# Evaluating line integral directly – part 1 (video)

Evaluating Line Integrals
Evaluating Line Integrals

## Multivariable calculus

• Stokes’ theorem intuition
• Green’s and Stokes’ theorem relationship
• Orienting boundary with surface
• Orientation and stokes
• Orientations and boundaries
• Conditions for stokes theorem
• Stokes example part 1
• Stokes example part 2
• Stokes example part 3
• Stokes example part 4
• Stokes’ theorem
• Evaluating line integral directly – part 1
• Evaluating line integral directly – part 2

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?(7 votes)
• 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.(7 votes)
• At, If a question provides us with a surface integral, we could therefore evaluate the surface integral by calculating the line integral? Is this the concept of stokes theorem? What would be the equation in this case? 0:35(3 votes)
• Atsir said only one parameter has to be used.What if the circle is not essentially a unit circle (eg x^2+ y^2=4)?. Then r can’t be considered as 1. Thanks in advance!! 1:07(1 vote)
• if it wrong? there should be a minus sign before y^2 sinTheta(1 vote)
• Look closely, you can see negative times negative equals positive.(1 vote)