5.2 Double integrals over general regions (Page 7/12)

Calculus volume 3 Page 7 / 12

Another important application in probability that can involve improper double integrals is the calculation of expected values. First we define this concept and then show an example of a calculation.

Definition

In probability theory, we denote the expected values $E (X)$ and $E (Y),$ respectively, as the most likely outcomes of the events. The expected values $E (X)$ and $E (Y)$ are given by

E (X) = \iint_{S} x f (x, y) d A and E (Y) = \iint_{S} y f (x, y) d A,

where $S$ is the sample space of the random variables $X$ and $Y .$

Finding expected value

Find the expected time for the events ‘waiting for a table’ and ‘completing the meal’ in [link] .

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

\begin{matrix} E (X) & = \iint_{S} x \frac{1}{600} e^{− x / 15} e^{− y / 40} d A = \frac{1}{600} \int_{x = 0}^{x = \infty} \int_{y = 0}^{y = \infty} x e^{− x / 15} e^{− y / 40} d A \\ = \frac{1}{600} lim_{(a, b) \to (\infty, \infty)} \int_{x = 0}^{x = a} \int_{y = 0}^{y = b} x e^{− x / 15} e^{− y / 40} d x d y \\ = \frac{1}{600} (lim_{a \to \infty} \int_{x = 0}^{x = a} x e^{− x / 15} d x) (lim_{b \to \infty} \int_{y = 0}^{y = b} e^{− y / 40} d y) \\ = \frac{1}{600} ({(lim_{a \to \infty} (−15 e^{− x / 15} (x + 15))) |}_{x = 0}^{x = a}) ({(lim_{b \to \infty} (−40 e^{− y / 40})) |}_{y = 0}^{y = b}) \\ = \frac{1}{600} (lim_{a \to \infty} (−15 e^{− a / 15} (x + 15) + 225)) (lim_{b \to \infty} (−40 e^{− b / 40} + 40)) \\ = \frac{1}{600} (225) (40) \\ = 15. \end{matrix}

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

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The joint density function for two random variables $X$ and $Y$ is given by

f (x, y) = {\begin{matrix} \frac{1}{600} (x^{2} + y^{2}) & if 0 \leq x \leq 15, 0 \leq y \leq 10 \\ 0 & otherwise \end{matrix}

Find the probability that $X$ is at most $10$ and $Y$ is at least $5 .$

$\frac{55}{72} \approx 0.7638$

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Key concepts

A general bounded region $D$ on the plane is a region that can be enclosed inside a rectangular region. We can use this idea to define a double integral over a general bounded region.
To evaluate an iterated integral of a function over a general nonrectangular region, we sketch the region and express it as a Type I or as a Type II region or as a union of several Type I or Type II regions that overlap only on their boundaries.
We can use double integrals to find volumes, areas, and average values of a function over general regions, similarly to calculations over rectangular regions.
We can use Fubini’s theorem for improper integrals to evaluate some types of improper integrals.

Key equations

Iterated integral over a Type I region
$\iint_{D} f (x, y) d A = \iint_{D} f (x, y) d y d x = \int_{a}^{b} [\int_{g_{1} (x)}^{g_{2} (x)} f (x, y) d y] d x$
Iterated integral over a Type II region
$\iint_{D} f (x, y) d A = \iint_{D} f (x, y) d x d y = \int_{c}^{d} [\int_{h_{1} (y)}^{h_{2} (y)} f (x, y) d x] d y$

In the following exercises, specify whether the region is of Type I or Type II.

The region $D$ bounded by $y = x^{3},$ $y = x^{3} + 1,$ $x = 0,$ and $x = 1$ as given in the following figure.

A region is bounded by y = 1 + x cubed, y = x cubed, x = 0, and x = 1.

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Find the average value of the function $f (x, y) = 3 x y$ on the region graphed in the previous exercise.

$\frac{27}{20}$

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Find the area of the region $D$ given in the previous exercise.

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The region $D$ bounded by $y = sin x, y = 1 + sin x, x = 0, and x = \frac{π}{2}$ as given in the following figure.

A region is bounded by y = 1 + sin x, y = sin x, x = 0, and x = pi/2.

Type I but not Type II

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Find the average value of the function $f (x, y) = cos x$ on the region graphed in the previous exercise.

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Find the area of the region $D$ given in the previous exercise.

$\frac{π}{2}$

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The region $D$ bounded by $x = y^{2} - 1$ and $x = \sqrt{1 - y^{2}}$ as given in the following figure.

A region is bounded by x = negative 1 + y squared and x = the square root of the quantity (1 minus y squared).

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Find the volume of the solid under the graph of the function $f (x, y) = x y + 1$ and above the region in the figure in the previous exercise.

$\frac{1}{6} (8 + 3 π)$

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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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