- #1
thepatient
- 160
- 0
I'm not getting the answer from the back of the book for some reason. Is the book wrong or am I wrong?
calculate \intf · 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
\int_{C}\widehat{F}*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
\int -sint(5+4cost)dt= 1/4 \int 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 \intf · 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
\int_{C}\widehat{F}*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
\int -sint(5+4cost)dt= 1/4 \int 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?
Last edited:
