Stokes theorem application

[bmxicle]

Homework Statement

Let $$F(x, y, z) = \left ( e^{-y^2} + y^{1+x^2} +cos(z), -z, y \right)$$
Let s Be the portion of the paraboloid $$y^2+z^2=4(x+1)$$ for $$0 \leq x \leq 3$$
and the portion of the sphere $$x^2 + y^2 +z^2 = 4$$ for $$x \leq 0$$

Find $$\iint\limits_s curl(\vec{F}) d \vec{s}$$

Homework Equations

I plotted the surfaces in matlab, including what i think should be the boundary.

The Attempt at a Solution

$$\vec{G} = curl(\vec{F}) = \left (2, -sin(z), h(x, y) \right)$$ where h(x, y) is some derivative i don't want to do.
$$div(G) = 0$$

So If we close the surface into a solid by adding the circle z^2 + y^2 = 16 so we have:

$$\iint\limits_s \vec{G} \cdot d\vec{s} + \iint\limits_{z^2 + y^2 = 16} \vec{G} \cdot d\vec{s} = \iiint\limits_E div(\vec{G}) = 0$$
$$\iint\limits_{z^2 + y^2 = 16} \vec{G} \cdot d\vec{s} = \iint\limits_{z^2 + y^2 = 16} \vec{G} \cdot \vec{n}ds$$ but we know that n is just the unit vector (1, 0, 0) so we have:

$$\iint\limits_{z^2 + y^2 = 16} \vec{G} \cdot d\vec{s} = \iint\limits_{z^2 + y^2 = 16} 2 ds = 16\pi*2 = 32\pi$$

Thus this gives the original integral as $$-32\pi$$

I have the solution to the question and they don't match so I'm wondering where my reasoning is going wrong.

 

[mighty2000]

Thank you for sharing your attempt at solving this problem. I can see that you have correctly calculated the curl of the vector field and found that it is equal to (2, -sin(z), h(x, y)). However, I believe there may be an error in your approach to solving the integral.

Firstly, the integral you have written is incorrect. It should be \iint\limits_s \vec{G} \cdot d\vec{s} instead of \iint\limits_s \vec{F} \cdot d\vec{s}. This is because we are calculating the surface integral of the curl of the vector field, not the original vector field itself.

Secondly, I'm not sure why you have added the circle z^2 + y^2 = 16 to the surface. This circle does not form part of the given surfaces and therefore should not be included in the integral. The boundaries of the surface are already defined by the paraboloid and the sphere, so there is no need to add an additional boundary.

Finally, when calculating the surface integral, we need to consider the orientation of the surface. In this case, the surface is oriented in the positive direction, so the unit normal vector is pointing outwards. Therefore, the correct integral should be:

\iint\limits_s \vec{G} \cdot d\vec{s} = \iint\limits_s (2, -sin(z), h(x, y)) \cdot \vec{n} ds = \iint\limits_s 2ds = 2 \iint\limits_s ds

This integral can then be evaluated over the two given surfaces separately to find the final result. I hope this helps clarify your understanding of the problem.