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I'm struggling with the following integral: $$\int \limits_{S_t(x)} y_1^2 + y_2^2 + y_3^2 \, dA(y),$$ where $S_t(x) = \{y \in \mathbb{R}^3: \lvert y - x\rvert = t \}$.

I understand the integral above as average value of the integrand over $S_t(x)$. However I don't know how to compute it.

I would appreciate any tips.

Hendrra
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    Have you studied the Stokes formula ? – Didier May 19 '20 at 11:25
  • @DIdier_, no. Can you provide me a link or something, please? – Hendrra May 19 '20 at 11:26
  • https://en.wikipedia.org/wiki/Stokes%27_theorem

    Here, it would say that your integral in the same as an intergral on the ball of radius $t$ and center $x$ of another function. It's the generalization of $\int_a^b f = F(b) - F(a)$

     – Didier May 19 '20 at 11:29
  • Here you can just use the Otrogradsky formulation, that is $\int_S (F\cdot n) = \int_B (\nabla\cdot F)$. You then have to find what $F$ is. – Didier May 19 '20 at 11:48
  • @DIdier_ is $F$ the antiderivative of my integrand? – Hendrra May 19 '20 at 11:58
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    No, $F$ should be a vector valued function and $n$ stands for the normal to your sphere. For simple purpose I take $x$ as the origin of the space, so $y_1^2 + y_2^2 + y_3^2 = || y||^2 = ||y||\times y \cdot n$ where $n$ is the normal to your sphere. Thus $F(y) = ||y|| \times y$ here. Edit : sorry I used the notation $F$ in two different comments but they do not refer to the same thing. – Didier May 19 '20 at 12:01

2 Answers2

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Translating the integral we get that

$$\int_{S_t(x)} y^2\:dA(y) = \int_{S_t(0)} (y+x)^2 \:dA(y)$$

Then use the fact that $dA(y) = t^2d\Omega$

$$\int_{S_t(0)} y^2+2y\cdot x + x^2 \:dA(y) = \int_{S^2} t^4 + x^2t^2\:d\Omega + \int_{S_t(0)} 2y\cdot x \:dA(y) = 4\pi t^2(x^2+t^2)$$

where the second integral vanished because the integrand was odd.

Ninad Munshi
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Alternative solution:

With substitutions $y_1 = x_1 + t\sin \phi \cos \theta$, $y_2 = x_2 + t\sin \phi \sin \theta$ and $y_3 = x_3 + t\cos \phi$ for $\phi \in [0, \pi]$ and $\theta \in [0, 2\pi)$. Then $\mathrm{d} A(y) = t^2 \sin\phi \ \mathrm{d}\phi \ \mathrm{d}\theta$.
See: http://math.mit.edu/~jorloff/suppnotes/suppnotes02/v9.pdf

We have \begin{align} I &= \int_0^\pi \int_0^{2\pi} [(x_1 + t\sin \phi \ \cos \theta)^2 + (x_2 + t\sin \phi\ \sin \theta)^2 + (x_3 + t\cos \phi)^2] t^2 \sin\phi \ \mathrm{d}\theta \ \mathrm{d}\phi\\ &= 4\pi t^2 (x_1^2 + x_2^2 + x_3^2 + t^2). \end{align}

River Li
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