Question

Find the work done by F= (x^2+y)i + (y^2+x)j +(ze^z)k over the following path from (4,0,0) to (4,0,4)

a. The line segment x=4 y=0 z=[0;4]

Find a scalar potential function f for ​F, such that

F=∇f.

The work done by F over the line segment is?

b. The helix r(t)= (4cost)i +(4sint)j +(2t/pi)k t=[0;2pi]

Find df/dt for F?

The Work done by F over the Helix?

c. The x axis from (4,0,0) to (0,0,0) followed by the line z=x, y=0 from (0,0,0) to (4,0,4)

What is the integral to comput the work done by F along the x-axis from followed by the line z=x?

What is the work done by F over the 2 curves

Answer 1

`\vec F(x,y,z)=(x^2+y)\,\vec\imath+(y^2+x)\,\vec\jmath+ze^z\,\vec k`

We want to find
`f(x,y,z)` such that
`\\abla f=\vec F`. This means

`(\partial f)/(\partial x)=x^2+y`

`(\partial f)/(\partial y)=y^2+x`

`(\partial f)/(\partial z)=ze^z`

Integrating both sides of the latter equation with respect to
`z` tells us

`f(x,y,z)=e^z(z-1)+g(x,y)`

and differentiating with respect to
`x` gives

`x^2+y=(\partial g)/(\partial x)`

Integrating both sides with respect to
`x` gives

`g(x,y)=\frac{x^3}3+xy+h(y)`

Then

`f(x,y,z)=e^z(z-1)+\frac{x^3}3+xy+h(y)`

and differentiating both sides with respect to
`y` gives

`y^2+x=x+(\mathrm dh)/(\mathrm dy)\implies(\mathrm dh)/(\mathrm dy)=y^2\implies h(y)=\frac{y^3}3+C`

So the scalar potential function is

`\boxed{f(x,y,z)=e^z(z-1)+\frac{x^3}3+xy+\frac{y^3}3+C}`

By the fundamental theorem of calculus, the work done by
`\vec F` along any path depends only on the endpoints of that path. In particular, the work done over the line segment (call it
`L`) in part (a) is

`\displaystyle\int_L\vec F\cdot\mathrm d\vec r=f(4,0,4)-f(4,0,0)=\boxed{1+3e^4}`

and
`\vec F` does the same amount of work over both of the other paths.

In part (b), I don't know what is meant by "df/dt for F"...

In part (c), you're asked to find the work over the 2 parts (call them
`L_1` and
`L_2`) of the given path. Using the fundamental theorem makes this trivial:

`\displaystyle\int_(L_1)\vec F\cdot\mathrm d\vec r=f(0,0,0)-f(4,0,0)=-\frac{64}3`

`\displaystyle\int_(L_2)\vec F\cdot\mathrm d\vec r=f(4,0,4)-f(0,0,0)=\frac{67}3+3e^4`