00:01
So in this question, we want to compute the line integral with respect to art length of the function f of x, y, z equals x times y, along the curve x equals y squared from the point zero zero one to the point nine three one in the plane z equals one.
00:17
So i am trying to find the line integral along c of x, y, d, s along this curve.
00:29
So the first thing i have to do is i have to parameterize the curve by letting y be the parameter t.
00:37
So their suggestion is let y equal t.
00:41
Now if y is equal to t and x is equal to y squared, then x should be equal to t squared.
00:50
And i'm in the plane z equals 1, so my z is equal to 1 this time.
00:56
Now, what are the t values that would correspond to 001 and 931? well, look at the y coordinate because y is equal to t.
01:09
So that first point corresponds to t equals 0, while that second point corresponds to t equals 3.
01:18
And so let's figure out our ds and then we'll be in pretty good shape.
01:23
So my ds is the square root, you may recall, of the square root of dx, dt being squared, plus dydt being squared, plus dz d t being squared, and then all of this is dt.
01:43
And so this time, what is my ds? well, it's the square root of dxdt is 2t, so i'm going to have to square that, plus dydt d .ydt is 1, that gets squared, plus dzdt, that's 0 being squared.
02:06
And so it looks like my ds this time is going to be the square root of 4t squared plus 1dt.
02:16
And so now we are ready to plug in and convert our line integral into the world of t.
02:23
So what am i getting? i'm getting an integral from 0 to 3 of x times y.
02:31
So that's t squared times t times ds.
02:36
My ds is the square root of 4 t squared plus 1 ds.
02:43
So that here we have the integral from 0 to 3 of t cubed times the square root of 4 t squared plus 1 ds...
