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Prove that $$\iint\limits_S f(ax+by+cz)\ ds=2\pi \int_{-1}^{1} f(u\sqrt{a^2+b^2+c^2})\ du,$$ where $f(u) \in C^{(1)}$, and $S$ is $x^2+y^2+z^2=1$.

My attempt. Let $u=\dfrac{ax+by+cz}{\sqrt{a^2+b^2+c^2}}$. Then $f(u\sqrt{a^2+b^2+c^2})= f(ax+by+cz)$. But I don’t know what should I do next.

Arctic Char
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jackson
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  • For a simpler example you may try $a=1, b=c=0$, so it reduces to $\iint_S f(x) ds = 2\pi \int_{-1}^1 f(u) du$. Can you do this first? (Some sort of Fubini Theorem I suppose) – Arctic Char Dec 26 '18 at 04:00
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    Consider orthogonal mapping: $u$ be the same as you do, $v=a_1x+b_1y+c_1z$ and $w=a_2x+b_2y+c_2z$. Then use spherical coordinate. –  Dec 26 '18 at 04:41

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