00:01
We are trying to calculate the surface integral of y, ds.
00:08
And in this case, since we have y in terms of x and z, y equals x squared plus z squared, we can use the formula that the surface integral of f of x, g of x, z, z, multiplied by the y partial with respect to x squared, plus the y partial with respect to z squared plus 1 d a we can use this formula to calculate the surface integral so we want our function to be in terms of x and z our integral to be in terms of x and z so instead of writing this is the integral of y we're going to write this as the integral of x squared plus z squared as shown above y equals x squared plus z squared and this will be multiplied by the x partial, which is 2x.
01:08
So 2x squared will be 4x squared, plus the z partial, which will also be 4 z squared, plus 1 d .a.
01:22
Now we are given that the bounds are from 0 to 1.
01:26
However, as we have x squared and x squared and 4x squares and 4 z squares, we can rewrite this in polar.
01:34
Doing so we'll give the integral from 0 to 1 integral from 0 to 1 of x squared plus z squared same as above however this is going to translate from 0 to 1 to 0 to 2 pi and 0 to 2 for the drd theta of r squared multiplied by the square root of 1 plus 4 r squared as x squared plus z squared is equal to r squared if we go to polar.
02:10
And then this integral will be in terms of dr d theta multiplied by r.
02:20
Solving for this, we're going to want to do a u sub for the e square root as the 4r squared is very hard to deal with.
02:28
So setting u equal to 1 plus 4r squared, we then know that du is equal to 8r.
02:38
And we want to get rid of this r squared right here with the use up.
02:43
So we're going to solve for r square using u equals 1 plus 4 r square...