00:01
Okay, so let's get started by finding our region of integration.
00:06
Well, we already know that we are working in the first octant, so we need x, y and z greater than or equal to 0.
00:16
Perfect.
00:17
Perfect.
00:18
Now, we also know that z must be less than or equal to 8 minus 2x squared minus y squared and greater than or equal to y squared.
00:31
Ok, perfect.
00:33
Let's keep in mind that since z is greater than or equal to a positive quantity, we don't need to write this.
00:42
That is we don't need to rewrite z greater than or equal to zero okay that being said now what do we need to do well we need to find the intersection between the graph of the function z equal to this one and the graph of the function z equal to this one this way we are gonna find we are gonna find the planar region of integration in the xy plan because we have a lower bound for z and an upper bound for z but we don't have any relation between x and y so we need to find this relation between x and y okay now let's find the intersection between the graph of z equal to y squared and z equal to 8 minus 2 x squared minus y squared well the intersection is just this one equal to this one so y squared equal to 8 minus 2 x squared minus y squared and well this is equivalent to x squared plus y squared equal to 4 perfect.
02:03
Okay, so this is the relation between x and y, just a relation defining a circle.
02:10
Perfect.
02:11
So, the intersection of our region of integration with the x -y plane, the intersection of our region of integration with the x -y plane is just this, this part of of our circle with center the origin and radius 2 because we need to keep in mind that x and y are greater than or equal to zero perfect okay now we have found our region of integration we know that z is bounded by these two functions and x and y belong to this quarter of a circle with radius 2 okay, now let me call r our region of integration.
03:04
Let's compute the integral of x over r.
03:08
This one can be written as an integral, a double integral with respect to, okay, let me call this region of integration a.
03:18
A.
03:19
So an integral over a of an integral with respect to z, so from y squared to 8 minus 2x squared minus y squared of x in dz and da, the infinitesimal element of area.
03:40
So we obtain an integral over a of well this integral here is easy to compute here we have 8 minus 2x squared minus 2y squared in the a perfect now to compute the o multiplied by x okay this one now to compute this integral here we are gonna use polar coordinates so we are gonna have an integral with respect to to the angle theta from 0 to pi halves and integral with respect to the radius from 0 to 2 of okay this one is r cosine of theta this one is 8 minus 2 r squared perfect the da is r, dr, d, theta.
04:42
Perfect...