First we plot the region [latex]D[/latex] (Figure 1); then we express it in another way.

Figure 1. The function [latex]f[/latex] is continuous at all points of the region [latex]D[/latex] except [latex](0,0)[/latex].

The other way to express the same region [latex]D[/latex] is

[latex]{D} = {\left \{{(x,y)}{:} \ {0} \ {\leq} \ {y} \ {\leq} \ {1,y^2} \ {\leq} \ {x} \ {\leq} \ {y} \right \}}[/latex]

Thus we can use Fubini’s theorem for improper integrals and evaluate the integral as

[latex]\large{\displaystyle\int_{y=0}^{y=1}\displaystyle\int_{x=y^2}^{x=y}\frac{e^y}y{dx}dy}.[/latex]

Therefore, we have

[latex]\displaystyle\int_{y=0}^{y=1}\displaystyle\int_{x=y^2}^{x=y}\frac{e^y}y{dx}dy=\displaystyle\int_{y=0}^{y=1}\frac{e^y}y{x}\bigg|_{x=y^2}^{x=y}dy=\displaystyle\int_{y=0}^{y=1}\frac{ey}y(y-y^2)dy=\displaystyle\int_0^1(ey-ye^y)dy=e-2.[/latex]

The region [latex]R[/latex] is the first quadrant of the plane, which is unbounded. So

[latex]\begin{align} \underset{R}{\displaystyle\iint}{xy}{e^{{-x^2}{-y^2}}}{dA}&=\displaystyle\lim_{(b,d)\to(\infty,\infty)}\displaystyle\int_{x=0}^{x=b}\left(\displaystyle\int_{y=0}^{y=d}{xy}{e^{{-x^2}{-y^2}}}dy\right)dx=\displaystyle\lim_{(b,d)\to(\infty,\infty)}\displaystyle\int_{x=0}^{x=b}xe^{-x^2}dx\displaystyle\int_{y=0}^{y=d}{y}{e^{-y^2}}dy \\ &=\displaystyle\lim_{(b,d)\to(\infty,\infty)}\frac14\left(1-e^{-b^2}\right)\left(1-e^{-d^2}\right)=\frac14 \end{align}[/latex]

Thus, [latex]\underset{R}{\displaystyle\iint}{xy}{e^{{-x^2}{-y^2}}}{dA}[/latex] is convergent and the value is [latex]\frac{1}{4}[/latex].

Waiting times are mathematically modeled by exponential density functions, with [latex]m[/latex] being the average waiting time, as

[latex]{f(t)}\left\{\begin{matrix} 0 & \text{if } t < 0,\\ {\frac{1}{m}}{e^{-t/m}} & \text{if } t \geq 0. \end{matrix}\right.[/latex]

If [latex]X[/latex] and [latex]Y[/latex] are random variables for ‘waiting for a table’ and ‘completing the meal,’ then the probability density functions are, respectively,

[latex]{{f_1}(x)} = \left\{\begin{matrix} 0 & \text{if } x < 0,\\ {\frac{1}{15}}{e^{-x/15}} & \text{if }x \geq 0. \end{matrix}\right. \text{and} \ {{f_2}(y)} = \left\{\begin{matrix} 0 & \text{if } y < 0,\\ {\frac{1}{40}}{e^{-y/40}} & \text{if }y \geq 0. \end{matrix}\right.[/latex]

Clearly, the events are independent and hence the joint density function is the product of the individual functions

[latex]{f(x,y)} = {f_1}{(x)}{f_2}{(y)} = \left\{\begin{matrix} 0 & \ \ \ \ \ \ \ \ \ \text{if } x < 0 \ \text{or} \ y < 0, \\ {\frac{1}{600}}{e^{-x/15}}{e^{-y/60}} & \text{if } x,y \geq 0. \end{matrix}\right.[/latex]

We want to find the probability that the combined time [latex]X+Y[/latex] is less than 90 minutes. In terms of geometry, it means that the region [latex]D[/latex] is in the first quadrant bounded by the line [latex]x+y=90[/latex] (Figure 2).

Figure 2. The region of integration for a joint probability density function.

Hence, the probability that [latex](X, Y)[/latex] is in the region [latex]D[/latex] is

[latex]{P} = {({X+Y} \ {\leq } \ 90)} = {P}{((X,Y)\in{D})} = {\iint\limits_D}{f(x,y)}{dA} = {\iint\limits_D}{\frac{1}{600}}{e^{-x/15}}{e^{-y/40}}{dA}.[/latex]

Since [latex]x+y=90[/latex] is the same as [latex]y=90-x[/latex], we have a region of Type I, so
[latex]\hspace{6cm}\begin{align} {D} &= {\left \{{(x,y)}{\mid}{0} \ {\leq} \ {x} \ {\leq} \ {90,0} \ {\leq} \ {y} \ {\leq} \ {90-x} \right \}}, \\ P(X + Y\leq90)&=\frac1{600}\displaystyle\int_{x=0}^{x=90}\displaystyle\int_{y=0}^{y=90-x}e^{-x/15}e^{-y/40}dxdy \\ &=\frac1{600}\displaystyle\int_{x=0}^{x=90}\displaystyle\int_{y=0}^{y=90-x}e^{-(x/15+y/40)}dxdy=0.8328. \end{align}[/latex]

Thus, there is an 83.2% chance that a customer spends less than an hour and a half at the restaurant.

Using the first quadrant of the rectangular coordinate plane as the sample space, we have improper integrals for [latex]E(X)[/latex] and [latex]E(Y)[/latex]. The expected time for a table is

[latex]\begin{align} E(X)&=\underset{S}{\displaystyle\iint}x\frac1{600}e^{-x/15}e^{-y/40}dA=\frac1{600}\displaystyle\int_{x=0}^{x=\infty}\displaystyle\int_{y=0}^{y=\infty}xe^{-x/15}e^{-y/40}dA \\ &=\frac1{600}\displaystyle\lim_{(a,b)\to(\infty,\infty)}\displaystyle\int_{x=0}^{x=a}\displaystyle\int_{y=0}^{y=b}xe^{-x/15}e^{-y/40}dxdy \\ &=\frac1{600}\left(\displaystyle\lim_{a\to\infty}\displaystyle\int_{x=0}^{x=a}xe^{-x/15}dx\right)\left(\displaystyle\int_{y=0}^{y=b}e^{-y/40}dy\right) \\ &=\frac1{600}\left((\displaystyle\lim_{a\to\infty}(-15e^{-x/15}(x+15)))\Bigg|_{x=0}^{x=a}\right)\left((\displaystyle\lim_{b\to\infty}(-40e^{-y/40}))\Bigg|_{y=0}^{y=b}\right) \\ &=\frac1{600}\left((\displaystyle\lim_{a\to\infty}(-15e^{-a/15}(x+15)+225)\right)\left((\displaystyle\lim_{b\to\infty}(-40e^{-b/40}+40)\right) \\ &=\frac1{600}(225)(40) \\ &=15. \end{align}[/latex]

A similar calculation shows that [latex]E(Y)=40[/latex]. This means that the expected values of the two random events are the average waiting time and the average dining time, respectively.