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As the title says, I'm looking for the marginal densities of $f (x,y) = c \sqrt{1 - x^2 - y^2}, x^2 + y^2 \leq 1$.

So far I have found $c$ to be $\frac{3}{2 \pi}$. I figured that out through converting $f(x,y)$ into polar coordinates and integrating over $drd\theta$, which is why I'm stuck on the marginal densities portion. I know that $f_x(x) = \int_{-\infty}^\infty f(x,y)dy$, but I'm not sure how to solve that without getting a big messy integral, and I know the answer isn't supposed to be a big messy integral. Is it possible to instead find $F(x,y)$, and then take $\frac{dF}{dx}$ to find $f_x(x)$? That seems like the intuitive way to do it but I can't seem to find anything in my textbook that states those relationships, so I didn't want to make the wrong assumptions.

Also, I apologize if this is inappropriate to this forum. I wasn't sure if I should ask this here or at the math stackexchange since I feel like my problem might have more to do with integration than statistics, but I might also be misunderstanding the relationship between marginal density and joint density.

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@kwak I'm not sure why changing the title was necessary...the "homework" tag should be sufficient. – Shane Oct 11 '10 at 14:10
@Shane:> ok changed back to the original. – user603 Oct 11 '10 at 15:18

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Geometry helps here. The graph of $f$ is a spherical dome of unit radius. (It follows immediately that its volume is half that of a unit sphere, $(4 \pi /3)/2$, whence $c=3/(2 \pi)$.) The marginal densities are given by areas of vertical cross-sections through this sphere. Obviously each cross-section is a semicircle: to obtain the marginal density, find its radius as a function of the remaining variable and use the formula for the area of a circle. Normalizing the resulting univariate function to have unit area turns it into a density.

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Ahh, it's sort of coming back to me from multivariable calculus. I remember doing problems like that. How do I find the radius as a function of the remaining variable? It still seems like I'm going to have some sort of monster integral left over. – Jarrod Oct 11 '10 at 22:13
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Let the remaining variable be y. Then x^2 <= 1-y^2 describes the region over which you have to integrate. Evidently the radius equals Sqrt(1 - y^2), whence the cross-sectional area equals \pi * (1 - y^2)/2. That's a pretty simple formula :-). (Remember, the theme here is geometry, not calculus...) – whuber♦ Oct 11 '10 at 22:29
Oh, right. That crossed my mind, but it seemed too simple. I guess I was determined for it to be complicated. Thanks! – Jarrod Oct 11 '10 at 22:40
I forgot to ask: how does c figure into this? – Jarrod Oct 11 '10 at 22:49
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In my opinion, Whuber's answer deserves to be upvoted for two reasons. First it answers the question asked, second as a model for how we could in the future handle (explicitly stated) homework questions: this type of answers actually contributes to the learning process and could be a better policy with respect to homework question than that adopted at MO/SO. – user603 Oct 12 '10 at 8:50
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