Integrate f over the given curve. f(x, y)=x+y, C: x^2+y^2=4 in the first quadrant from (2,0) to

step 1 Okay, folks, so we're going to be doing problem number 29 here. So for this problem, we have a c

Integrate f over the given curve. f(x, y)=x+y, C: x^2+y^2=4...

Key Concepts

Curve Parameterization

Parameterizing a curve means expressing the coordinates of points on the curve as functions of one or more parameters. This technique simplifies the process of integration by transforming the integral into an integral with respect to the parameter, which is particularly useful when dealing with curves defined implicitly.

Arc Length Differential

The arc length differential, typically denoted as ds, represents an infinitesimal segment of the curve. It is computed from the derivatives of the parameterization and is used to properly account for the contribution of each segment to the total line integral, ensuring that the integration is performed with respect to the actual distance along the curve.

Line Integrals

A line integral of a scalar function along a curve involves integrating the function over the path weighted by a differential element, often the arc length. This concept is essential in calculus and physics for computing quantities such as work done by a force or mass distributed along a wire.

Transcript

00:01 Okay, folks, so we're going to be doing problem number 29 here.

00:06 So for this problem, we have a circle, and we're going to be integrating a function f along the circle in the first quadrant from 2 -0 to 0 -2.

00:20 So before we get started doing the integral, let's just get a few preliminary steps out of the way.

00:27 So for a circle you have this equation, obviously you have x squared plus y square equals 4.

00:33 And from this, you can solve for y, and that is going to be root, root 4 minus x squared.

00:43 Well, you know, because we're integrating along the first quadrant, so y is always going to be positive.

00:49 And y prime, which is a derivative with respect to the variable x, is going to be negative x over root for a, minus x squared.

01:00 So that's y prime.

01:01 And from this, you can get ds.

01:04 And ds is going to be a positive number because that's because ds is a is the length of like a really, really small triangle, which can be written as dx squared plus d y squared.

01:21 So this little length here is what we call ds.

01:25 Excuse me.

01:26 My handwriting is really horrible.

01:28 Ds.

01:29 So this is dx.

01:30 And, and and this is d .y.

01:31 So d .s is always going to be a positive number because it's the length of a, of the, of this side.

01:41 I forgot what this side is called.

01:44 So d .s is always going to be a positive number because it's a length, okay, and you cannot have a negative length.

01:49 So now that we're clear on the fact that d .s is going to be a positive number, we're going to plug in.

01:59 What we're going to do is we're going to divide out the x inside of the square root.

02:03 And multiply it back in on the outside...

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