How do you find the surface area of the part of the circular paraboloid #z=x^2+y^2# that lies inside the cylinder #x^2+y^2=1#?

1 Answer
Mar 24, 2015

I assume the following knowledge; please ask as separate question(s) if any of these are not already established:

  1. Concept of partial derivatives
  2. The area of a surface, #f(x,y)#, above a region R of the XY-plane is given by #int int_R sqrt((f_x')^2 + (f_y')^2 +1) dx dy# where
    #f_x'# and #f_y'# are the partial derivatives of #f(x,y)# with respect to #x# and #y# respectively.
  3. In converting the integral of a function in rectangular coordinates to a function in polar coordinates: #dx dy rarr (r) dr d theta#

If #z = f(x,y) = x^2 + y^2#
then #f_x' = 2x# and #f_y'= 2y#

The Surface area over the Region defined by #x^2+y^2 = 1#is given by
#S =int int_R sqrt(4x^2 + 4y^2 + 1) dx dy#

Converting this to polar coordinates (because it is easier to work with the circular Region using polar coordinates)
#S = int_(theta = 0)^(2pi) int_(r=0)^1 (4 r^2+1)^(1/2) (r) dr d theta#

#= int_(theta=0)^(2pi) ((4r^2+1)^(3/2))/(12) |_(r=0)^1 d theta#

#= int_(theta=0)^(2pi) (5sqrt(5)-1)/(12) d theta#

#= (5sqrt(5) -1)/(12) theta |_(theta=0)^(2pi)#

#= (5sqrt(5)-1)/6 pi#