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)?

  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ 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

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 $$ ds = \|r_{\theta} \times r_{\varphi}\|\, d\theta\, d\varphi. $$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.