Find the following average values. The average of the...

Key Concepts

Volume of a Region

Understanding and computing the volume of a region is fundamental when averaging a function over a space. The volume serves as the normalizing factor to ensure that the average value is properly scaled by the total size of the region.

Coordinate Transformations

In multivariable integration, it is often advantageous to transform coordinates (such as switching from Cartesian to cylindrical or spherical coordinates) to simplify the integrals, especially when the region of integration has symmetry that matches the new coordinate system.

Average Value of a Function Over a Region

This concept involves finding the mean of a function's values over a given domain by integrating the function over the region and then dividing by the total measure (area, volume, etc.) of the region. It generalizes the idea of averaging discrete values to continuous functions defined over geometric domains.

Triple Integration

Triple integration is the process of integrating a function over a three-dimensional region. It is used to calculate quantities like volume, mass, or averages of functions defined on a 3D domain, by successively integrating with respect to three variables.

Transcript

00:01 So we are given a solid paraboloid d with z between 0 and 4 minus x squared minus y squared, and asked to find the average value of the squared distance between the origin and points in this paraboloid.

00:13 So we know that the average value function is f bar equals 1 over the volume of the solid times the triple integral over d of the function f of xyz dv.

00:31 So first, let's go ahead and find this function.

00:35 So we're trying to find the squared distance between the origin and points in the paraboloid.

00:44 We know that the distance between a point and the origin is the square root of x squared plus y squared plus z squared.

00:56 And the squared distance is this whole quantity squared, which is just equal to x squared plus y squared plus z squared.

01:08 Now we have this f figured out.

01:10 Now let's figure out d, which will be used both in finding the volume of the solid and in this integral.

01:19 So we're already given the bounds for z.

01:23 Let's look at when z equals 0.

01:27 So when z equals 0, this equation becomes 0 equals 4 minus x squared minus y squared, or x squared plus y squared equals 4 which is the equation of a circle with radius 2.

01:43 So now we can write out d.

01:48 We'll have d is the set x, y, z with the same z bound, z between 0 and 4 minus x squared minus y squared.

02:06 Y will take between negative root 4 minus x squared and positive root 4 minus x squared and then for x we'll take from negative 2 to positive 2 because 2 is the radius of this circle all right now we know d let's first evaluate this second part of the function and then we can go back and evaluate the volume and then combine them to find the total average value let's go ahead and set up our first integral this will be the integral over d of f of x, y, z, dv.

02:57 So this is equal to.

02:59 We'll put x on the very outside, y in the middle, and then z on the very inside.

03:06 So have the integral from negative 2 to 2, of the integral from negative root 4 minus x squared, to positive root 4 minus x squared, of the integral from 0, to 4 minus x squared minus y squared of x squared plus y squared plus z squared d z d y d x so let's evaluate this very center integral we'll copy over the first two integration symbols the integral of x squared plus y squared plus z squared with respect to z is z times x squared plus y squared we're going to lump these together.

04:06 They're both act as constants, plus the z squared turns into one -third times z cubed.

04:17 And we'll evaluate this from z -equal -0 to z -equal 4 minus x squared minus y squared.

04:26 D -y -d -x.

04:30 So this is equal to copy over the first two integration symbols again.

04:43 Z, which is 4 minus x squared minus y squared times x squared plus y squared plus one -third times four minus x squared minus y squared cubed and then both of these terms are zero when z equals zero, d y d x.

05:12 Now this is going to be a little tricky to evaluate.

05:17 So what we can do is convert it to polar.

05:22 As we know that for the remaining x and y, we're just integrating over this circle, x squared plus y squared equals 4, which is a perfect candidate for conversion to polar.

05:34 So as a reminder for polar coordinates, we have x squared plus y squared equals r squared, x equal to cosine theta, y is equal to sine theta, and dx, y, d .y is equal to r, dr, d theta.

05:59 Alright, let's go ahead and convert this integral from cartesian into polar coordinates.

06:05 So we will put theta on the outside and r on the inside.

06:09 So for theta, we'll go over the whole circle, 0 to 2 pi.

06:13 For r we know that this circle has a radius of 2 so we'll go from 0 to 2.

06:23 Now to convert this inside we have 4 minus the quantity x squared plus y squared which is r squared times x squared plus y squared which is r squared plus 1 3 3 times 4 minus r squared again cubed and this whole quantity will be multiplied by r, d -theta.

06:54 Now when this is all simplified out, you have the integral from 0 to 2 pi, the integral from 0 to 2 of 4 r cubed minus r to the 5th plus 1 3 3 minus r squared, cubed, cubed, cubed, d r d squared, cubed, times r, dr d theta.

07:22 Alright, now we can evaluate this integral.

07:25 So this is equal to the going to go from 0 to 2 pi...

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