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## Multivariable calculus

### Course: Multivariable calculus>Unit 5

Lesson 9: Proof of Stokes' theorem

# Stokes' theorem proof part 2

Figuring out a parameterization of our surface and representing dS. Created by Sal Khan.

## Want to join the conversation?

• At is dxdy = dS, where dS is not a vector but represents a small patch on the surface S?

Never mind, I understand now that |rx X ry| dxdy = dS, which makes perfect sense since that will be a point in dxdy times the area of the corresponding "patch" on S, so dS represents changes in the area of "patches" on S with respect to dxdy.
• Is there a better or (at least) another way to see if the vector ds is pointing in the right direction other than the right-hand rule Pr. Khan begins to mention at ?
• A vector that points toward increasing x crossed with a vector that points toward increasing y will always result in a vector that points toward increasing z.