This is an example of a surface integral problem where we're given the equation
of a plane and we're asked to find the surface integral, which we're also given,
that lies in the first octant. Because it's simple for us to solve for C in this case,
this is actually a pretty simple problem. We're just going to use the equation that f of XYZ ds
equals double integral f of XY g of XY times the square root partial Z partial X squared
plus partial Z partial Y squared plus 1 dA. If we wanted to we could draw what the plane looks
like. It's just going to be when X and Y are 0, Z equals 4. 1, 2, 3, 4. X equals 4 as well and
y equals 1. You don't really have to but that's the surface that we're integrating over. If for
some reason you're bored like me and you wanted to graph it in MATLAB, it would look like that. Okay,
so this is a pretty simple problem. We're just going to solve for Z from the plane equation,
so 4 minus X minus 4Y. And then we can plug that in like the equation says, we're plugging
in X and Y for Z. It's going to be the double integral 4 minus X minus 4Y and then we take the
partial derivative of Z with respect to X. That's going to be -1 squared plus -4 squared plus 1 dA.
And then when we imagine when Z equals zero, we have X plus 4Y equals 4. If we wanted to, we could
solve for X. That seems pretty simple. That's going to be 4 minus 4Y. X goes from 0 to 4Y. You
can change this to dx dy and-- sorry 4 minus 4Y and Y goes from 0 to 1. Okay, so then let's just
solve the double integral. That's going to be 1 plus 16 plus 1. That's going to be the square root
of 18, 0 to 1, 0 to 4 minus 4Y, 4 minus X minus 4Y dx dy. Integrate with respect to X first. That's
going to be 4X minus X squared over 2 minus 4XY. It's going to be 0 to 4 minus 4Y. Plug that in,
we get 4 minus 4Y minus 4 minus 4Y squared over 2 minus 4, 4 minus 4Y, Y, and then from 0 to 1.
I won't bore you with this single integral but you basically distribute,
expand this binomial, and then distribute and simplify. And also don't forget that
there's a square root 18 in the front. The integral part will give you 8 over
3 and that's times the square root of 18 and that's your final answer.
Here we have another example of a surface integral problem where we're given that the surface in this
case is a part of a cone above the Z plane. And if we're familiar with surface equations, we can kind
of tell just by looking at the equation that this is going to be a downward facing cone that starts
at Z equals 1 and goes to Z equals 0 on the X axis and the radius will just go from 0 to 1.
I have that plotted there. We're going to use a similar equation to last time in that the surface
integral equals ... times the square root partial Z partial X squared plus partial Z partial Y
squared plus 1 except now we're going to switch to cylindrical coordinates because that will make it
a lot easier. And instead of X and Y, we're going to have theta and r as the variables. If we switch
to cylindrical coordinates here, we're just going to get Z equals 1 minus r. When we switch this
to cylindrical coordinates, we're going to get r squared cosine squared theta plus r squared sine
squared theta plus 5 and then we plug in the Z. That's 1 minus r squared. When we simplify that,
that just gives us r squared plus 5 times 1 minus r squared. When we change the square root,
we're going to do partial Z partial theta squared plus partial Z partial r squared plus 1. When we
evaluate that, we're going to get 0 squared plus 1 squared plus 1. And it's going to give
us square root 2. Now we're ready to plug in everything into our double integral. We're
going to have the square root of 2 out front and then since we switch to cylindrical coordinates,
that's going to give us r dr dtheta as the integrand ranges. r goes from 0 to 1 like we
said before and theta goes from 0 to 2pi. Next we want to expand this binomial. We're going to do
1 minus r times 1 minus r. That gives us 1 minus 2r plus r squared times 5 plus r squared and then
that whole thing is still times r. I'm going to go ahead and distribute the 5. That's 5 minus 10 r
plus 5 r squared plus r squared. That gives us 6 r squared plus 5 minus 10 r times r. That gives us 6
r cubed plus 5 r minus 10 r squared. And now we're finally ready to take the integral from 0 to 1 of
r. Let's do that part first. We're going to get 6 r to the 4 divided by 4 plus 5 r squared over
2 minus 10 r cubed over 3 and that's from 0 to 1. When we plug in 1, we get 6 fourths plus 5 halves
minus 10 thirds minus 0 and that's times 2pi and times square root of 2. When we simplify these
fractions, we get two thirds. It's going to be 2 times two-thirds times pi times square root of
2. The simple answer is four-thirds times pi times the square root of 2 and that's your final answer.