# Whats the difference between surface integral and volume integral?

If the line integral can be interpreted as the area under the curve $C$ on a scalar field, then the surface integral would be the volume under a surface on a scalar field right?

If so, whats the difference between a surface and volume integrals?

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How do you interprete the length of a circle $\int_{S^1} ds$ ? – Damien L Feb 4 '13 at 16:22
@DamienL I can't understand your question... isnt that a line integral? – Tiago Costa Feb 4 '13 at 16:27
It is, but you cannot interpret it has the area under the graph. – Damien L Feb 4 '13 at 16:28
@DamienL yes, you can. (after a variable transformation of course) – example Feb 4 '13 at 16:29
@DamienL From wikipedia "The line integral over a scalar field $f$ can be thought of as the area under the curve $C$ along a surface $z = f(x,y)$, described by the field." – Tiago Costa Feb 4 '13 at 16:30

## 1 Answer

Yes, you are right. The two dimensional integral $\int\! f(x,y)\,dx\,dy$ can be thought of as finding the volume to the body that is described by the function $f(x,y)$ and $z=0$ (and the integration intervals). Any one or two dimensional integral can be visualized like that: $x$ and $y$ axis for the arguments of the integrand and $z$ axis for the value of the integrand at position $x,y$. This fails for three dimensional integrals because we would need a fourth dimension to visualize the value of the integrand.

Where is the difference? Well the integrated structure has different dimensions for surface and volume integrals. The Riemannian sum corresponding to a surface integral devides the surface into small squares (or other shape) and sums the value for those squares, while the volume integrals acts on a body and devides it into small cubes (or other 3-dimensional shape) and sums the values for those cubes.

When I first worked with three dimensional integrals I found it useful to "misuse" the average of a function to imagine what the result of the integral will be (be warned that this is very subjective of course). Imagine a integral of the function $f(x,y,z)$ over some three dimensional shape $S$ with volume $V$. The average of the function $f$ over this shape is given as $$\left<f\right>_S = \frac{1}{V} \int_S\!f(x,y,z)\,dx\,dy\,dz$$ therefore the integral itself is given as the product of the average value of the function $f$ times the volume of the integrated shape $$\int_S\!f(x,y,z)\,dx\,dy\,dz = \left<f\right>_S \cdot V .$$ Maybe this helps to get a feeling for those integrals.

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