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I want to evaluate the following surface integral: $\iint_{\partial\Omega}(x^3,z,y)\cdot n\ ds$ where $\partial\Omega=\{(x,y,z)\in\mathbb{R}^3:x^2+y^2+z^2=4\}$ and the normal vector is directed outside.

I parametrized the surface like this: $x=2\sin\varphi\cos\theta, y=2\sin\varphi\sin\theta, z=2\cos\varphi$ so I get as a normal vector: $-r_{\theta}\times r_{\varphi}=-(-2\sin\varphi\sin\theta,2\sin\varphi)\times (2\cos\varphi\cos\theta, 2\cos\varphi\sin\theta,-\sin\varphi)=(4\sin^2\varphi\cos\theta,4\sin^2\varphi\sin\theta,4\sin\varphi\cos\varphi)$

thus $\iint_{\partial\Omega}(x^3,z,y)\cdot n\ ds=$

$\int_{\theta=0}^{\theta=2\pi}\int_{\varphi=0}^{\varphi=\pi/2} (8\sin^3\varphi\cos^3\theta,2\cos\varphi,2\sin\varphi\sin\theta)\cdot(4\sin^2\varphi\cos\theta,4\sin^2\varphi\sin\theta,4\sin\varphi\cos\varphi)d\varphi d\theta =\int_{\theta=0}^{\theta=2\pi}\int_{\varphi=0}^{\varphi=\pi/2}(32\sin^5\varphi\cos^4\theta+16\sin^2\varphi\cos\varphi\sin\theta)d\varphi d\theta=\frac{256}{20}\pi=\frac{64}{5}\pi $.

I have two questions about this:

1) Is this correct?

2) I thought of another way to compute this; since a normal vector is also (in cartesian coordinates): $n=(x,y,z)$ I thought about using it in spherical coordinates, i.e. $n=(2\sin\varphi\cos\theta, 2\sin\varphi\sin\theta,2\cos\varphi)$ and I get for the surface integral: $\int_{\theta=0}^{\theta=2\pi}\int_{\varphi=0}^{\varphi=\pi/2} (8\sin^3\varphi\cos^3\theta,2\cos\varphi,2\sin\varphi\sin\theta)\cdot (2\sin\varphi\cos\theta,2\sin\varphi\sin\theta,2\cos\varphi)d\varphi d\theta =\frac{9\pi^2}{4}$.

Can someone explain to me why I get a different result (i.e. why are the two normal vectors I've found not equivalent for the purpose of evaluting the surface integral)?

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    $\begingroup$ Well for one you have the wrong normal for your first one as that is the inner directed normal, but that's just a matter of a minus sign $\endgroup$ – Triatticus Jan 22 '17 at 15:58
  • $\begingroup$ @Triatticus corrected, thanks. $\endgroup$ – lorenzo Jan 22 '17 at 16:04
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    $\begingroup$ I think the thing that is incorrect is the conversion of ds, it doesn't directly turn I to angular variables there should be Jacobian factors and such $\endgroup$ – Triatticus Jan 22 '17 at 16:07
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There are multiple questions. Roughly in the order they're asked:

  1. Your parametrization covers the sphere if $0 \leq \varphi \leq \pi$. As written (with $0 \leq \varphi \leq \pi/2$) you get only the hemisphere $z \geq 0$.

  2. The cross product $r_{\theta} \times r_{\varphi}$ is inward-pointing (as Triatticus notes).

    I haven't checked the numerical evaluation of your integral, but using the correct limits of integration causes more cancellation than is currently present. (The integral of $2yz$ over the entire sphere vanishes.)

  3. In your second attempt, using the Cartesian form of the normal, you've

    • Used a normal vector of length $2$ (rather than a unit normal);

    • Written $ds = d\theta\, d\varphi$, while the surface element is $$ ds = \|r_{\theta} \times r_{\varphi}\|\, d\theta\, d\varphi. $$

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