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# Evaluating line integral directly - part 1

Showing that we didn't need to use Stokes' Theorem to evaluate this line integral. Created by Sal Khan.

## Want to join the conversation?

• why didnt we have to use another parameter, 'r' in this?
• Because for this side of Stokes' theorem we are only taking a line integral and not a surface integral. A line integral only requires a parametrization in one variable since it is the integral across a curve and not a surface, which requires two variables for its parametrization. The curve parametrization and the surface parametrization are actually related- the curve parametrization (with respect to f instead of theta) is (cos f)*i + (sin f)*j +(2-sin f)*k, whereas the surface parametrization (with respect to r and f instead of r and theta) is (r*cos f)*i + (r*sin f)*j + (2-r*sin f)*k. The surface parametrization is different from the curve parametrization in that it multiplies all theta terms by r. Doing so creates the surface out of the curve- as Sal explained in previous videos, for any value of theta we can take different values of r and include every point on the interior of the curve by adding r as a parameter.