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How can I calculate the distribution of $X_1, $if $X_1$ denote the first coordinate of a point chosen at random from the $3$-dimensional ball of radius $\sqrt{3}$?

Someone could give me some hints pls?

Rosa Maria Gtz.
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  • The probability $X_1 = x$ is the area of the cross-section at $x$ divided by the volume of the sphere. – Doug M Oct 12 '17 at 20:53
  • The distance from the cross section to the end of the sphere is $r-y$ and $r=\sqrt{3}$ so we have

    $$\sqrt{3}-y.$$

    If $x^2 + y^2 = r^2$ then

    \begin{align} y^2 = r^2 - x^2\ y = \sqrt{r^2 - x^2}\ y = \sqrt{3 - x^2}\ \end{align} The area of a circle is determined by $\pi r^2$ so we can say the area function for a cross section that crosses the $x$-axis is

    \begin{align} A(x) = \pi \sqrt{3 - x^2}^2\ A(x) = \pi (3 - x^2)\ \end{align} Is it correct the area of the cross section?

     – Rosa Maria Gtz. Oct 12 '17 at 21:06
  • Or is just $$ P(X_1=x)= \frac{A}{V}=\frac{\pi r^2}{\frac{4\pi r^3}{3}}=\frac{3}{4r} ?$$ – Rosa Maria Gtz. Oct 12 '17 at 21:13
  • The first one is right. $P(X = x) = \frac{A}{V} = \frac {\pi r^2}{\frac {4}{3} \pi R^3} = \frac {3-x^2}{4 \sqrt 3}$ – Doug M Oct 12 '17 at 21:24
  • So, the distribution is $\frac{3-x^2}{4r}$? – Rosa Maria Gtz. Oct 16 '17 at 15:02

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