# Geometrically, why is this surface integral $0$?

Question is in Paul's Math Online Notes Q3: http://tutorial.math.lamar.edu/Classes/CalcIII/SurfaceIntegrals.aspx

It says: Evaluate $\int \int_S y \: dS$ where $S$ is the portion of the cylinder $x^2 + y^2 = 3$ between $z=0$ and $z=6$.

I know that this integral calculates the surface area of this cylinder weighted by the function $f(x,y,z) = y$, but why does this equal zero? I don't see any parts of the surface that would cause any negative values to cancel the positive ones.

Fix a value of $0 < z < 6$ and imagine you walk around the circle $x^2 + y^2 = 3$. At every location during your walk you add the value of $y$, eventually you will get to the other side of the circle where you must add the value the same value but now with oposite sign. So on that particular value of $z$ the integral goes to zero, now repeat for all values of $z$, the result is also zero!
• Would this only work with odd powers of $y$ (and $x$)? – Twenty-six colours May 7 '17 at 12:58
• Works for odd functions of $y$ (and $x$) – caverac May 7 '17 at 13:00