Let $S$ denote the surface of the cylinder $x^2+y^2=4,-2\le z\le2$ and consider the surface integral $$\int_{S}(z-x^2-y^2)\,dS$$ How can one use geometry and symmetry to evaluate the integral without resorting to a parametrization of $S$?
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$\begingroup$Write $\displaystyle\int_S(z-x^2-y^2)\,dS = \int_Sz\,dS - \int_S(x^2+y^2)\,dS$
For the first integral, the integrand is odd with respect to $z$ and the surface $S$ is symmetric about $z = 0$. Hence, this integral is $0$.
For the second integral, the integrand is constant on the surface, so the integral is simply this constant multiplied by the surface area. The surface area of a cylinder is easy to compute. Note that $S$ is the lateral surface of the cylinder, which does not include the top or bottom circles.
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