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This link below has a surface integral (example 4, surface S3) for which I don't know why it evaluates to zero.

http://tutorial.math.lamar.edu/Classes/CalcIII/SurfaceIntegrals.aspx

To me, surface S3 is a disk in the xy-plane, having a radius of sqrt(3), so I expected its area to be 3*pi.

Yet, the worked example for S3 yields zero, so there is something amiss with my understanding.

Thanks for reading,

Chris

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    The integral isn't $\int dA$, which would give the area. It's $\int y\,dA$. – 2017-01-12
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    I think my misunderstanding was that I regarded _any_ surface integral as ∫dA. My knowledge doesn't extend far beyond integral 101. Thank you! – 2017-01-12
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    Still, what area does that integral ∫ydA yield? How can I visualize it? – 2017-01-13
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    Integrals don't always give areas. You seem stuck in one-variable calculus where the definite integral $\int f(x)\,dx$ is introduced as the (signed) "area under the curve" $y=f(x)$. The situation changes once you move to higher dimensions. The integral $\int_R f(x,y)\,dx\,dy$ of a function $f(x,y)$ over a region $R$ in the $xy$-plane gives the (signed) *volume* under the surface $z=f(x,y)$. This is one way you can view your integral $\int y\,dA$, since your region is in the $xy$-plane. – 2017-01-13
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    I've had some (read: little) exposure to functions (and integrals) with two parameters, but nothing like the link above. So, you meant to say that surface integrals can also yield a volume over some region? – 2017-01-14
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    No. You need to read a book: no point in trying to learn this from comments on the internet. There are many kinds of surface integrals. The surface integral of a real-valued function can be thought of as giving the mass of the surface if you think of the function as the mass density per unit area. You can also integrate vector-valued functions over surfaces. These are usually interpreted as "flux" integrals. – 2017-01-14

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The center of gravity of $S_3$ is the origin. So, $\int_{S_3} y\text{d}S=0$ as well as $\int_{S_3}z\text{d}S=0$.