00:01
So in this question, i want to evaluate the line integral where c is the given curve.
00:04
I have the integral along c of xyz squared ds, where c is the line segment from negative 350 to negative 164.
00:17
So the first thing i have to do is i have to parameterize this line segment.
00:23
So what is the vector traveling from negative 350 to negative 164? such a vector travels two units in the x direction, one unit in the y direction, and four units in the z direction.
00:41
And so parameterizing this line segment, i am getting x equals negative three plus two t, y equals five plus t, and z just equals four t, where my t's they are ranging from 0 to 1.
01:02
So now, i'll be able to replace my x, y, and z.
01:06
But i also need my ds.
01:10
Remember that my ds is the square root of dx, dt, being squared, plus d ydt being squared, plus dzdt being squared, all of this dt.
01:30
Well, let's see.
01:31
My dx, dt this time, that's 2 being squared.
01:36
My d -y -d -t, that's 1 being squared, and my d -z -d -t, that's 4 being squared, all of this d -t.
01:48
And so i have, my ds is the square root of 4 plus 1 plus 16 dt.
01:58
It is the square root of 21 dt.
02:05
And so now i'm going to substitute into my in.
02:09
I'm going to have an interval.
02:11
I'm converting everything into t's.
02:13
So my t's range from 0 to 1.
02:16
My x, that was negative 3 plus 2t, times my y.
02:24
My y was 5 plus t, times my z being squared, so times 4t being squared.
02:35
This is times ds.
02:37
My ds is the square root of 21 d3.
02:43
And so let's see what's happening here.
02:48
I'm going to go ahead and i'm going to multiply this out.
02:51
I have integral 0 to 1 up.
02:54
I've got negative 15 minus 3t plus 10t plus 2t squared.
03:05
I'm multiplying this by 16t squared times the square root of 21dt.
03:15
My root 21, i pull that through.
03:18
Root 21 times the integral from 0 to 1.
03:22
Now let's see...