In Exercises 23-26, sketch the region of integration and convert each polar integral or sum of i

In Exercises 23-26, sketch the region of integration and...

Key Concepts

Jacobian Determinant

When changing variables in double integrals, the area element transforms according to the Jacobian determinant. In the case of polar to Cartesian conversion, the differential area element in polar form, r dr d?, appears as the product of the Jacobian (r) and the differentials of the polar coordinates.

Coordinate Transformation

Converting an integral from polar to Cartesian coordinates involves using the relationships x = r cos? and y = r sin?, along with the fact that r² = x² + y². This transformation is crucial when the integral or the region is originally expressed in polar form, and a reformulation in terms of x and y is desired.

Region of Integration

The region of integration must be carefully translated between coordinate systems. For example, an integration over r from 0 to 1 and ? from 0 to ?/2 in polar coordinates corresponds to a quarter of a unit disk in the first quadrant in Cartesian coordinates, described by the inequalities x² + y² ? 1 with x ? 0 and y ? 0.

Polar Coordinates

Polar coordinates represent points in the plane using a radius and an angle (r, ?) instead of the usual Cartesian coordinates (x, y). This system is especially useful for regions with circular symmetry or when dealing with circles, arcs, and sectors.

Cartesian Coordinates

Cartesian coordinates describe points in the plane using x and y values. They are the most common coordinate system for many integration problems and are especially useful when the region of integration or the function’s formulation is better suited to horizontal and vertical boundaries.

Transcript

00:01 So they want us to change this from a polar integral to a cartesian interval.

00:08 So first let's go ahead and draw what our bounds are going to be.

00:11 So over here, notice that our radius goes from 0 to 1.

00:14 So let's write that down.

00:18 And then theta is supposed to rotate from 0 to pi half.

00:25 So if we were to sketch this now, i'll do that down here.

00:34 So our radius rotates from 0.

00:38 To pi half, so we're in this region.

00:42 So our lower bound is supposed to rotate from zero to pi half.

00:46 So, i mean, since our radius is zero at this point, we're not going anywhere.

00:49 And then we rotate out to one for our radius, and then we just swing it from zero to pi half.

00:59 And so now this, all right here, will be our region of integration.

01:09 So this is really just a circle of radius 1 in the first quadrant.

01:15 So we know that this equation should be, well, first x squared plus y squared is equal to 1.

01:22 And if we're going to be in the first quadrant, we can rewrite this in two ways.

01:28 But we'll figure those out once we decide how we're going to use our bounds for this.

01:37 At least we have this for right now.

01:39 But now let's go ahead and start doing our conversion.

01:43 So first, so i'm going to just leave the bounds empty for a second.

01:49 So to convert from polar to regular coordinates, we'll need to do the following.

01:55 We'll first need to divide all of this by r.

01:58 So actually, i'll do this off on the side over here.

02:05 So r cubed, sine theta, cosine, theta.

02:10 So we first divide all this by r.

02:12 Because normally we would multiply by r to go from cartesian to polar, so we just divide by r to go back.

02:21 Now we have r squared side of theta, cosine of theta.

02:27 And at this point, we can make the substitutions, because normally we would say, okay, well, x is going to be r cosine theta.

02:37 Y is going to be r sine theta here, not y.

02:46 We'll notice that this here is exactly what we have down there, because we can rewrite this as r -sign theta, r -cosin -theta, and then we'll look at that.

02:57 This is y, and this is x.

03:00 So we can rewrite this here as x times y, and then i'll just put da for right now, and then i'll just put r down here for our region.

03:12 So now, depending on, if we'd rather integrate with respect to x or y, we'll have this in two ways.

03:20 So let's actually go ahead and first do where we integrate with respect to x.

03:26 So if we do that, so we'd have xy d x, d y.

03:32 So in this is the case, we're going to want to solve this function in terms of y, or in terms of x.

03:44 So let's go ahead and do that...

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