If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

### Course: Multivariable calculus>Unit 4

Lesson 1: Line integrals for scalar functions

# Line integral example 2 (part 2)

Part 2 of an example of taking a line integral over a closed path. Created by Sal Khan.

## Want to join the conversation?

• I thought parametrization for C2 and C3 may be a overkill. C2 would be bounded by y axis (0 to 2) and curve z=y^2. So the area of this portion of curtain would be def Int (y^2) for these bounds (0-2) -this would be 8/3. For area of C3, the curtain area will be bounded by a line z=x and the x axis (from 0-2); For this portion one does not even need calculus as it is a right angled triangle with base of 2 and height of 2 units and area will be 2. But it is satisfying to see the math is consistent-well that is why it is math!
• Why is the direction of the curve/contour so important?

(So important that Sal needs to preserve it by using y=2-t instead of the simple y=t)
• In this case it isn't, but it's good practice for vector fields.
x=t. Could y=t and x=2-t? I hope this is not a stupid question.
• I know this is a late reply 8 years after, but you will still get the same result if you set it as y = t. The reason why he uses 2-t is because the contour is going backward.
• Would it not just be easier to take the integrals of curves 2 and 3 with respect to y and x respectively? Seems an awful lot of hassle to go through parameterising them in terms of t.
• Some integrals are easier to do than others. Transforming the variables is sometimes necessary to get an integral which is solvable symbolically.
(1 vote)
• Why is x equal to t? I was expecting 2-t for x as well.
• t is just being treated as a counter from 0 to 2, either putting x = t or x = 2-t would've worked. x = 2-t would add the bars from 2 to 0 while x = t would add them from 0 to 2.
• At , is there a particular reason, conceptual or otherwise, to parametrize the equation as x=0, y=2-t, 0≦t≦2? Why shouldn't we integrate y^2 from 0 to 2? Would this mess us up in later maths?
• no in this they are conceptually doing the same thing
(1 vote)
• would someone please explain why when on the y axis 2-t is y but when along the x axis x=t is it
• Sal moves around the walls in a counter clockwise fashion: He starts at (2,0) follows the semi circle to (0,2) moves down the y axis to (0,0) then along the x axis back to (2,0). Since the movement down the y axis is in the negative direction, he uses negative t.
(1 vote)
• here is the problem i noticed .i understand the different parameters for C2 however the directions are switched .as shown in sal's example letting y=2-t yields a final area of 8/3 .however when one uses the parameters y=t where t goes from 2 to 0 ,the integral yields an answer of -8/3 .so these two approaches are clearly not equivalent ,unless we are simply interested in the absolute values ,which makes sense for this example .but in more abstract problems this would obviously lead to different answers .so my question is which approach best avoids this problem? .when is direction important in other words ?