the integral, we have:

W2 = โซ (5(3)(3t)(3 dt) + 4(3t)^2(dt))
= โซ (45t^2 dt + 36t^2 dt)
= โซ 81t^2 dt
= 27t^3 + C.

Evaluating the integral from t = 0 to t = 1, we get:

W2 = (27(1)^3 - 27(0)^3) = 27.

3) The particle moves from (3, 3, 1) to (0, 3, 1). The displacement vector for this segment is dr = -dx i. Substituting into the line integral formula, we have:

W3 = โซ F ยท dr = โซ(z^2 (-dx)) = โซ-dx = -x.

Since x goes from 3 to 0, W3 = -(0-3) = 3.

4) The particle moves from (0, 3, 1) back to the origin. The displacement vector for this segment is dr = -dy j - dz k.

W4 = โซ F ยท dr = โซ(-5xy dy - 4y^2 dz).

Similarly to step 2, we need to express x, y, and z in terms of a parameter. Let's use t as the parameter and parametrize the path from (0, 3, 1) back to the origin as:
y = 3, z = 1 - t, x = 3t.

dy = 0, dz = -dt, dx = 3dt.

Substituting these values into the integral, we have:

W4 = โซ (-5(3t)(3) dt - 4(3)^2(0 - dt))
= โซ (-45t dt + 0)
= -22.5t^2 + C.

Evaluating the integral from t = 0 to t = 1, we get:

W4 = -(22.5(1)^2 - 22.5(0)^2) = -22.5.

To get the total work done, we sum up the individual line integrals:

Total work done = W1 + W2 + W3 + W4
= 0 + 27 + 3 - 22.5
= 7.5.

Therefore, the work done by the force field on the particle as it moves along the given path is 7.5 units.

## Here is some of the work so far that I have had helped with:

A (3, 0, 0), B (3, 3, 1), C (0, 3, 1)
the plane containing these points has normal vector
n = <3 -3, 0 -3, 0 -1> x <3 -0, 3 -3, 1 -1> = <0, -3, -1> x <3, 0, 0> = <0, -3, 9>
then curl F = del x F = <d/dx, d/dy, d/dz> x <z^2, xy, 4y^2> = <8y -0, -(0 -2z), 5y -0> = <8y, 2z, 5y>
by Stokes theorem, the line integral around the path on which the particle allegedly travels is the surface integral of the dot product of the curl F and the normal to that surface
the region of integration is the projection of the plane onto the xy-plane, a triangle of vertices
(3, 0), (3, 3), (0, 3)
which corresponds to the region bounded by the lines
y = 3
x = 0
y = -x +3
x = 3
n dot curl F = <0, -3, 9> dot <8y, 2z, 5y> = 0 -6z +45y
need to solve for z in terms of x, y, using the equation of the tangent plane containing each point
choosing point A (3, 0, 0)
n dot <x -x0, y -y0, z -z0> = 0
<0, -3, 9> dot <x -3, y, z> = 0 -3y +9z = 0
9z = 3y
z = y/3
evaluate the double integral of
-6(y/3) +45y dy dx from [3, -x +3] [3, 0]
= 43y^2 /2 dy dx from [3, -x +3] [3, 0]
= 43*9/2 -(43/2)(-x +3)(-x +3) dx [3, 0]
= 43*9/2 -(43/2)(x^2 -6x +9) dx [3, 0]
= 43*9/2 x -(43/2)(x^3 /3 -6x^2 /2 +9x) [3, 0]
= 43*9/2 *3 -(43/2)(27/3 -6*9/2 +27)
= 387