- #1
bmxicle
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Homework Statement
Let [tex] F(x, y, z) = \left ( e^{-y^2} + y^{1+x^2} +cos(z), -z, y \right) [/tex]
Let s Be the portion of the paraboloid [tex] y^2+z^2=4(x+1) [/tex] for [tex] 0 \leq x \leq 3 [/tex]
and the portion of the sphere [tex] x^2 + y^2 +z^2 = 4 [/tex] for [tex] x \leq 0 [/tex]
Find [tex] \iint\limits_s curl(\vec{F}) d \vec{s} [/tex]
Homework Equations
I plotted the surfaces in matlab, including what i think should be the boundary.
The Attempt at a Solution
[tex] \vec{G} = curl(\vec{F}) = \left (2, -sin(z), h(x, y) \right) [/tex] where h(x, y) is some derivative i don't want to do.
[tex] div(G) = 0 [/tex]
So If we close the surface into a solid by adding the circle z^2 + y^2 = 16 so we have:
[tex] \iint\limits_s \vec{G} \cdot d\vec{s} + \iint\limits_{z^2 + y^2 = 16} \vec{G} \cdot d\vec{s} = \iiint\limits_E div(\vec{G}) = 0[/tex]
[tex] \iint\limits_{z^2 + y^2 = 16} \vec{G} \cdot d\vec{s} = \iint\limits_{z^2 + y^2 = 16} \vec{G} \cdot \vec{n}ds[/tex] but we know that n is just the unit vector (1, 0, 0) so we have:
[tex] \iint\limits_{z^2 + y^2 = 16} \vec{G} \cdot d\vec{s} = \iint\limits_{z^2 + y^2 = 16} 2 ds = 16\pi*2 = 32\pi [/tex]
Thus this gives the original integral as [tex]-32\pi[/tex]
I have the solution to the question and they don't match so I'm wondering where my reasoning is going wrong.