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Let S be an octant of sphere of radius $2$ centered at the origin. Precisely we take the surface $z=f(x,y)=\sqrt{4-x^2-y^2}$ and restrict to $x\geq 0$ and $y\geq 0$ we want to compute the surface integral over S of the function $z^2$.

To solve it we can use symmetry which shows that:

$$\int(x^2)=\int(y^2)=\int(z^2)$$

But I haven’t understood why? Please geometrical explanation would be very appreciated Thank you in advance for your help!

Math
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Farouk Baly
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  • Welcome to MSE. You have to puts $ signs around the MathJax commands, or the commands won't work. $y^2$ comes out as $y^2$. – saulspatz Dec 16 '19 at 19:41
  • Would it help to write $x^2+y^2+z^2=4$ and think about interchanging the variables? – Ted Shifrin Dec 16 '19 at 19:41
  • You can simply change variables in each of them to transform the integral it into each of the other integrals. The important thing is that the constraints on each variable are effectively the same, they contribute the same to the sum of the squares and all have nonnegativity constraints. This wouldn't work if e.g. you had the upper hemisphere, which would distinguish z from x and y. – Ian Dec 16 '19 at 19:42
  • Since its a sphere, it's axially symmetrical, meaning, no matter the order of axes it remains same, that is why we can the write that statement. – Mike Karter Dec 16 '19 at 19:54

1 Answers1

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Note, with the symmetry and spherical coordinates, the surface integral is,

$$S=\int_Az^2dA=\frac13\int(x^2+y^2+z^2)dA=\frac13\int_0^{\pi/2}\int_0^{\pi/2}r^4\sin\theta d\theta d\phi=\frac{8\pi}3$$

Quanto
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