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I understand that we can use the following equation:

$$ S=\iint\limits_D {\sqrt {{({f_x})^2}+{({f_y})^2}+1}} \,dA \\$$

to calculate the surface area of a surface given by $z=f(x,y)$ bound by the 2-D region D. So (please correct me if I'm wrong) to my understanding, we can find the surface area of a 3-D shape (be it a plane, paraboloid or sphere) that is restricted by this region D.

My question is can we have a triple integral surface area equation? The reason I ask this is because I'm confused as to how I would apply cylindrical/spherical coordinates on the double integral equation if spherical/cylindrical coordinates require 3 parameters.

This question came about because I had a problem where I needed to find the surface area of a sphere which was above a cone. I did this by finding the intersection of the two shapes (which was a circle) and treating that as my region D, then applying the equation above using polar coordinates and treating the sphere as my surface.

However, I really wanted to try using spherical coordinates to find the surface area in the above question but I wasn't sure how to do that in a double integral, so I wanted to understand the theory behind the surface area equation using triple integrals first (if it exists). Any help would be appreciated, thank you!

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The triple integral of that equation would end up giving you the volume of the part of the sphere. Converting to polar coordinates and evaluating the double integral that way would be the most efficient way to solve your problem.

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Surfaces are two dimensional manifolds embedded in three dimensional space.

To evaluate the surface area you project it down to the $x-y$ plane and find a double integral because the projection region $D $ is only two dimensional.

Volumes are three dimensional objects so they require triple integrals.

I have not seem triple integrals used to find surface areas.

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If you are using spherical coordinates to find a triple integral, you are going to find that one of the variables (usually $\rho$) has no freedom to move. The upper bound equals the lower bound. If you attempted a triple integral you would get 0 volume.

In more abstract transformations you say...

$x = f(s, t)\\y = g(s,t)\\z = h(s,t)$

$dS = (\frac {\partial f}{\partial s},\frac {\partial g}{\partial s},\frac {\partial h}{\partial s})\times (\frac {\partial f}{\partial t},\frac {\partial g}{\partial t},\frac {\partial h}{\partial t}) \ ds\ dt$

And the surface integral is

$\iint \|dS\|\ ds\ dt$

Back to spherical. If you define $\rho$ as a function of $\theta, \phi$ and calculate $\|dS\|$ as described above, you get the same $\rho^2\sin\phi$ as you had as your Jacobean for a triple integral.

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