Solution to section 15.6, #10 of Grossman, Calculus

Calculate the surface integral of x^2 + y^2 + z^2 over the part of the plane
x - y = 4 that lies inside the cylinder y^2 + z^2 == 4.

Solution. This problem might seem easier if the plane were a function
z = f(x,y), since then we know from p. 997, eq. (12) that the surface area element is
d sigma = Sqrt[1 + (\partial f/\partial x)^2 + (\partial f/\partial y)^2] dx dy .
There is no reason, however, that we couldn't regard x as a function of y and z instead:

x = f(y,z) =
Here:

In[1]:=

  X[y_,z_] := y + 4

In[2]:=

  FunnyFactor[f_,y_,z_] := Sqrt[1 + (D[f,x])^2 + (D[f,y])^2]

In[3]:=

  FunnyFactor[X[y,z],y,z]

Out[3]=

  Sqrt[2]

That's not so funny!

What is the domain for the variables y and z? It is the interior of a circle of radius 2:
y^2 + z^2 == 4
So wouldn't it be nice if the variables were x and y so we could use polar
coordinates? Well, why not use them anyway, with
z := r Cos[theta], y = r Sin[theta] ?
Sure. The integral then runs for theta from 0 to 2 Pi and for r from 0 to 2.
The integrand has to be a function of r and theta, so we substitute for x, y, and z:

In[4]:=

  X[y,z]^2 + y^2 + z^2

Out[4]=

   2          2    2
  y  + (4 + y)  + z

In[5]:=

  % /. {y -> r Sin[theta], \
                               z -> r Cos[theta]}

Out[5]=

   2           2    2           2                     2
  r  Cos[theta]  + r  Sin[theta]  + (4 + r Sin[theta])

In[6]:=

  Simplify[%]

Out[6]=

          2    2
  32 + 3 r  - r  Cos[2 theta] + 16 r Sin[theta]
  ---------------------------------------------
                        2

In[7]:=

  Integrand[r_,theta_] = %

Out[7]=

          2    2
  32 + 3 r  - r  Cos[2 theta] + 16 r Sin[theta]
  ---------------------------------------------
                        2

In[8]:=

  Integrate[Integrand[r,theta], {r,0,2}, {theta,0,2 Pi}]

Out[8]=

  72 Pi