Question

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`?

Answer 1

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`