Calc 3 question: verifying Green's theorem?

Question

I have a Calc 3 homework due this evening, and one question is causing me trouble. The question is: Verify that Green's Theorem is true for the vector field F = (y^2-7y)i + (2xy+2x)j where C is the circle x^2+y^2=1. (C is the curve to integrate on.)

I set my vector function r=cos(t)i + sin(t)j where 0 < t < 2pi. Then dr= (-sint(t)dt, cos(t)dt). F = (cos^2(t)-7cos(t), -2sin(t)cos(t)-2sin(t)). Is this all right?

I evaluated the line integral from 0-2pi of F dot dr, and it came out to be zero. Then I evaluated the double integral from Green's theorem (dQ/dx) - (dP/dy), with the limits being 0 < y < 1 and 0 < x < sqareroot(1-y^2), and I got 9pi/4. If I'm verifying Green's theorem, the answers from these two integrals should be equal, but they're not. What did I do wrong?

Thanks so much.

Answer 1

"I set my vector function r=cos(t)i + sin(t)j where 0 < t < 2pi."
This is correct.
"Then dr= (-sint(t)dt, cos(t)dt)."
This is correct.
"F = (cos^2(t)-7cos(t), -2sin(t)cos(t)-2sin(t))."
NO! This is wrong - F=. Remember, you are substituting the x and y coordinates of r, not dr. This completely changes the calculation.
You unfortunately did not post the details of your integration, so I cannot tell you whether you made any mistakes there. Here is the correct integration:

[0, 2π]∫F·dr
[0, 2π]∫·<-sin t, cos t> dt
[0, 2π]∫-sin³ t + 7 sin² t + 2 cos² t sin t + 2 cos² t dt
[0, 2π]∫-sin t(1-cos² t) + 7 sin² t + 2 cos² t sin t + 2 cos² t dt
[0, 2π]∫-sin t + 7 sin² t + 3 cos² t sin t + 2 cos² t dt
[0, 2π]∫-sin t + 7 - 7 cos² t + 3 cos² t sin t + 2 cos² t dt
[0, 2π]∫-sin t + 7 + 3 cos² t sin t - 5 cos² t dt
[0, 2π]∫-sin t + 7 + 3 cos² t sin t - 5 (cos (2t) + 1)/2 dt
[0, 2π]∫-sin t + 9/2 + 3 cos² t sin t - 5/2 cos (2t) dt
cos t + 9/2t - cos³ t - 5/4 sin (2t)|[0, 2π]
(1 + 9π - 1 - 0) - (1 + 0 - 1 - 0)
9π

For the double integral, your limits are wrong. If you're using the integral with dx dy, the limits should be [-1, 1] and [-√(1-y²), √(1-y²)]. The way you did it omits three quarters of the circle. Here's the integration I did, this one using polar coordinates (since it's easier):

∬dQ/dx - dP/dy dA
∬(2y+2) - (2y-7) dA
∬ 9 dA
[0, 2π]∫[0, 1]∫9r dr dθ
[0, 2π]∫9/2r²|[0, 1] dθ
[0, 2π]∫9/2 dθ
9/2θ|[0, 2π]
9π

Which is the same as we got for the line integral.

Answer 2

Im sorry, I dont recall what the Greens theorem is, but Im trying to understand your question, though. Perhaps I still can help you.

F(t) seems right to me, so does dr.

F*dr = (cos^2 t - 7 cos t)(-sin t) +(-2sint cost - 2 sint) cos t

F*dr = -sint cos^2 t + 7 cost sint - 2 sint cos^2 t - 2 sint cost = -3 sint cos^2 t + 5 sint cost

Lets integrate that:

int [0-2pi](-3 sin t cos^2 t + 5 sint cost) dt=
= int [0-2pi] (3 u^2 -5 u) du, where u = cos x, and du = - sint dt

int [0-2pi] (u^2 -5u)du = (u^3/3-5/2 u^2) [0-2pi]= cos^3 t- 5 cos^2 t [0-2pi] = 1-5 - (1-5) = 0

OK.

I wont check the other integral, I dont know what Q and P are, but probably the problem here is that you arent in the theorem hypothesis. Try to check that.

Anabel

Answer 3

A steady drive is one that doesn't rely on time or function, it is a vector F whose course is tangent to the closed curve. So you've a line vital of F*(d^3x) which you'll convert to an subject vital, no? That will have to get you began...