Verifying Stokes' theorem for F = yi + zj + xk on a paraboloid

[sunnyday11]

Homework Statement

Verify Stoke's Theorem for F = y i + z j + x k and S the paraboloid z=1-(x^2+y^2) with z bigger or equal to 0 oriented upward, and the curve C which is the boundary of S.

Homework Equations

Stoke's Theorem
line interal = flux of curl of F (don't know how to use the symbol function)

The Attempt at a Solution

For the line integral, I use r(t) = cos (t) i + sin (t) j , and the solution i got was -pi but I could not get the flux of curl of F to match.

Curl F = -(i+j+k)
double integral of (-2x-2y-1)dxdy
==> double integral of (-2r cost(t) - 2r sin(t) drdt with r [0,1] and t [0, 2pi]


Thank you very much!

 

[HallsofIvy]

Homework Statement

Verify Stoke's Theorem for F = y i + z j + x k and S the paraboloid z=1-(x^2+y^2) with z bigger or equal to 0 oriented upward, and the curve C which is the boundary of S.

Homework Equations

Stoke's Theorem
line interal = flux of curl of F (don't know how to use the symbol function)

The Attempt at a Solution

For the line integral, I use r(t) = cos (t) i + sin (t) j , and the solution i got was -pi but I could not get the flux of curl of F to match.

Curl F = -(i+j+k)
double integral of (-2x-2y-1)dxdy
==> double integral of (-2r cost(t) - 2r sin(t) drdt with r [0,1] and t [0, 2pi]

I agree with you up to here but changing
$$\int\int (-2x- 2y+ 1)dxdy$$
to polar coordinates gives
$$\int\int (-2rcos(t)- 2r sin(t)+ 1)(rdrdt)$$


That is, you seem to have forgotten the "+ 1" in the integrand. And the differential in polar coordinates is rdrdt, not just "drdt".

Thank you very much!