Orientation of the boundary must be the same too get

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Unformatted text preview: f not careful here) Thus if you are dealing with a complicated surface you can simplify the problem by picking a simple surface with the same boundary. Example: Calculate where S is the closed surface of and and These two surfaces share the common boundary C: We have a Right-Hand Rule that gives me the orientation of the boundary curve. Notice these two surfaces have opposite orientation and so the work done in one direction cancels out the work done in the other direction. Example: Use Stoke’s Theorem to find with the cylinder where C is the space curve which is the intersection of the plane . Also given: Assume the orientation of curve C is counterclockwise as viewed from above I used the right-hand rule to figure out which way the normal vector is pointing It is easy to find since it is normal to the plane Thus Curl and need to be calculated: Curl Using what we learned earlier: Example: Verify Stoke’s Theorem: plane The curve C was parameterized by Now using the surface integral: Where Example: S is the portion of the paraboloid lying above the xy- Verify Stoke’s Theorem: and below the plane z = 1. S is the portion of the paraboloid lying above the xy-plane The curve C was parameterized by Now using the surface integral: Where NOTE!! points outward when the z-component is NEGATIVE so I must use instead!!!! This was using the paraboloid as the surface. Would I have the same answer if I used the plane z = 1 ? Of course. Net Flux = Rate Out – Rate In Circulation: A measure of the tendency of the fluid to rotate or circulate around the curve C. Other Forms of Green’s Theorem: Gauss’s Theorem in the Plane: This gives us the FLUX outward across the closed curve C Stoke’s Theorem in the Plane: This gives us the CIRCULATION along the closed curve C Divergence Theorem or Gauss’s Theorem: This lifts Green’s Theorem up one dimension This gives us the FLUX outward across the closed surface Stoke’s Theorem: This is a generalization of Green’s Theorem to Surfaces This gives the CIRCULATION along the open surface S = the circulation along the boundary of the surface What this implies is the above surface integral will lead to the same result regardless of the surface used as long as the boundary of the surfaces are the same! Thus if you are dealing with a complicated surface you can simplify the p...
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