5.1 Double integrals over rectangular regions (Page 6/12)

Calculus volume 3 Page 6 / 12

Evaluate the integral $\iint_{R} x e^{x y} d A$ where $R = [0, 1] \times [0, ln 5] .$

$\frac{4 - ln 5}{ln 5}$

Applications of double integrals

Double integrals are very useful for finding the area of a region bounded by curves of functions. We describe this situation in more detail in the next section. However, if the region is a rectangular shape, we can find its area by integrating the constant function $f (x, y) = 1$ over the region $R .$

Definition

The area of the region $R$ is given by $A (R) = \iint_{R} 1 d A .$

This definition makes sense because using $f (x, y) = 1$ and evaluating the integral make it a product of length and width. Let’s check this formula with an example and see how this works.

Finding area using a double integral

Find the area of the region $R = {(x, y) | 0 \leq x \leq 3, 0 \leq y \leq 2}$ by using a double integral, that is, by integrating 1 over the region $R .$

The region is rectangular with length 3 and width 2, so we know that the area is 6. We get the same answer when we use a double integral:

A (R) = \int_{0}^{2} \int_{0}^{3} 1 d x d y = \int_{0}^{2} [{x |}_{0}^{3}] d y = \int_{0}^{2} 3 d y = 3 \int_{0}^{2} d y = 3 {y |}_{0}^{2} = 3 (2) = 6 .

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We have already seen how double integrals can be used to find the volume of a solid bounded above by a function $f (x, y)$ over a region $R$ provided $f (x, y) \geq 0$ for all $(x, y)$ in $R .$ Here is another example to illustrate this concept.

Volume of an elliptic paraboloid

Find the volume $V$ of the solid $S$ that is bounded by the elliptic paraboloid $2 x^{2} + y^{2} + z = 27,$ the planes $x = 3$ and $y = 3,$ and the three coordinate planes.

First notice the graph of the surface $z = 27 - 2 x^{2} - y^{2}$ in [link] (a) and above the square region $R_{1} = [−3, 3] \times [−3, 3] .$ However, we need the volume of the solid bounded by the elliptic paraboloid $2 x^{2} + y^{2} + z = 27,$ the planes $x = 3$ and $y = 3,$ and the three coordinate planes.

This figure consists of two figures marked a and b. In figure a, in xyz space, the surface z = 20 minus 2x2 minus y2 is shown for x and y from negative 3 to positive 3. The shape looks like a sheet that has been pinned at the corners and forced up gently in the middle. In figure b, in xyz space, the surface z = 20 minus 2x2 minus y2 is shown for x and y from 0 to positive 3. The surface is the upper corner of the figure from part a, and below the surface is marked the solid S. — (a) The surface $z = 27 - 2 x^{2} - y^{2}$ above the square region $R_{1} = [−3, 3] \times [−3, 3] .$ (b) The solid S lies under the surface $z = 27 - 2 x^{2} - y^{2}$ above the square region $R_{2} = [0, 3] \times [0, 3] .$

Now let’s look at the graph of the surface in [link] (b). We determine the volume V by evaluating the double integral over $R_{2} :$

\begin{matrix} V & = \iint_{R} z d A = \iint_{R} (27 - 2 x^{2} - y^{2}) d A \\ = \int_{y = 0}^{y = 3} \int_{x = 0}^{x = 3} (27 - 2 x^{2} - y^{2}) d x d y & Convert to iterated integral. \\ = \int_{y = 0}^{y = 3} {[27 x - \frac{2}{3} x^{3} - y^{2} x] |}_{x = 0}^{x = 3} d y & Integrate with respect to x . \\ = \int_{y = 0}^{y = 3} (64 - 3 y^{2}) d y = 63 y - {y^{3} |}_{y = 0}^{y = 3} = 162 . \end{matrix}

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Find the volume of the solid bounded above by the graph of $f (x, y) = x y sin (x^{2} y)$ and below by the $x y$ -plane on the rectangular region $R = [0, 1] \times [0, π] .$

$\frac{π}{2}$

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Recall that we defined the average value of a function of one variable on an interval $[a, b]$ as

f_{ave} = \frac{1}{b - a} \int_{a}^{b} f (x) d x .

Similarly, we can define the average value of a function of two variables over a region R . The main difference is that we divide by an area instead of the width of an interval.

Definition

The average value of a function of two variables over a region $R$ is

f_{ave} = \frac{1}{Area R} \iint_{R} f (x, y) d x d y .

In the next example we find the average value of a function over a rectangular region. This is a good example of obtaining useful information for an integration by making individual measurements over a grid, instead of trying to find an algebraic expression for a function.

Calculating average storm rainfall

The weather map in [link] shows an unusually moist storm system associated with the remnants of Hurricane Karl, which dumped 4–8 inches (100–200 mm) of rain in some parts of the Midwest on September 22–23, 2010. The area of rainfall measured 300 miles east to west and 250 miles north to south. Estimate the average rainfall over the entire area in those two days.

A map of Wisconsin and Minnesota that shows many cities with numbers affixed to them. The highest numbers come in a narrow band, and the map is colored accordingly. The map effectively looks like a contour map, but instead of elevations, it uses these numbers. — Effects of Hurricane Karl, which dumped 4–8 inches (100–200 mm) of rain in some parts of southwest Wisconsin, southern Minnesota, and southeast South Dakota over a span of 300 miles east to west and 250 miles north to south.

Place the origin at the southwest corner of the map so that all the values can be considered as being in the first quadrant and hence all are positive. Now divide the entire map into six rectangles $(m = 2 and n = 3),$ as shown in [link] . Assume $f (x, y)$ denotes the storm rainfall in inches at a point approximately $x$ miles to the east of the origin and y miles to the north of the origin. Let $R$ represent the entire area of $250 \times 300 = 75000$ square miles. Then the area of each subrectangle is

Δ A = \frac{1}{6} (75000) = 12500 .

Assume $(x_{i j}^{*}, y_{i j}^{*})$ are approximately the midpoints of each subrectangle $R_{i j} .$ Note the color-coded region at each of these points, and estimate the rainfall. The rainfall at each of these points can be estimated as:

At $(x_{11}, y_{11})$ the rainfall is 0.08.

At $(x_{12}, y_{12})$ the rainfall is 0.08.

At $(x_{13}, y_{13})$ the rainfall is 0.01.

At $(x_{21}, y_{21})$ the rainfall is 1.70.

At $(x_{22}, y_{22})$ the rainfall is 1.74.

At $(x_{23}, y_{23})$ the rainfall is 3.00.

Another version of the previous storm map, but this time with lines drawn for x = 100, 200, and 300 and for y = 125 and 250. There is a dot in the center of each of the resulting rectangles. — Storm rainfall with rectangular axes and showing the midpoints of each subrectangle.

According to our definition, the average storm rainfall in the entire area during those two days was

\begin{matrix} f_{ave} & = \frac{1}{Area R} \iint_{R} f (x, y) d x d y = \frac{1}{75000} \iint_{R} f (x, y) d x d y \\ ≅ \frac{1}{75,000} \sum_{i = 1}^{3} \sum_{j = 1}^{2} f (x_{i j}^{*}, y_{i j}^{*}) Δ A \\ ≅ \frac{1}{75,000} [f (x_{11}^{*}, y_{11}^{*}) Δ A + f (x_{12}^{*}, y_{12}^{*}) Δ A \\ + f (x_{13}^{*}, y_{13}^{*}) Δ A + f (x_{21}^{*}, y_{21}^{*}) Δ A + f (x_{22}^{*}, y_{22}^{*}) Δ A + f (x_{23}^{*}, y_{23}^{*}) Δ A] \\ ≅ \frac{1}{75,000} [0.08 + 0.08 + 0.01 + 1.70 + 1.74 + 3.00] Δ A \\ ≅ \frac{1}{75,000} [0.08 + 0.08 + 0.01 + 1.70 + 1.74 + 3.00] 12500 \\ ≅ \frac{5}{30} [0.08 + 0.08 + 0.01 + 1.70 + 1.74 + 3.00] \\ ≅ 1.10. \end{matrix}

During September 22–23, 2010 this area had an average storm rainfall of approximately 1.10 inches.

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Questions & Answers

Three charges q_{1}=+3\mu C, q_{2}=+6\mu C and q_{3}=+8\mu C are located at (2,0)m (0,0)m and (0,3) coordinates respectively. Find the magnitude and direction acted upon q_{2} by the two other charges.Draw the correct graphical illustration of the problem above showing the direction of all forces.

Kate Reply

To solve this problem, we need to first find the net force acting on charge q_{2}. The magnitude of the force exerted by q_{1} on q_{2} is given by F=\frac{kq_{1}q_{2}}{r^{2}} where k is the Coulomb constant, q_{1} and q_{2} are the charges of the particles, and r is the distance between them.

Muhammed

What is the direction and net electric force on q_{1}= 5µC located at (0,4)r due to charges q_{2}=7mu located at (0,0)m and q_{3}=3\mu C located at (4,0)m?

Kate Reply

what is the change in momentum of a body?

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what is a capacitor?

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Capacitor is a separation of opposite charges using an insulator of very small dimension between them. Capacitor is used for allowing an AC (alternating current) to pass while a DC (direct current) is blocked.

Gautam

A motor travelling at 72km/m on sighting a stop sign applying the breaks such that under constant deaccelerate in the meters of 50 metres what is the magnitude of the accelerate

Maria Reply

please solve

Sharon

8m/s²

Aishat

What is Thermodynamics

Muordit

velocity can be 72 km/h in question. 72 km/h=20 m/s, v^2=2.a.x , 20^2=2.a.50, a=4 m/s^2.

Mehmet

A boat travels due east at a speed of 40meter per seconds across a river flowing due south at 30meter per seconds. what is the resultant speed of the boat

Saheed Reply

50 m/s due south east

Someone

which has a higher temperature, 1cup of boiling water or 1teapot of boiling water which can transfer more heat 1cup of boiling water or 1 teapot of boiling water explain your . answer

Ramon Reply

I believe temperature being an intensive property does not change for any amount of boiling water whereas heat being an extensive property changes with amount/size of the system.

Someone

Scratch that

Someone

temperature for any amount of water to boil at ntp is 100⁰C (it is a state function and and intensive property) and it depends both will give same amount of heat because the surface available for heat transfer is greater in case of the kettle as well as the heat stored in it but if you talk.....

Someone

about the amount of heat stored in the system then in that case since the mass of water in the kettle is greater so more energy is required to raise the temperature b/c more molecules of water are present in the kettle

Someone

definitely of physics

Haryormhidey Reply

how many start and codon

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what is field

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physics, biology and chemistry this is my Field

ALIYU

field is a region of space under the influence of some physical properties

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what is ogarnic chemistry

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determine the slope giving that 3y+ 2x-14=0

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Another formula for Acceleration

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a=v/t. a=f/m a

IHUMA

innocent

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pratica A on solution of hydro chloric acid,B is a solution containing 0.5000 mole ofsodium chlorid per dm³,put A in the burret and titrate 20.00 or 25.00cm³ portion of B using melting orange as the indicator. record the deside of your burret tabulate the burret reading and calculate the average volume of acid used?

Nassze Reply

how do lnternal energy measures

Esrael

Two bodies attract each other electrically. Do they both have to be charged? Answer the same question if the bodies repel one another.

JALLAH Reply

No. According to Isac Newtons law. this two bodies maybe you and the wall beside you. Attracting depends on the mass och each body and distance between them.

Dlovan

Are you really asking if two bodies have to be charged to be influenced by Coulombs Law?

Robert

like charges repel while unlike charges atttact

Raymond

What is specific heat capacity

Destiny Reply

Specific heat capacity is a measure of the amount of energy required to raise the temperature of a substance by one degree Celsius (or Kelvin). It is measured in Joules per kilogram per degree Celsius (J/kg°C).

AI-Robot

specific heat capacity is the amount of energy needed to raise the temperature of a substance by one degree Celsius or kelvin

ROKEEB

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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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