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learning surface integrals now.
I have a surface which is composed of three surfaces, one of them is the unit circle on the x-y plane. How would I find the surface integral of this first surface (this unit circle)? Isn't it just the area of a circle? If so, how would I do this the "surface integral" method? As in, finding a parametrization for this surface and evaluating the integral?
The parametrization to me looks like $$r(u,v) = \cos u i + \sin u j + 0k$$ which, when finding the cross product of the partials, obviously evaluates to zero... meaning the integral (i.e. surface area) is zero? Why is this not correct?

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Ok, a couple of things:

  1. How would I find the surface integral of this first surface (this unit circle)? Isn't it just the area of a circle?

No, the integral depends on the function you're integrating. If the function is $1$ then the integral will be the area of the circle, but for almost any other function there is no reason for the integral to be the area of the circle. This is similar to asking in 1D integrals if computing the integral over some interval gives you its length. Well, for $f(x)=1$, $$\int_a^b f(x)dx = b-a $$ But for almost any other function this isn't correct because the function gives different weight to different parts of the interval/area. For a uniform weight of 1 the sum of the weights will be the area.

  1. About the parameterization. You used only one parameter so you didn't parameterize the whole circle but only its edge. You forgot about the "fill". What you should do is:

$$ r(u,v) = v\cos u \hat{i} + v \sin u \hat{j} + 0 \hat{k} $$

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  • $\begingroup$ Ah very nice. Thanks. I didn't realise that the length from the origin should change. While on the topic of 1., is this why the volume of a sphere (using triple coordinates) has the integrand as 1 as well? $\endgroup$ – Twenty-six colours May 7 '17 at 10:43
  • $\begingroup$ Of course, you should think about the integral of a function as a sum over a specific region of space with a different weight to different parts of it (times the 'n dimensional infinitesimal volume' - length in 1D , area in 2D, volume in 3D) , so when you add together all the 'n dimensional infinitesimal volume' with a uniform weight of 1, you just get the total area/length/volume. $\endgroup$ – Ofek Gillon May 7 '17 at 11:29

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