Solving Stokes' Theorem: Find \int_{\partial S} F \cdot ds

[mkkrnfoo85]

Here is the problem:

S is the ellipsoid x^2+y^2+2z^2=10

and F is a vector field F=(sin(xy),e^x,-yz)

Find: \int \int_S ( \nabla \mbox {x} F) \cdot dS


So, I know that Stokes' Theorem states that:
\int \int_S ( \nabla \mbox {x} F) \cdot dS = \int_{\partial S} F \cdot ds
where \partial S equals the boundary of the ellipsoid. How do you find \partial S? My professor just told me that any closed surface has no boundary and therefore the answer is 0, but would someone show me how I can show this? And can someone tell me under what conditions does the answer become 0 using Stokes' Theorem? Thanks a lot.

 

[jdavel]

Stokes' theorem is derived by dividing a bounded surface up into a bunch of small flat surfaces and then noticing that the line integrals around all these surfaces cancel each other out except at the boundary of the big surface. Think about that and what it means if the surface you're splitting up has no boundary.

 

[thepatientmental]

To solve this problem, we can start by finding the boundary of the ellipsoid \partial S. The boundary of a surface is defined as the set of points where the surface ends or meets another surface. In this case, the ellipsoid is a closed surface, meaning it has no boundaries. This can be seen by visualizing the ellipsoid and noticing that it has no edges or boundaries where it meets another surface. Therefore, \partial S = \emptyset and the integral becomes:

\int \int_S ( \nabla \mbox {x} F) \cdot dS = \int_{\emptyset} F \cdot ds = 0

This shows that the answer is indeed 0, as your professor stated.

Stokes' Theorem is applicable when the surface is closed, meaning it has no boundaries or edges. In this case, the surface is a closed ellipsoid and therefore the theorem can be used. In general, Stokes' Theorem can be used for any closed surface, whether it is a sphere, cylinder, or any other shape with no boundaries.

In summary, to solve this problem we first determine that the boundary of the ellipsoid is \emptyset. Then, using Stokes' Theorem, we can rewrite the integral as \int_{\emptyset} F \cdot ds = 0. This shows that the answer is 0 and also shows the conditions under which Stokes' Theorem can be used - for closed surfaces with no boundaries.