Scalar Surface Integral ∫∫xy dS, S is the triangular region (1,0,0), (0,2,0), (0,0,2)
Автор: Jonathan Walters
Загружено: 2019-10-19
Просмотров: 19114
Evaluate the Surface Integral ∫∫xy dS, S is the triangular region (1,0,0), (0,2,0), (0,0,2).
The first thing to do for any surface integral is find the parameterization.
Then if it's a scalar surface integral we take the magnitude of the normal vector |s_u x s_v|. This cross product acts like our jacobian for the change of variables.
In this problem our bounds are derived from the plane equation. The region in the xy-plane here is a triangle for which we set up type I bounds.
Lastly we substitute in our parameterization to a function, multiply by our Jacobian, then integrate.
As always, if you have any questions, let me know!
Thanks for watching!
-dr. dub
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