# 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?

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, $f \left(x , y\right)$, above a region R of the XY-plane is given by $\int {\int}_{R} \sqrt{{\left({f}_{x} '\right)}^{2} + {\left({f}_{y} '\right)}^{2} + 1} \mathrm{dx} \mathrm{dy}$ where
${f}_{x} '$ and ${f}_{y} '$ are the partial derivatives of $f \left(x , y\right)$ with respect to $x$ and $y$ respectively.
3. In converting the integral of a function in rectangular coordinates to a function in polar coordinates: $\mathrm{dx} \mathrm{dy} \rightarrow \left(r\right) \mathrm{dr} d \theta$

If $z = f \left(x , y\right) = {x}^{2} + {y}^{2}$
then ${f}_{x} ' = 2 x$ and ${f}_{y} ' = 2 y$

The Surface area over the Region defined by ${x}^{2} + {y}^{2} = 1$is given by
$S = \int {\int}_{R} \sqrt{4 {x}^{2} + 4 {y}^{2} + 1} \mathrm{dx} \mathrm{dy}$

Converting this to polar coordinates (because it is easier to work with the circular Region using polar coordinates)
$S = {\int}_{\theta = 0}^{2 \pi} {\int}_{r = 0}^{1} {\left(4 {r}^{2} + 1\right)}^{\frac{1}{2}} \left(r\right) \mathrm{dr} d \theta$

$= {\int}_{\theta = 0}^{2 \pi} \frac{{\left(4 {r}^{2} + 1\right)}^{\frac{3}{2}}}{12} {|}_{r = 0}^{1} d \theta$

$= {\int}_{\theta = 0}^{2 \pi} \frac{5 \sqrt{5} - 1}{12} d \theta$

$= \frac{5 \sqrt{5} - 1}{12} \theta {|}_{\theta = 0}^{2 \pi}$

$= \frac{5 \sqrt{5} - 1}{6} \pi$