Joint PDF of $Y=\min\{{X_1,X_2,X_3}\}$ and $Z=\max\{{X_1,X_2,X_3}\}$ 
Suppose $X_1, X_2,$ and $X_3$ are independent and uniformly distributed in $[0,1]$.
(a) Find the joint PDF of $Y=\min\{{X_1,X_2,X_3}\}$ and $Z=\max\{{X_1,X_2,X_3}\}$.
  (b) Determine $E(Z|Y)$.
  (c) Verify the law of total variance in this setting by explicitly computing all terms. 

(a) Using the answer below the joint PDF is 
$$f_X,_Y(x,y)=6(z-y)$$ I understand how the marginal pdfs are obtained below, but I am still struggling with finding the bounds of integration for the joint pdf.
(b) Using the solution from part (a) I was able to find $E(Z|Y)$. 
$$E(Z|Y)=\int_{-\infty}^\infty z f_{Z|Y} (z,y)dz$$
The conditional can be expressed as the joint divided by the marginal of Y, both of which have been determined and so the integral becomes
$$=\frac{1}{3(1-y^2)}\int_y^1z(6(z-y))dz$$
$$E(Z|Y)=\frac{2}{3}+\frac{y}{3}$$
(c) I must verify the formula $$var(z)=E(var(Z|Y))+var(E(Z|Y))$$
using part (b) $var(E(Z|Y))=var(\frac{2}{3}+\frac{y}{3})=\frac{1}{9}var(Y
)$
here I am uncertain about $var(Y)$ I know that $Y=\min\{{X_1,X_2,X_3}\}$ but how do I find the var of this? For a single uniform random variable I know that the variance will be $\frac{1}{12}$ but since there are 3 $(X_1,X_2,X_3)$ should I multiple $\frac{1}{12}$ by 3? 
now for $E(var(Z|Y))$, I am not sure about how to find the $var(Z|Y)$
If someone could please help clarify my questions, it would be very helpful to me. 
 A: The following may get you started, by finding the density functions of $Y$ and $Z.$
For $Z = \max(X_1, X_2, X_3),$ we have the CDF
$$F_Z(t) = P(Z \le t) = P(X_1 \le t)P(X_2 \le t)P(X_3 \le t) = t^3,$$
for $0 <t <1.$ [If the maximum is less than $t,$ then so must each of the three independent $X_i$'s be less than $t.$] So $f_Z(t) = 3t^2,$ which is the density function of the distribution
$\mathsf{Beta}(3,1).$
Similarly, for the minimum, $F_Y(t) = (1-t)^3,$ for $0 < t < 1,$ and
$Y \sim \mathsf{Beta}(1,3).$
For illustration, the plot below shows histograms of simulations of 100,000 realizations of $Y$ and of $Z,$ along with the appropriate Beta densities.

A: I'll provided a way, might not be the most straight-forward/easy one, but should be able to solve it. Plus it starts from direct definitions, and could apply generically to statistics other than max/min as well.
$$F_{Y,Z}(y,z)=\mathbb P((Y\le y)\cap(Z\le z))$$
$$=\mathbb P(((X_1 \le y)\cup (X_2 \le y) \cup (X_3 \le y)) \cap ( (X_1 \le z) \cap (X_2 \le z) \cap (X_3 \le z)))$$
Now for the set:
$$((X_1 \le y)\cup (X_2 \le y) \cup (X_3 \le y)) \cap ( (X_1 \le z) \cap (X_2 \le z) \cap (X_3 \le z))$$
$$=\bigcup_{i \ne j \ne k \in \{1,2,3\}}((X_i \le \min(y,z))\cap(X_j \le z) \cap (X_k \le z))$$
Then use the fact that $\mathbb P(A \cup B \cup C)= \mathbb P(A) + \mathbb P(B) + \mathbb P(C) - \mathbb P (A \cap B) - \mathbb P (C \cap B) - \mathbb P (A \cap C) + \mathbb P(A\cap B \cap C)$ 
to cut them into probability of intersections of events that's from the different $X_1,X_2,X_3$, and since the three r.v. are independent, the events from $\sigma(X_1), \sigma(X_2), \sigma(X_3)$ respectively are independent, then you could break up those intersection events into multiplication, and find the distribution.
