I tried solving the question below in two ways, using Stokes and then using the Divergence Theorem. But I got different answers.
Q. Let \(\displaystyle S\) be the bucket shaped surface consisting of the cylindrical surface \(\displaystyle y^2\, +\, z^2\, =\, 9\) between \(\displaystyle x\, =\, 0\) and \(\displaystyle x\, =\, 5,\) and the disk inside \(\displaystyle yz\)-plane of radius \(\displaystyle 3\) centered at the origin. (The bucket \(\displaystyle S\) has a bottom, but no lid.) Orient \(\displaystyle S\) in such a way that the unit normal points outward. Compute the flux of the vector field \(\displaystyle \triangledown\, \times\, \vec{G}\) through \(\displaystyle S,\) where \(\displaystyle \vec{G}\, =\, \langle\, x,\, -z,\, y\, \rangle.\)
What is wrong?
1) Using Stokes
Take the capping surface as the disk at x = 5 with radius 3 centered on the x-axis. The curve is
counter-clockwise as the normal vector points outward. Normal vector is <1,0,0>
curl G = <2,0,0>
Flux is
double integral of curl G dot n dS = double integral of <2,0,0> dot <1,0,0< dS
= 2(area of disk) = 2(9pi) = 18pi
2) Using the divergence theorem
Close the boundary with the disk at x = 5 with radius 3 centered on x-axis. Call this surface S1
S2 is the surface we want. S1 is oriented outwards because of the divergence theorem, so its normal vector is <1,0,0>
S = S1 U S2 is closed
Flux through S = triple integral of div (curl G) dV = 0 as the div(curl G) = 0 for any "well-behaved" vector field
Flux through S1:
double integral of curl G dot n dS = double integral of <2,0,0> dot <1,0,0< dS
= 2(area of disk) = 2(9pi) = 18pi
Flux through S = flux through S1 + flux through S2
0 = 18pi + flux through S2
flux through S2 = -18pi
I got different answers. I think my orientation of one of the surfaces is wrong. Why?
Q. Let \(\displaystyle S\) be the bucket shaped surface consisting of the cylindrical surface \(\displaystyle y^2\, +\, z^2\, =\, 9\) between \(\displaystyle x\, =\, 0\) and \(\displaystyle x\, =\, 5,\) and the disk inside \(\displaystyle yz\)-plane of radius \(\displaystyle 3\) centered at the origin. (The bucket \(\displaystyle S\) has a bottom, but no lid.) Orient \(\displaystyle S\) in such a way that the unit normal points outward. Compute the flux of the vector field \(\displaystyle \triangledown\, \times\, \vec{G}\) through \(\displaystyle S,\) where \(\displaystyle \vec{G}\, =\, \langle\, x,\, -z,\, y\, \rangle.\)
What is wrong?
1) Using Stokes
Take the capping surface as the disk at x = 5 with radius 3 centered on the x-axis. The curve is
counter-clockwise as the normal vector points outward. Normal vector is <1,0,0>
curl G = <2,0,0>
Flux is
double integral of curl G dot n dS = double integral of <2,0,0> dot <1,0,0< dS
= 2(area of disk) = 2(9pi) = 18pi
2) Using the divergence theorem
Close the boundary with the disk at x = 5 with radius 3 centered on x-axis. Call this surface S1
S2 is the surface we want. S1 is oriented outwards because of the divergence theorem, so its normal vector is <1,0,0>
S = S1 U S2 is closed
Flux through S = triple integral of div (curl G) dV = 0 as the div(curl G) = 0 for any "well-behaved" vector field
Flux through S1:
double integral of curl G dot n dS = double integral of <2,0,0> dot <1,0,0< dS
= 2(area of disk) = 2(9pi) = 18pi
Flux through S = flux through S1 + flux through S2
0 = 18pi + flux through S2
flux through S2 = -18pi
I got different answers. I think my orientation of one of the surfaces is wrong. Why?
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