Evaluate the given integral by changing to polar coordinates. ∬D x^2 y d A, where D is the top h

Evaluate the given integral by changing to polar...

Key Concepts

Double Integrals

Double integrals provide a means to integrate functions over two-dimensional regions. By switching to a more suitable coordinate system, such as polar coordinates for radially symmetric regions, the evaluation of these integrals can become more straightforward, both in terms of the integrand's form and the integration limits.

Setting Limits of Integration

Changing the coordinate system requires redefining the region of integration in terms of the new variables. For circular or semicircular regions, the limits for r are defined by the radius of the circle, and the limits for ? correspond to the angular span covering the region, which simplifies the process of setting up the integral.

Polar Coordinates

Polar coordinates are a two-dimensional coordinate system that represents points with a distance from the origin (r) and an angle (?). This system is particularly useful when dealing with circular or sector-shaped regions since it simplifies the integration limits and often the integrand.

Jacobian Determinant

When converting an integral from Cartesian to polar coordinates, the Jacobian determinant accounts for the change in the area element. Specifically, the differential area element dA in Cartesian coordinates becomes r dr d? in polar coordinates, where r is the Jacobian multiplier.

Transcript

00:02 We are given an integral of a function over a region and the description of this region.

00:11 We are asked to use polar coordinates to find this integral.

00:21 The integral is the double integral over the region d of the function x squared y, da, where d is the top half of a disk with a center at the center at the, the origin and the radius of 5.

00:42 So we can draw our region d using polar coordinates as follows.

00:52 We have a line theta equals pi over 2 and line theta equals 0.

00:59 And we know that the maximum value of r is going to be 5.

01:07 And we have that this is the top half of a disk.

01:12 So we're only looking at theta from 0.

01:17 To pi and since it's a disk we're looking at all radiuses less than or equal to five and greater than equal to zero so our region d looks something like this this blue shaded region which is a semicircle and we see that in polar coordinates the description of d is a set of all r's and theetas such that r lies between zero and five and data lies between 0 and pi.

02:16 If we treat our function x squared y as some function of x and y, call it f of x, y, we see that x squared is a continuous function, and y is a continuous function on d, and therefore f is continuous on the region d.

02:44 And so, since the region d is simple, it follows that the double integral of x squared, y over the region d is well defined and is given by the integral from zero to pi integral from zero to five of polar coordinates x is r cosine theta so this is r cosine theta and y is r sign theta and then for polar coordinates da is going to be r, dr, d -theta, where the r is added in the change of variable.

03:40 This is equal to and since we have that the limits on the integrals are constant values, we can separate these integrals using fabini's theorem, and write them as product of integrals.

03:59 So this is the same as the integral from 0 to pi of cosine square theta, sine theta, d theta, times the integral from 0 to 5 of r to the fourth, dr, and using substitution for the first integral, say u equals cosine of theta, then we have that du is going to be negative sine of theta, d theta...

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