Find the mean and the variance of an F random variable with $r_1$ and $r_2$ degrees of freedom. 
First find $E(U), E(\frac{1}{V}), E(U^2),E(\frac{1}{V^2})$.

When I consider finding $E(U)$ I feel as though integrating over the pdf of the F distribution multiplied by $u$ will leave me with a spare $u$. Is there a better strategy to consider? I know that the F distribution is made up of two independent Chi-Square distributions, perhaps I should calculate the expected values separately?
 A: $\newcommand{\E}{\operatorname{E}}$
$$
F = \frac{U/r_1}{V/r_2}
$$
where $U\sim\chi^2_{r_1}$ and $V\sim\chi^2_{r_2}$ and $U,V$ are independent.
\begin{align}
& \E(F) = \overbrace{\E\left( \frac{U/r_1}{V/r_2} \right) = \E(U/r_1)\E\left( \frac 1{V/r_2} \right)}^{\text{because of independence}} \\[8pt]
= {} &  \frac {r_2}{r_1} \E(U)\E\left(\frac1V\right) = \frac{r_2}{r_1} r_1 \E\left(\frac 1 V\right) = r_2\E\left(\frac 1 V\right)
\end{align}
Here I have assumed you know that $\E(\chi^2_{r_1})=r_1$.  To find $\E\left(\frac 1 V\right)$, evaluate
$$
\int_0^\infty \frac 1 v f(v)~dv
$$
where $f$ is the $\chi^2_{r_2}$ density, i.e.
$$
\frac 1 {\Gamma(r_2/2)} \int_0^\infty \frac 1 v \left(\frac v 2\right)^{(r_2/2)-1} e^{-v/2}\frac{dv}2.
$$
Letting $w=v/2$, this becomes
\begin{align}
& \frac 1 {\Gamma(r_2/2)} \int_0^\infty \frac 1 {2w} w^{(r_2/2)-1} e^{-w}~dw \\[8pt]
= {} & \frac 1 {2\Gamma(r_2/2)} \int_0^\infty w^{(r_2/2)-2} e^{-w}~dw \\[8pt]
= {} & \frac{\Gamma((r_2/2)-1)}{2\Gamma(r_2/2)} = \frac{1}{2\left( \frac{r_2}2-1 \right)} = \frac{1}{r_2-2}.
\end{align}
In a similar way, one finds $\E(F^2)= \E\left(\frac{U^2}{V^2}\right)$.  Finally, $\operatorname{var}(F) = \E(F^2)-(\E(F))^2$.
A: In answer to your doubt:

When I consider finding E(U) I feel as though integrating over the pdf of the F distribution multiplied by u will leave me with a spare u.

Well here's the 'direct' way which I tried:
If $U\sim F\left(r_1,r_2\right)$, then,
$$\begin{align}
E(U)=&\int_0^\infty \frac{u\left(\frac{r_1}{r_2}\right)^{r_1/2}u^{\left(r_1/2\right)-1}}{B\left(\frac{r_1}{2},\frac{r_2}{2}\right)\left(1+\frac{r_1}{r_2}u\right)^{\left(r_1+r_2\right)/2}}du\\
=&\frac{\left(\frac{r_1}{r_2}\right)^{r_1/2}}{B\left(\frac{r_1}{2},\frac{r_2}{2}\right)}\int_0^\infty \frac{u^{\left(r_1/2\right)}}{\left(1+\frac{r_1}{r_2}u\right)^{\left(r_1+r_2\right)/2}}du\\
\end{align}$$
Now, substitute $\frac{r_1}{r_2}u=t$. Then,  we're getting:
$$\begin{align}
E(U)=&\frac{\left(\frac{r_1}{r_2}\right)^{r_1/2}}{B\left(\frac{r_1}{2},\frac{r_2}{2}\right)}\int_0^\infty \frac{\left(\frac{r_2}{r_1}t\right)^{r_1/2}}{\left(1+t\right)^{\left(r_1+r_2\right)/2}}\left(\frac{r_2}{r_1}\right)dt\\
=&\frac{\left(\frac{r_2}{r_1}\right)}{B\left(\frac{r_1}{2},\frac{r_2}{2}\right)}\int_0^\infty \frac{t^{r_1/2}}{\left(1+t\right)^{\left(r_1+r_2\right)/2}}dt\\
=&\frac{\left(\frac{r_2}{r_1}\right)B\left(\frac{r_1}{2}+1,\frac{r_2}{2}-1\right)}{B\left(\frac{r_1}{2},\frac{r_2}{2}\right)}\int_0^\infty \underbrace{\frac{t^{\left(r_1/2\right)+1-1}}{B\left(\frac{r_1}{2}+1,\frac{r_2}{2}-1\right)\left(1+t\right)^{\overline{\left(r_1/2\right)+1}+\overline{\left(r_2/2\right)-1}}}}_{\text{pdf of a Beta distribution}}dt\\
=&\frac{\left(\frac{r_2}{r_1}\right)B\left(\frac{r_1}{2}+1,\frac{r_2}{2}-1\right)}{B\left(\frac{r_1}{2},\frac{r_2}{2}\right)}\\\end{align}$$
Now, if I use the relationship of the Beta function with the Gamma function, we'll end up getting something like this:
$$\begin{align}
E(U)=&\left(\frac{r_2}{r_1}\right)\frac{\Gamma\left(\frac{r_1}{2}+1\right)\Gamma\left(\frac{r_2}{2}-1\right)}{\Gamma\left(\frac{r_1+r_2}{2}\right)}\frac{\Gamma\left(\frac{r_1+r_2}{2}\right)}{\Gamma\left(\frac{r_1}{2}\right)\Gamma\left(\frac{r_2}{2}\right)}\\
=&\left(\frac{r_2}{r_1}\right)\frac{\Gamma\left(\frac{r_1}{2}+1\right)\Gamma\left(\frac{r_2}{2}-1\right)}{\Gamma\left(\frac{r_1}{2}\right)\Gamma\left(\frac{r_2}{2}\right)}\\
=&\left(\frac{r_2}{r_1}\right)\left(\frac{r_1}{2}\right)\frac{1}{\frac{r_2}{2}-1}\\
=&\frac{r_2}{r_2-2}\\
\end{align}$$
This gives us the expectation of $U$ if $U\sim F\left(r_1,r_2\right)$.
Similarly, you can obtain $E\left(U^2\right)$ and ultimately find the $Var(U)=E\left(U^2\right)-(E(U))^2$.
I would still recommend Michael Hardy's method as that is far more efficient than this one.
P.S. Pardon me for any formatting errors that might've crept while writing the answer.
