00:02
In this question, we're asked to evaluate this line integral on the plane over the curve c, which is given here.
00:13
And as you can see, it's made up of this portion of a circle and this straight line.
00:22
So first of all, note that the integral we have here is just an abbreviated form of the integral with respect to x over c plus the integral with respect to y over c.
00:37
However, due to the formulas, or what these actually are in terms of the parameter t, because of that, this integral can just be written, like both of them we'll see can be written together in one integrand.
00:57
So the formula for a line integral over a two -dimensional scalar field f of x, y with respect to x over a curve c in the plane, is this.
01:19
Where x of t, y of t, is a vector function that parametrizes the curve c, and for values of t between a and b.
01:33
And then the same thing would hold if we were to change this and this to a y.
01:45
And as such, this sum of things with respect to x and with respect to y can be written like this, with one integrand, the one with respect to x, being multiplied by x prime, and the one with respect to y, which i call g here, multiplied by y prime.
02:10
Okay, so clearly we have to find x of t, y of t, x prime of t, and y prime of t.
02:17
So let's find that for our case.
02:22
Now, note that since c is made up of two quite different pieces, let's split c up into a c1, representing this circle, and a c2, representing this line segment.
02:42
Then we can parametrize each independently.
02:45
As we know, circles can be parametrized by trigonometric functions.
02:55
The usual way we do that is write x as a function of, or as a multiple of cosine, and y as a multiple of sine of t.
03:08
That way, x starts at one, or a multiple of one, and y starts at zero.
03:16
And this multiple is just the radius of the circle.
03:22
So we start here, and by the time, as t increases, we go along the circle, the unit circle, multiplied by this radius to make it the radius whatever circle, in this case two.
03:38
And so there's our parametrization.
03:46
And then x prime and y prime can just be found by differentiating with respect to t, as we know.
03:54
As for the line segment, we have a useful formula that is the initial point multiplied by one minus t, plus the final point multiplied by t.
04:22
And so x is negative t, and y is two minus t, for t ranging between zero and one...