5.2 Double integrals over general regions (Page 6/12)

Calculus volume 3 Page 6 / 12

It is very important to note that we required that the function be nonnegative on $D$ for the theorem to work. We consider only the case where the function has finitely many discontinuities inside $D .$

Evaluating a double improper integral

Consider the function $f (x, y) = \frac{e^{y}}{y}$ over the region $D = {(x, y) : 0 \leq x \leq 1, x \leq y \leq \sqrt{x}} .$

Notice that the function is nonnegative and continuous at all points on $D$ except $(0, 0) .$ Use Fubini’s theorem to evaluate the improper integral.

First we plot the region $D$ ( [link] ); then we express it in another way.

The line y = x is shown, as is y = the square root of x. — The function $f$ is continuous at all points of the region $D$ except $(0, 0) .$

The other way to express the same region $D$ is

D = {(x, y) : 0 \leq y \leq 1, y^{2} \leq x \leq y} .

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

\int_{y = 0}^{y = 1} \int_{x = y^{2}}^{x = y} \frac{e^{y}}{y} d x d y .

Therefore, we have

\int_{y = 0}^{y = 1} \int_{x = y^{2}}^{x = y} \frac{e^{y}}{y} d x d y = \int_{y = 0}^{y = 1} \frac{e^{y}}{y} {x |}_{x = y^{2}}^{x = y} d y = \int_{y = 0}^{y = 1} \frac{e y}{y} (y - y^{2}) d y = \int_{0}^{1} (e y - y e^{y}) d y = e - 2 .

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As mentioned before, we also have an improper integral if the region of integration is unbounded. Suppose now that the function $f$ is continuous in an unbounded rectangle $R .$

Improper integrals on an unbounded region

If $R$ is an unbounded rectangle such as $R = {(x, y) : a \leq x \leq \infty, c \leq y \leq \infty},$ then when the limit exists, we have $\iint_{R} f (x, y) d A = lim_{(b, d) \to (\infty, \infty)} \int_{a}^{b} (\int_{c}^{d} f (x, y) d y) d x = lim_{(b, d) \to (\infty, \infty)} \int_{c}^{d} (\int_{a}^{b} f (x, y) d y) d y .$

The following example shows how this theorem can be used in certain cases of improper integrals.

Evaluating a double improper integral

Evaluate the integral $\iint_{R} x y e^{− x^{2} - y^{2}} d A$ where $R$ is the first quadrant of the plane.

The region $R$ is the first quadrant of the plane, which is unbounded. So

\begin{matrix} \iint_{R} x y e^{− x^{2} - y^{2}} d A & = lim_{(b, d) \to (\infty, \infty)} \int_{x = 0}^{x = b} (\int_{y = 0}^{y = d} x y e^{− x^{2} - y^{2}} d y) d x = lim_{(b, d) \to (\infty, \infty)} \int_{y = 0}^{y = d} (\int_{x = 0}^{x = b} x y e^{− x^{2} - y^{2}} d y) d y \\ = lim_{(b, d) \to (\infty, \infty)} \frac{1}{4} (1 - e^{− b^{2}}) (1 - e^{− d^{2}}) = \frac{1}{4} \end{matrix}

Thus, $\iint_{R} x y e^{− x^{2} - y^{2}} d A$ is convergent and the value is $\frac{1}{4} .$

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Evaluate the improper integral $\iint_{D} \frac{y}{\sqrt{1 - x^{2} - y^{2}}} d A$ where $D = {(x, y) x \geq 0, y \geq 0, x^{2} + y^{2} \leq 1} .$

$\frac{π}{4}$

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In some situations in probability theory, we can gain insight into a problem when we are able to use double integrals over general regions. Before we go over an example with a double integral, we need to set a few definitions and become familiar with some important properties.

Definition

Consider a pair of continuous random variables $X$ and $Y,$ such as the birthdays of two people or the number of sunny and rainy days in a month. The joint density function $f$ of $X$ and $Y$ satisfies the probability that $(X, Y)$ lies in a certain region $D :$

P ((X, Y) \in D) = \iint_{D} f (x, y) d A .

Since the probabilities can never be negative and must lie between $0$ and $1,$ the joint density function satisfies the following inequality and equation:

f (x, y) \geq 0 and \iint_{R^{2}} f (x, y) d A = 1 .

Definition

The variables $X$ and $Y$ are said to be independent random variables if their joint density function is the product of their individual density functions:

f (x, y) = f_{1} (x) f_{2} (y) .

Application to probability

At Sydney’s Restaurant, customers must wait an average of $15$ minutes for a table. From the time they are seated until they have finished their meal requires an additional $40$ minutes, on average. What is the probability that a customer spends less than an hour and a half at the diner, assuming that waiting for a table and completing the meal are independent events?

Waiting times are mathematically modeled by exponential density functions, with $m$ being the average waiting time, as

f (t) = {\begin{matrix} 0 & if t < 0, \\ \frac{1}{m} e^{− t / m} & if t \geq 0. \end{matrix}

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

f_{1} (x) = {\begin{matrix} 0 & if x < 0, \\ \frac{1}{15} e^{− x / 15} & if x \geq 0. \end{matrix} and f_{2} (y) = {\begin{matrix} 0 & if y < 0, \\ \frac{1}{40} e^{− y / 40} & if y \geq 0. \end{matrix}

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

f (x, y) = f_{1} (x) f_{2} (y) = {\begin{matrix} 0 & if x < 0 or y < 0, \\ \frac{1}{600} e^{− x / 15} e^{− y / 60} & if x, y \geq 0. \end{matrix}

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

The line x + y = 90 is shown. — The region of integration for a joint probability density function.

Hence, the probability that $(X, Y)$ is in the region $D$ is

P (X + Y \leq 90) = P ((X, Y) \in D) = \iint_{D} f (x, y) d A = \iint_{D} \frac{1}{600} e^{− x / 15} e^{− y / 40} d A .

Since $x + y = 90$ is the same as $y = 90 - x,$ we have a region of Type I, so

\begin{matrix} D & = & {(x, y) | 0 \leq x \leq 90, 0 \leq y \leq 90 - x}, \\ P (X + Y \leq 90) & = & \frac{1}{600} \int_{x = 0}^{x = 90} \int_{y = 0}^{y = 90 - x} e^{− x / 15} e^{− y / 40} d x d y = \frac{1}{600} \int_{x = 0}^{x = 90} \int_{y = 0}^{y = 90 - x} e^{− x / 15} e^{− y / 40} d x d y \\ = & \frac{1}{600} \int_{x = 0}^{x = 90} \int_{y = 0}^{y = 90 - x} e^{− (x / 15 + y / 40)} d x d y = 0.8328. \end{matrix}

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

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Source: OpenStax, Calculus volume 3. OpenStax CNX. Feb 05, 2016 Download for free at http://legacy.cnx.org/content/col11966/1.2

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