Question

Using f(x) = yx^2 +yz
, calculate the line integral F(dr) where C is the curve given by R = 2ti − 3sin(π/2t)j + t^2k, for
0 ≤ t ≤ 1

Answer 1

For this, do I not just use the function, parameterized into t, and evaluated from 0 to 1?

Answer 2

F = grad f. Therefore I should be using f dot dr, take the derivative of R and convert f to spherical parametric coordinates?

Answer 3

Hmmm, so let's think about this for a moment.

Answer 4

It appears that \(f(x)\) is a scalar function, and not a vector field.

Answer 5

So they are asking for: \[
\int_Cf~ds
\]

Answer 6

If F was a vector field and I used partial integration to determine what f(x,y,z) was, should f(x,y,z) be a vector field also?

Answer 7

If \(\mathbf F\) is a vector field. then what is \(f\)? Is it \(\nabla f = \mathbf F\)?

Answer 8

\[F = (2xy)i + (x^2 + z)j + (y)k\]

Answer 9

I believe so yes

Answer 10

FTC for line integrals: \[
\nabla f = \mathbf F \implies \int_C\mathbf F~d\mathbf r = f(C_f) - f(C_i)
\]Where \(C_f\) and \(C_i\) are the final and initial point of the curve.

Answer 11

\[
C_i=\mathbf r(0),\quad C_f=\mathbf r(1)
\]

Answer 12

So R is given to me to evaluate at the endpoints. Then I use these values for my integration?

Answer 13

Sorry. I'm not quite sure what to do with the R vector given or why it was given exactly.

Answer 14

You don't have to integrate. Essentially \(f\) is the anti-derivative.

Answer 15

Yes

Answer 16

I take \[0 \le t \le 1 \] and evaluate R? Then use these for my anti-derivative?

Answer 17

Uhm. I kind of already gave the answer. Does it make any sense? Which part is confusing?

Answer 18

Ci=r(0),Cf=r(1) is the part that is confusing. I evaluate R at 0 and 1 and use those values for my anti-derivative? Plus, I got 0 for both => my integral is 0?

Answer 19

What did you get for \(\mathbf r(0)\) and \(\mathbf r(1)\)?