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Lecture 6: Surface Integrals

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- Recall, area is a vector

- Vector area of this surface is
which has sensible magnitude and direction

- Or, by projection

z

(0,1,1)

(1,0,1)

y

(0,1,0)

x

(1,0,0)

z

z

x

y

y

(1,0)

x

- Example from Lecture 3 of a scalar field integrated over a small (differential) vector surface element in plane z=0
- Example

(0,1)

Do x first

Then do y

- But answer is a vector ; get positive sign by applying right-hand rule to an assumed path around the edge (z>0 is “outside” of surface)

Example: volume flow of fluid, with velocity field , through a surface element yielding a scale quantity

Consider an incompressible fluid

“What goes into surface S through S1 comes out through surface S2”

S1 and S2 end on same rim

and depends only on rim not surface

- Consider a simple fluid velocity field
- Through the triangular surface
- More complicated fluid flow requires an integral

(0,1)

y

(1,0)

x

(see second sheet of this Lecture)

want to calculate

z

y

x

because

Note S need not be planar!

Note also, project onto easiest plane

- Surface Area of some general shape

- and S is the closed surface given by the paraboloid S1
and the disk S2

- On S1 :

Exercises for student (in polars over the unit circle, i.e. projection of both S1 and S2 onto xz plane)

- On S2 :
- So total “flux through closed surface” (S1+S2) is zero