Calculate Line Integral of Vector Field f(x,y) over Curve C | Homework Problem

[stratusfactio]

Homework Statement

calculate the integral f · dr for the given vector field f(x, y) and curve C:
f(x, y) = (x^2 + y^2) i; C : x = 2 + cos t, y = sin t, 0 ≤ t ≤ 2π (2pi)

Homework Equations

Would the vector F simply be <(x^2+y^2),0> since there is no j component?
The solution is 4pi and I'm getting zero.

The Attempt at a Solution

integral of C = F · dr
F = <((2+cos t)^2 + (sin t)^2),o> = <(5 + 4 cos t), 0>
dr = <-sin t, cos t>

Integral of C [0, 2pi] <(5 + 4 cos t), 0> · <-sin t, cos t> = 0 :(

I'm thinking that my error lies in the vector I'm using for F.

 

[HallsofIvy]

If the problem is exactly as you stated, then the correct answer is 0.

 

[thepatient]

That's right, the F vector function only has a i component. It would be equivalent as writing it as F(x,y) = <x^2 + y^2, 0>

BTW I posted this same exact problem. :P where did you find this?

 

[thepatient]

https://www.physicsforums.com/showthread.php?t=515786

 

[stratusfactio]

^Haha. I'm self teaching myself Multivariable Calculus using this online book: http://www.mecmath.net/calc3book.pdf in conjuction with Youtube's UCBerkely Multivariable Calc lectures.

It's just weird because I did all the steps and analyzed each step and can't see where I went wrong...we may be right because I see you got 0 too, sometimes the books make errors. I just don't see how we can get 4pi when we're evaluating an integral involving trig. when it's going to give us a rational number.

 

[thepatient]

I was doing the same with that same book. :P There were a few other errors too in other parts. I think the book just needs more revising.