00:01
So in this question, we want to compute the line integral along c of y squared dx minus x squared dy around the triangle whose vertices in the xy plane are at 1 -0, so 1 -0, followed by 0 ,0, and then negative 1 -0, in the xy plane.
00:35
So my region looks something like this.
00:39
Now, what we're going to use is something called greens theorem, which tells me that if i'm integrating around a closed curve, which i am this time, if i'm computing the line integral of a closed curve, that the line integral along c of pdx plus qdy is supposed to be equal to the double integral over the region enclosed of my partial of q with respect to x minus my partial of p with respect to y da.
01:22
And so what do we have here? we've got the double integral over r of my partial of q with respect to x.
01:31
My q is negative x squared.
01:35
So its derivative with respect to x is negative 2x, minus my partial of p with respect to y.
01:46
My partial of y squared with respect to y, that's just 2y, da.
01:56
So we're left with the double interval over r of negative 2x minus 2y d a.
02:02
Now, i think i want to set this up in the dx, d, d, y direction.
02:10
So i'm going to set this up in the dx, dy direction.
02:17
Well, let's see here.
02:18
What do i have? here on the left, i have y equals, y equals slope is 1, x plus 1.
02:29
Solving for x, i'm getting x equals y, minus 1 and over here on the right this was y equals negative x plus 1 so we have y minus 1 equals negative x where x equals 1 minus y so if i'm integrating with respect to x looking at a line parallel to the x -axis that line enters my region when x equals y minus y 1 and it exits my region when x equals 1 minus y and i am doing this from y equals 0 to y equals 1 and so let's see what this gives us now it's just a matter of not doing something silly i have the integral from 0 to 1 when i integrate with respect to x i've got negative x squared minus 2xy, and this is being evaluated from y minus 1 to 1 minus y, d .y.
03:50
And so, what do i do? well, i'm going to plug in my top limit of integration first.
03:57
I'm going to plug in.
03:58
Sorry about that weird little mark there.
04:00
Let me redo that.
04:01
So i'm going to plug in my 1 minus y to begin.
04:06
In...