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I'm struggling with surface integrals, and I still do not have much confidence with the parameterization of functions. This is the exercise I would like to solve:

Calculate the surface integral of the function

$$f = (x-1)^2 + (y-2)^2$$

extended to the surface $P \equiv (1+\rho \cos \vartheta, 2 + \rho \sin \vartheta, 4 - \rho^2)$ with $\vartheta \in [0, 2\pi], \rho \in [\sqrt{2}, \sqrt{3}]$

without going into detail in the steps and in the final result, how should I proceed? The first thing I should do is find the parameterization of the surface, right? And can I retrieve it?

Thanks in advances

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user3204810
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1 Answers1

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Your surface $S$ is already given in a parametrized form, i.e. your parametrization is $$ P(u,v) = (1+u\cos v, 2+u\sin v, 4-v^2), \quad u\in [\sqrt 2, \sqrt 3], \quad v \in [0,2\pi]. $$ Now you should compute your surface area element $\mathrm dA$, which is defined as $$ \mathrm dA = |\partial_u P \times \partial_v P| \, \mathrm du \, \mathrm dv, $$ and the sought integral then becomes $$ \int_S f \, \mathrm dA = \int_{0}^{2\pi} \int_{\sqrt 2}^{\sqrt 3} f(P(u,v)) |\partial_u P \times \partial_v P| \, \mathrm du \,\mathrm dv. $$ Can you proceed?

MSDG
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