00:01
If s is given by this, then our double integral will be expressed over a surface s to be equal to this, where d is the projection of the surface on the x, y plane.
00:21
So, giving that, given that we have our s, such that z is equal to 2 divided by 3.
00:35
You have x to the power 3 divided by 2 plus y to the power 3 on 2 then it implies that a double integral over the surface s of y the s is going to be equal to the integral over the projection d you have y and at the s is going to be 2 divided by 3 times 3 divided by 2 you have x the partial derivative with respect to x that will give us 1 on 2 all squared plus the partial derivative of this which is 2 divided by 3 times 3 divided by 2 you have y to the power 1 on 2 all squared plus 1 plus 1 and this is 2.
01:39
Whole its square roots.
01:44
Okay.
01:45
So this is going to give us, this is equal to the integral d, you have d, y.
01:59
Then this whole thing will give us square roots of x plus y plus 1, d .a.
02:12
So giving that, so from here, given that our interval for x and y x is from 0 to 1 and y and y is from 0 to 1 implies that so then this is going to be equal to the double integral from 0 0 to 1 you have y square root of x plus y plus one the x the y the y so you first integrate with with respect to x and what's integrating we will treat y as a constant so then this is going to be equal to the integral to one and this will give us 2 divided by 3 y you have x plus y plus 1 3 or 2 the interval from 0 to 1 the y and this is equal to 3 2 so you have 2 divided by 3 the interval from 0 to 1 you have y here and this gives us y plus two to the power three on two minus y you have y plus one to the power three on two the y so now we integrate with respect to why so to do so to do so you realize that we can apply there's some rule here so that we have three we have two divided by three this integral to one of y y plus two three on two the y you have minus three two on three two on three your integral from zero to one why you have y plus one one three onto the y so let's substitute if you parametized so we substitute we substitute we substitute y equal to t squared minus two in the first integral so for this then and and y equal to b squared minus one for the second was integral so this goes here and this goes there as well so we do not change you have to note that they you have to note that since you are using this substitution then the interval as well will change the change of limits from to 1 will change as well...