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I'm not getting the answer from the back of the book for some reason. Is the book wrong or am I wrong?

calculate [itex]\int[/itex]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

itex]\int[/itex]f · dr

r(t) = <x(t), y(t)>

dr = r'(t) dt

[itex]\int[/itex][itex]_{C}[/itex][itex]\widehat{F}[/itex]*dr

F = <x^2+y^2, 0>

C: x=2+cost, y = sint, 0<=t<=2pi

r(t) = <2+cost, sint>

r'(t) = <-sint, cost>

F(r(t)) * r'(t) = <(2+cost)^2 + sin^2(t), 0> * <-sint, cost> =

-sint(4 + 4cos(t) + cos^2(t) + sin^2(t)) = -sint(5+4cost)

Letting u = 5+ 4cost

du/4 = -sintdt

[itex]\int[/itex] -sint(5+4cost)dt= 1/4 [itex]\int[/itex] udu = 1/8 (5+4cost)^2 |t = 0..2pi =

81/8 -81/8 = 0

But the answer in the back of the book says 4pi, what did I do wrong?

Since the path taken is a closed curve, I also tried green's theorem to verify if it's right or not, but I got -2 2/3 using greens theorem. Green's theorem does apply in this case, does it not?

## Homework Statement

calculate [itex]\int[/itex]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

## Homework Equations

itex]\int[/itex]f · dr

r(t) = <x(t), y(t)>

dr = r'(t) dt

## The Attempt at a Solution

[itex]\int[/itex][itex]_{C}[/itex][itex]\widehat{F}[/itex]*dr

F = <x^2+y^2, 0>

C: x=2+cost, y = sint, 0<=t<=2pi

r(t) = <2+cost, sint>

r'(t) = <-sint, cost>

F(r(t)) * r'(t) = <(2+cost)^2 + sin^2(t), 0> * <-sint, cost> =

-sint(4 + 4cos(t) + cos^2(t) + sin^2(t)) = -sint(5+4cost)

Letting u = 5+ 4cost

du/4 = -sintdt

[itex]\int[/itex] -sint(5+4cost)dt= 1/4 [itex]\int[/itex] udu = 1/8 (5+4cost)^2 |t = 0..2pi =

81/8 -81/8 = 0

But the answer in the back of the book says 4pi, what did I do wrong?

Since the path taken is a closed curve, I also tried green's theorem to verify if it's right or not, but I got -2 2/3 using greens theorem. Green's theorem does apply in this case, does it not?

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