Question

Evaluate the integral of the vector field F(x,y,z)=3xi+4yj-zk along the path c(t) = (sin(t),cos(t),t), for t between 0 and 3pi/2.

Answer 1

I calculate the derivative of the parameterized curve c(t) and got (cos(t),-sin(t),1) and I found F(c(t)) and multiplied that with c'(t) using the dot product. I'm having trouble solving the integral from 0 to 3pi/2 of 3sin(t)cos(t) - 4cos(t)sin(t) - t.

Answer 2

my Ta mentioned using the trig identity sin2x = 2sinxcosx but I'm still not getting the correct answer.

Answer 3

*to compute the integral

Answer 4

I can help you with the integral

Answer 5

you want to integrate
3sin(t)cos(t) - 4cos(t)sin(t) - t from 0 to 3pi/2 ?

Answer 6

what should the integral be equal to. I get
-9pi^2/8 - 1/2

Answer 7

That's what I got except it was +1/2. I probably made an error somewhere. Let me check.

Answer 8

Yeah, that was correct. I'm not sure where I messed up.

Answer 9

your expression
3sin(t)cos(t) - 4cos(t)sin(t) -t = - sin t cos t - t

Answer 10

Oh yeah, probably so. Damn negative signs -___-. Thanks so much.

Answer 11

Your welcome :) if you leave this question open, other people might comment

Answer 12

also it might help if you wrote out the calculations you got, c(t), F(c(t)) , etc, someone may spot an error