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Difference between double and surface integrals? Purpose of surface integrals?

  1. May 29, 2012 #1
    I'm preparing for a vector calculus course in the fall and I've been self studying some topics.

    I've taken multivariable calculus and I'm familiar with using double integrals, how to solve them and how to use them to find volume.

    From what I've read so far, I'm familiar with how to SOLVE a surface integral by projecting the surface onto the xy plane.

    This just turns it into a double integral right? And what does the double integral of this projection give you (surface area????)? I'm just trying to figure out the PURPOSE of surface integrals and how they differ from double integrals.

    Sorry if my question is silly, but I haven't taken vector calculus yet and I'm trying to get ready.
  2. jcsd
  3. May 29, 2012 #2
    What happens if the solid object you are integrating has a variable projection on the xy-plane though?

    Triple integrals are how continuum mechanics in 3 dimensions is built, so it's a crucial subject to understand.

    In any case, you can usually go back and forth double and triple integrals, using the divergence theorem or (more generally) the Stokes theorem. Keep in mind that there are applications where you can get 4 or more integrals though:wink: A great example is if the object is also moving (time is added).
  4. May 29, 2012 #3


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    A "double integral" can be used in applications that have nothing to do with "surfaces".

    It is not necessary that a surface be given in terms of a function x and y at all, For example, we can always write a surface as x= f(u,v), y= g(u,v), z= h(u,v)- that is, as parametric equations in the parameters u and v. In that case, our integral would be with respect to the parameters x and y.

    We can write any point in the surface as the vector equation [itex]\vec{r}(u,v)= x\vec{i}+ y\vec{j}+ z\vec{k}= f(u,v)\vec{i}+ g(u,v)\vec{j}+ h(u,v)\vec{k}[/itex] so that the derivatives [itex]\vec{r}_u= f_u\vec{i}+ g_u\vec{j}+ h_u\vec{k}[/itex] and [itex]\vec{r}_v= f_v\vec{i}+ g_v\vec{j}+ h_v\vec{k}[/itex] are tangent vectors and their cross product is normal to the surface.

    The integral of a vector function, [itex]\vec{F}(u,v)[/itex], on the surface, would be the dot product of [itex]\vec{F}[/itex] with that cross product, dudv. And the integral of a scalar function, f(u,v), on that surface, would be the product of f(u,v) with the length of that cross product, dudv.
  5. May 29, 2012 #4
    Surface integrals are useful if you want to "sum up" the values of a function over all the points on a surface. That is what the integral of a function over a surface represents.

    The other thing is that your surface could be more complicated than the graph of a function. Consider the surface of a donut. How are you going to project that to the xy plane in order to calculate a surface integral? You have to use a parametrization like Halls describes.

    Here is a common application. You have a vector field (such as an Electric field, a velocity field for a fluid, a Force field, among other possiblilities). One thing that is commonly calculated for various reasons is the flux of that vector field across a given surface. If the field is the velocity field V of a fluid, then the flux represents the total amount of fluid crossing the surface per unit time. That is calculated by a surface integral.

    Finally, surface integrals have a meaning independent of double integrals. Just because you use double integrals to calculate surface integrals doesn't make that an essential feature of surface integrals. If anything double integrals are a special case of surface integrals where the surface is a flat region in the plane.
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