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I am trying to evaluate a surface integral

$$\int \int_S z \space dS$$

where

$S$ is the surface

$$x^2+y^2+z^2=a^2, \space z \in [0,a]$$

I am a little confused on what to do here, because although my first instinct was to use spherical coordinates I only have two integrands. If I try to use $x=a\cos(\theta), \space y=a\sin(\theta), z=z$ then the equation does not work out.

It seems as though I am missing something very basic but I cannot get my mind around it.

May I have some assistance, please?

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  • $\begingroup$ What went wrong in both cases? $\endgroup$ Commented Feb 7, 2021 at 19:32

3 Answers 3

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If you are parametrizing in spherical coordinates,

For a sphere of radius $a$, $|r'_{\theta} \times r'_{\phi}|$ is simply $a^2 \sin \phi$. If you do not remember it, you can parametrize as below and take derivative wrt $\theta$, $\phi$ and do the cross product.

$r(\theta, \phi) = (a \cos\theta \sin\phi, a\sin\theta \sin\phi, a \cos\phi)$

$|r'_{\theta} \times r'_{\phi}| = a^2 \sin \phi$

So your integral is $\displaystyle \int_0^{2\pi} \int_0^{\pi/2} a \cos\phi \cdot a^2 \sin \phi \ d\phi \ d\theta$

Similarly if you parametrize in cylindrical coordinates, it will be

$r(\rho,\theta) = (\rho \cos\theta, \rho \sin \theta, \sqrt{a^2-\rho^2})$

Now can you find $|r'_{\rho} \times r'_{\theta}|$ and integrate in cylindrical coordinates? Let me know if you have questions.

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  • $\begingroup$ Ah, thank you so much for the clarification! For some reason I was thinking that 'a' had to be a parameter and was having trouble, fitting in one more variable. $\endgroup$
    – hyg17
    Commented Feb 7, 2021 at 23:42
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I think are you on the right track to evaluating this directly, so I will instead show you a fast way to do the problem. The unit normal vector for a sphere is given by $$\frac{1}{a}(x,y,z)$$ You can either recognize this formula straight away or derive it by taking the gradient of the sphere equation and normalizing it. With this we can rewrite this scalar surface integral as a vector one:

$$\iint_SzdS=\iint_S(0,0,a)\cdot ndS$$

Because the divergence of this vector field is $0$, this integral is equivalent to the intgeral of the same vector field over any other surface that shares the same boundary and orientation as this one by Stokes' theorem. So let's change the integral to be over a disk in $xy$ plane oriented up

$$\iint_{z=0\cap x^2+y^2\leq a^2} (0,0,a)\cdot (0,0,1)dS = \pi a^3$$

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  • $\begingroup$ Have you just invoked Stokes' theorem? $\endgroup$
    – ferhenk
    Commented Feb 7, 2021 at 19:54
  • $\begingroup$ @ferhenk yea because $\nabla\cdot F = 0$ implies that there exists another vector field $A$ such that $\nabla\times A = F$. From there it's just Stokes' theorem. $\endgroup$ Commented Feb 7, 2021 at 21:21
  • $\begingroup$ It's a bit overkill for such a straight-forward exercise, isn't it? $\endgroup$
    – ferhenk
    Commented Feb 7, 2021 at 21:32
  • $\begingroup$ @ferhenk only the theory seems like much when you have to explain it, but notice that I never parametrized, took a cross product, or integrated, ever. All I used was the area of a circle formula. $\endgroup$ Commented Feb 7, 2021 at 23:03
  • $\begingroup$ I actually have to learn Stoke's theorem some time soon, so perhaps I will come back to this problem and see its application. Thank you. $\endgroup$
    – hyg17
    Commented Feb 7, 2021 at 23:44
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If you decide to go with spherical coordinates, parametrize $S$ by $F: U \to \mathbb{R}^{3}$, where $$ F(\phi, \theta) = \left(\begin{array}{c} a\cos\phi \cos\theta\\ a\cos\phi \sin\theta\\ a\sin\phi \end{array} \right) $$ for an appropriate $U \subset \mathbb{R}^{2}$ (which I leave to you to determine). Compute $F_\phi := \frac{\partial F}{\partial \phi}$ and $F_\theta := \frac{\partial F}{\partial \theta}$ as well as $$ d\sigma = \left \|F_\phi \times F_{\theta} \right\| d\phi \, d\theta, $$ which is known as surface element. Then, $$ \int_{S} z \, d\sigma = \int_{U}a \sin \phi \left \|F_\phi \times F_{\theta} \right\| d\phi \, d\theta $$ by definition of the surface integral.

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  • $\begingroup$ Thank you so much! I thought something was off in my thoughts, and it was that I was for some reason thinking that 'a' had to be a parameter. $\endgroup$
    – hyg17
    Commented Feb 7, 2021 at 23:43

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