I've been trying different methods but keep getting the same answer for the following question:
Verify that Stokes’ Theorem is true for the given vector field F and surface S.$14.\; \vec F(x, y, z) = <-2yz, y, 3x>, \\ z = 5-x^2-y^2\\ z>1$
I first tried without Stokes' Theorem to get an initial value, and got this:
$\oint_c \vec F \cdot d\vec r = \oint_c -2yz\,dx + y\,dy + 3x\,dz = \oint_c -2yz\,dx + y\,dy = \iint (\frac{\partial}{\partial x}(y) - \frac{\partial}{\partial y}(-2yz))dA = \iint 2z\,dA = 2 \iint 5-x^2-y^2 dA = 2\iint 5-r^2 dA = 2 \int_0^{2\pi}\int_0^2 5r-r^3drd\theta = 24\pi$
With Stokes' Theorem, though, I got this:
$curl \,\vec F = <0, -2y-3, 2z>\\ \vec n = \frac{<-2x, -2y, 1>}{\sqrt{1+4x^2+4y^2}} \\ dS = dxdy\sqrt{1+4x^2+4y^2} \\ \oint_c \vec F \cdot d\vec r = \iint_S (curl\,\vec F) \cdot \vec n\,dS = \iint_S (2y(2y+3) + 2z)dxdy = 32\pi$
Where did I go wrong?
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$\begingroup$$\oint_c \vec F \cdot d\vec r = \oint_c -2yz\,dx + y\,dy + 3x\,dz = \oint_c -2yz\,dx + y\,dy = \iint (\frac{\partial}{\partial x}(y) - \frac{\partial}{\partial y}(-2yz))dA$
Okay you have used Green's theorem here which is the baby version of stokes theorem when your contour lies in a plane.
continuing.
$\iint 2z\ dA$
$A$ is the disk of radius $2,$ at $z = 1$
$8\pi$
If we don't use Green's theorem.
$x = 2 \cos t\\ y = 2\sin t\\ z = 1$
$\int_0^{2\pi} 8\sin^2 t + 4\sin t\cos t \ dt = 8\pi$
Stokes theorem.
$dS = (-\frac {\partial z}{\partial x},-\frac {\partial z}{\partial y}, 1) = (2x, 2y, 1)$
(it took me far too long to spot this error)
$\iint \nabla \times F dS = \iint -4y^2 - 6y + 2(5-x^2 - y^2) \ dy\ dx$
Converting to polar seem to make sense here, but is not entirely necessary.
$\int_0^{2\pi}\int_0^2 -4r^3 \sin^2\theta - 6r^2\sin\theta + 10r - 2r^2\ dr\ d\theta$
Lets integrate by $\theta$ first.
$\int_0^2 20\pi r -8\pi r^3 \ dr\\ 10 r^2 - \frac {8}{4} r^4|_0^2\pi\\ 8\pi$
By the way, one of the implications of Stokes theorem, is that divergence free force integrated over any surface with the same boundary has the same flux.
So the integral over the disk equals the integral over the cone.
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