Question

Consider F and C below. F(x, y, z) = y2 sin(z) i + 2xy sin(z) j + xy2 cos(z) k C: r(t) = t2 i + sin(t) j + t k, 0 ≤ t ≤ π (a) Find a function f such that F = ∇f. f(x, y, z) = (b) Use part (a) to evaluate C ∇f · dr along the given curve C.

Answer 1

a)
`f (x,y,z)= xy^2\sin(z)`

b)
`\int_C F \cdot dr =0`

Explanation:

Recall that given a function f(x,y,z) then
`\\abla f = ((\partial f)/(\partial x),(\partial f)/(\partial y),(\partial f)/(\partial z))`. To find f, we will assume it exists and then we will find its form by integration.

First assume that F =
`\\abla f`. This implies that

`(\partial f)/(\partial x) = y^2\sin(z)` if we integrate with respect to x we get that

`f(x,y,z) = xy^2\sin(z) + g(y,z)` for some function g(y,z). If we take the derivative of this equation with respect to y, we get

`(\partial f)/(\partial y) = 2xy\sin(z) + (\partial g)/(\partial y)`

This must be equal to the second component of F. Then

`2xy\sin(z) + (\partial g)/(\partial y)=2xy\sin(z)`

This implies that
`(\partial g)/(\partial y)=0`, which means that g depends on z only. So
`f(x,y,z) = xy^2\sin(z) + g(z)`

Taking the derivative with respect to z and making it equal to the third component of F, we get

`xy^2\cos(z)+(dg)/(dz) = xy^2\cos(z)`

which implies that
`(dg)/(dz)=0` which means that g(z) = K, where K is a constant. So

`f (x,y,z)= xy^2\sin(z)`

b) To evaluate
`\int_C F \cdot dr` we can evaluate it by using f. We can calculate the value of f at the initial and final point of C and the subtract them as follows.

`\int_C F \cdot dr = f(r(\pi))-f(r(0))`

Recall that
`r(\pi) = (\pi^2, 0, \pi)` so
`f(r(\pi)) = \pi^2\cdot 0 \cdot \sin(\pi) = 0`

Also
`r(0) = (0, 0, 0)` so
`f(r(0)) = 0^2\cdot 0 \cdot \sin(0) = 0`

So
`\int_C F \cdot dr =0`