Module 11 - Integration 4 (self Study)

SCHOOL OF MATHEMATICS MATHEMATICS FOR PART I ENGINEERING Self Study Course

MODULE 11

INTEGRATION IV

Module Topics 1. Double integration 2. Polar coordinates 3. Triple integration

A:

Work Scheme (if you use either edition of JAMES)

You have all integrated functions of a single variable, y = f (x) say. In the real world, however, most physical quantities depend on more than one variable and therefore it is important in many branches of engineering to be able to integrate these more general functions. The so-called multiple integrals are not covered in J. and so you should study the sections below. 1. Recall that in Module 3 we defined the definite integral of a function of one variable, f (x), by the limit Z

b

f (x) dx = lim lim

δxi →0 n→∞

a

n X

f (xi ) δxi ,

i=1

where a = x0 < x1 < x2 < ... < xn = b, δxi = xi − xi−1 and xi−1 ≤ xi ≤ xi . This integral represents the area between the curve y = f (x) and the x-axis and between x = a and x = b, as shown in Figure 11.1. .. ........ ......... .. ... .... .. ... ... ... ... ... ... ................................... ... ......... ..... ........ ... .. ....... . . . . . ... . ... . . ...... ... ... ...... . . ............. . . ... . ... . . . ..... . . . .. ...................... . . . . ... . . ... . . . ........................... .. ... . . ... . . . .......... ... ... .... . ......... .... ... .... ............ .. .. ... ........... ... ... ... ............ ... ... ... ........... ... ... ... ............ ... ... .. . ................................................................................................................................................................................................................................................................................................................................ . ... ... i ... ... ..

y

y = f (x)

0

a

x

b

x

Figure 11.1 2. Let us now define a double integral. Consider a region R in the xy-plane bounded by a closed curve, and subdivide this into a large number (N ) of small regions δS1 , δS2 ,... δSN by means of a grid of lines parallel to the two axes at distances δx, δy apart respectively. Then, except near the boundary of S, δSi = δx δy (see Figure 11.2).

–1–

.. ........................ ..................................... ........................ . .... ................................................................................................ .. ....... ... ... ........ .. . . . .. ... .... ... ... ... ... .. ... ... ... .. . ... . . ....................... .. ..................... ... ...... ... ... ........... ... ... ... ... ........ .............................. ... ....... ... ... ... ... ... ...... ... ... . . . . . ... . ... . . . . . . . . .. . . ..... ... ... ... .. . ... ... ... ... ........ ... ... ... .. ... . ... .. ... .... ... .. ... . ... . ..... .. ..... ... ... ..... .... ... . ... ... ... ... . . . . . . . . . ... . . . . . . . . ... ... .... ... .. .............. .......... ................................. .... ... .... .. ... ... ... ............... ... . .. ............ ... ... ..................................... ... .. ............ .. ... ......................................................................................................................................................................... ... . .. . ... ... .... .... ... ... .. ... ... ... ... ... ... . ... ... ... . ... .... . ... ... .. . .. . .. . . ... ... . . . . . ... . ... ... . . ... . ... ... ... ... . .. . . . . . . ... . . .... . . . . .................. . . ... .. . . . . . ... . . ...... . . . . . ... . . . . ... . . . ... . . . . . . . . . . . . ... . . . ... . . . .. .... . ........ .. . . . . . . . . . . ... . . . . ... . . . . .......... . ........ . ..... . . . . . . . ... . . . . . .................................................................................................................................................................................................................................................................................. . . . ... . . ... ..................................................................................................... . ..... ............. . . . . . ... . . . i ............................................................................................................................................................................................................................................................... . . . ... . . .. ... ........ ... .............................. ... ....... ...... . .. ....... ....... ......... ..................... .......... ... ........... ...... .................................................................... ....... . . . . . . ... . . . . . . ..... ...... ....... ... ...... ...... ....... ....... .... ....... ...... ....... ....... .. ....... ...... ....... ...... . . . . . . ... ............. . . . . .. ... . ............................................................................................................................................................................................................................................................................................................................................................................................ ..

z

z = f (x, y)

y

δy

δS = δxδy

R

0

x

δx

Figure 11.2 Let (xi , yi ) be a point inside the ith element of area δSi and let fi = f (xi , yi ), so that fi is the height of the surface above (xi , yi ). Then fi δSi is approximately equal to the volume of a thin vertical tube of rectangular cross-section constructed on δSi as base, the upper boundary being where this tube intersects the surface. Let f1 , f2 , ..., fN be the values taken by the function f at arbitrary points inside the elements of area δS1 , δS2 , ..., δSN respectively. Let VN = f1 δS1 + f2 δS2 + ... + fN δSN ≡

N X

fi δSi .

i=1

If VN tends to a limit as δx → 0, δy → 0 and so N → ∞, the limit is known as the double integral of f over the region R and its value is equal to the volume of the vertical cylinder constructed on R as its base, the upper boundary being where this cylinder cuts the surface f . The double integral is denoted by ZZ f (x, y) dx dy . R

ZZ

In the special case when f (x, y) = 1 for all x and y,

ZZ f (x, y) dx dy is written simply as

R

and two useful interpretations are:

dxdy R

(a) it is the volume of the cylinder of uniform height 1 based on the region R or (b) it is the area of the region R. 3. To evaluate a double integral you simply have to perform two successive single integrations. The difficult problem is getting the limits right . The first step is to draw the region R in the xy-plane over which the double integral is to be evaluated. This is easy when the region is rectangular, as illustrated in the following example. Z y=1 Z x=2 (1 + x + y) dx dy. Example 1 Evaluate y=0

x=1

The region of integration in this example is rectangular, since all limits are constants and hence independent of x and y. –2–

When a double integral is written in the form displayed in Example 1, with dx as the inner infinitesimal and dy as the outer, then the function f (x, y) = 1 + x + y is first integrated with respect to x and the resulting quantity then integrated with respect to y. Although the limits in Example 1 are written in their fullest form, you will often encounter integrals written Z 1Z 2 (1 + x + y) dx dy, where you can assume from the order of the infinitesimals and integrals which limits 0

1

apply to which variables. The situation is completely clear if you  introduce an extra set of brackets. For Z 1 Z 2 instance, the integral could be written (1 + x + y) dx dy. 0

1

The region of integration for Example 1 is easy to find, but in general you proceed as follows. Draw on a diagram the lines defined by the four limits appearing explicitly in the integral, i.e. x = 1, x = 2, y = 0 and y = 1 in the current example. For the inner integral y is constant and x varies between 1 and 2. Similarly, for the second integral y has outer limits y = 0 and y = 1. The region of integration is then the shaded area in Figure 11.3, with the thin strip showing the part of the region along which the first integral is taken. y

... ..... .......... .. ... .... .. ... ... . ...................................................................................................................................................... .. ... .... .... .... .... .. ....... .... .... .... ... ... ... .......................................................................... ... .................................................................... ... .. .. ... ... ... .. ...... ...... ...... ...... ....... .... . .. .... ...... ...... ...... ...... ... ...... ..... ..... ..... .... ... ... ..... ..... ...... ........ ... .. .... ... ... .... .. .. . ..................................................................................................................................................................................................... . ... ... ... ... ... ... ... ... ...

1

0

1

2

x

Figure 11.3 Let us now evaluate the double integral. First, consider the inner integral which, as stated above, is an integration with respect to x along a strip with y kept fixed. The calculation of this inner integral proceeds as follows: Z 1

2

x=2      1 5 x2 22 + xy + 2y − 1 + + y = + y. (1 + x + y) dx = x + = 2+ 2 2 2 2 x=1

Note that in evaluating the above integral y remained constant. The full calculation of the double integral then gives Z 0

1

Z 1

2

 Z (1 + x + y) dx dy = 0

1



5 +y 2





5y y 2 dy = + 2 2

1

 =

0

5 1 + 2 2

 − (0) = 3.

Integrals of the type appearing in Example 1 can have a number of interpretations. The one mentioned earlier is the calculation of volumes and for this situation you can think of the above answer, 3, as the volume of a solid body with its base on the xy-plane (i.e. z = 0), with height z = f (x, y) = 1 + x + y and its sides being the planes x = 1, x = 2, y = 0 and y = 1. Retaining the volume interpretation, the volume in Example 1 was calculated above by dividing it into thin slices parallel to the x-axis (i.e. parallel to the xz-plane). The volume of one of these slices was calculated through the first integral, over the variable x, and then the second integral is equivalent to adding together the volumes of all the thin slices to give the total volume of the solid. It is not difficult to see that the total volume could also have been divided into slices parallel to the y-axis (or yz-plane). In this case the first integral would give the volume of the thin slice parallel to the y-axis, with the second integral providing the sum of the volumes of these slices. Hence, a given double integral can also be evaluated, at least in principle, by changing the order of integration. During this process the limits will normally need to be changed too. –3–

When the first integral in Example 1 is with respect to y, the picture for the region of integration becomes that shown in Figure 11.4. y

... ...... .......... .... ... ... .. ...... ............................................................................... ... ...... ......... ..... ..... ... .... .. . . .. . ... ...... ........... .......... ....... ........ ... . . .. .. .. ... ........... ............. ...... ...... ... ... ... .. . ... .... ... ... . ... ... ...... ............... ...... ........... ... ........ ....... ... ..... ..... ... .. ... .. ...... .... ..... ... . ... .... ...... ............. ...... ....... .... . ................................................................................................................................................................................................. . .. ... ....

1

0

1

x

2

Figure 11.4 The function being integrated (i.e. the height of the solid if it is a volume being calculated) is unaltered. So the double integral can also be expressed Z

Z

2



1

(1 + x + y) dy dx. 1

0

Note the order of dx and dy, and the limits of integration, have been changed. The limits are normally determined from the figure. For this new double integral x is fixed in the inner integral and from the figure y varies along the thin strip between 0 and 1. The calculation of the integral now proceeds as follows: Z

2

Z



1

Z

2

(1 + x + y) dy dx = 1

0

1

Z

2

= 1





y2 y + xy + 2

y=1

Z dx =

y=0

1

2

   1 − 0 dx 1+x+ 2

  2   3x x2 3 3 1 + x dx = + + = 3. = (3 + 2) − 2 2 2 1 2 2

The final answer, obviously, must be the same as the value calculated earlier, but the details of the intermediate steps are usually very different. Changing the order of integration can be very useful in practice, since it is often easier to perform the calculation one way than the other. As you will see later, in some cases it is only possible to evaluate the integral one way. ***Do Exercise A Evaluate directly Z 2Z 1 (i) (y + ex ) dx dy, (ii) 0

0

***Do Exercise B

Z 0

π/2Z π/4 0

sin(2x − y) dx dy.

Evaluate Exercise A(i) after changing the order of integration.

4. As seen in Section 3 the evaluation of double integrals is relatively straightforward when the region of integration, R, is rectangular. Consider now the more usual situation in which R is non-rectangular. Suppose R is bounded by the two curves y = g1 (x) and y = g2 (x) which intersect where x = x1 and x = x2 .

–4–

... ....... ......... .... ... ... ............. ... ............. ........ ............ ... ............ .... .. . . . . . . . . . . ... . . . . . . ... . . ..... ........... .... .... ... 2 ...... ........... . . . . . . . . . . . ... . . ... . . . . . .. . ... .......... ... .. .... ... .............. ......... . . . . . . . . . . . ... . . . . . . . ...... ...... . . . . . . . . . . . . ... . . ... . . . . . ....... . ...... . . . . . . . . . . . . . . . ... . . . . . . . ... . . .................................................................. ... ... ... ... 1 . ... . ... . ... ... ... ... ... ... ... ... ... ... ... .. ... ... . ... ... ... ... . . . . . . ................................................................................................................................................................................................................................................................................................................................................................................................... . .. .... 1 2 .

y

B

y = g (x)

A R y = g (x)

x

x

0

x

Figure 11.5 For a given value of x, the variable y can take all values between g1 (x) and g2 (x). The region R is therefore that for which ) g1 (x) ≤ y ≤ g2 (x) (1) x1 ≤ x ≤ x2 Consider now the double integral

ZZ f (x, y) dx dy , R

which can represent, for instance, the volume under the surface z = f (x, y) above the region R. With the region described by equation (1), the integral can be written with the integration with respect to y being carried out first, followed by integration with respect to x, as: ) Z x=x2 (Z y=g2 (x) I= f (x, y) dy dx . (2) x=x1

y=g1 (x)

In other words, for a fixed x in the range x1 ≤ x ≤ x2 we evaluate the area AP QB in Figure 11.6: Z y=g2 (x) f (x, y) dy , AreaAP QB = y=g1 (x)

which is clearly a function of x only. Multiply this by dx to give the contribution to the volume from the thin slice AP QBA0 P 0 Q0 B 0 , and then integrate with respect to x to add the contributions from the various slices, so giving the total volume. .. ....... .......... ... 0 .... .................................................. .. ............................. . . . .......... . . . . . . . . . . ... . . . . . .......... .... ... .......... ...... .. ... ........... ................. ...... ........... ...... ... .. .. ...... ........... ... .... .......... .. ................... ...... .......... . . . . . . . . . . . . . . ... .................................. . . . . . . . . . ... ...... ...... . . . . . . . . . . . . . . . . . . . ... . . . . . . . ... . . . ...... .... .... ... ...... . . . . . . . . . . . . . . . . . . . ... . . . . . . ... . . . . .... .............. ...... ...... ... .... ... .......... ...... ............. ...... . . . . . . . . . . ..... . . . . . . . ... . . . . ... . . . . . ....... .... ..... . . . . . . . .. ........ . . . . . . . ... . . . . . ... . . . . . . ... . ...... .... ........ ... . . . . . . . . . .. . . . . . . ... . . . . . . ... . . . . ....... ...... ....................... . . . . . .. . . . . . ... . . ... . . . . . . . . ... ........................................ . . .. . . ... ... . . . . . ..... . . . . ... . ... .... . .... 0 .... .... ... ... ... .... .... 0 . . . .. ... ... . ... . . .. .... .... ....... . .. ... ... ... ... . . . . ........ .... ............ . . . . .. . . . . ... . ... ... ... . . .... . .... ......... ... . . . . . .. . . . . . . ... . ... ... ... . . . . . .... ... .... . . . . . . . . .. ... .... ... ... . . . ... . ................... . . . . . .... .. .... .... . . . . . . . . . . . . . . . . . . ... . . . .... .... . . . . . ... .............. . ..... . . . . .... .... ... . . . . . . ... . . . . . ... ...... .... . . ...... . . . . . . .. .. . . . . . . . . . . . . . . ... . . . . . . . . . . . .... .. ... ...... .............. ...... ... ....... .... .... ............... ..... ....... ............... ............ ... . .... ..... ...... ....... ................ .... ...... ...... .. ....... ........... ... .................. ... ................... ..................................... ....... ... ............ .... ............. ........................................ ....... . . . . . . . . . ... . ................. . . . . . . . .............. .. . .. .............. ... ... . ... ....... ... ........................................ ....... ....... ....... . ....... ....... ....... ... 0 ...... ...... . . ....... ....... ....... . . . . . . . . . . . . . . . ... . . . . . . . . . . ..... ...... ..... ...... ....... ... ....... ....... ....... ....... ....... .... ....... ....... ....... ....... ...... ....... ....... ....... ....... .. ....... ...... ...... ...... ...... . . . . ...... . . . . . . . . . . . . . . . . ... . . . . . . . . . . . . . ...... ...... ....... ....... ....... ... ...... ....... ...... ...... ...... ... ...... ....... ...... ....... ....... ...... ...... ...... ....... ... ............. ....... ...... ...... ...... . . . . . . . . . . ....... . . . . . . . . . . . . . .................................................................................................................................................................................................................................................................................................................................................................................... .

z

Q

P

P

y

A

0

x1

x

x + δx

B

Q

z = f (x, y)

B

y = g2 (x)

y = g1 (x)

A

x2

Figure 11.6 –5–

x

As in Section 3 the order of integration can be changed, but much more care is now needed to obtain the correct limits of integration. Consider again the region R in Figure 11.5. This region can equally well be specified by the inequalities ) h1 (y) ≤ x ≤ h2 (y) y1 ≤ y ≤ y2 where the equations y = g1 (x) and y = g2 (x) have been solved for x to give x = h2 (y) and x = h1 (y) respectively, see Figure 11.7. The actual curves are, of course, the same as before. The slices are now taken parallel to the xz-plane and the picture of the region of integration, the xy-plane, becomes .. ...... .......... ... .... . 2 ....... .... .... .... .... .... .... .... .... .... .... .... .... .... .... .... .... .... .... .... ....................................... . ... ... .......... ... ... ........................................ .................................................. .... 1 ......... . . . . . . . . . . .. . . . . . ......... ...... ... ........ ...... ... ........ ....... ........ ......... ... ....... .......... . . . . . . . . . . . . . . . . . ... . .... ............ 1 ........ .... .... .... .... .... .... ........................................... 2 .... ... ... ... ... ... ... ... ... . .. .......................................................................................................................................................................................................................................................................................................................... . .... .. .... ... .

y

y

x = h (y) R

y

x = h (y)

0

x

Figure 11.7

It follows that the double integral may be rewritten Z

y=y2

Z

x=h2 (y)

f (x, y) dx dy,

I= y=y1

x=h1 (y)

in which the integration with respect to x is performed first , keeping y constant and then, after filling in the limits in x, carrying out the integration with respect to y. The procedures for calculating a double integral in a non-rectangular region are illustrated in the following Examples. Example 2 The volume of a solid body of triangular cross-section with height z = xy 2 and bounded by the planes z = 0, x = 0, y = 0 and y = 1 − x is given by Z

x=1

Z

y=1−x

V = x=0

xy 2 dy dx .

y=0

Sketch the region in which the body intersects the xy-plane and evaluate the integral. Check the result by reversing the order of integration. Draw the lines x = 0, x = 1, y = 0 and y = 1 − x as in Figure 11.8, noting that for the purposes of finding this region the integrand xy 2 is irrelevant.

–6–

y

. ... .... ............. .... .... .. ... .. ... .... .. ... ..... ...... ... ... ..... ... ... ....... ... ... ..... .... ... .... ........ ... ..... . . .. . ... . . ... ....... ...... ... .... . . ... ..... ... . .. .... . . . ... ......... ... .. . ... . . . . ... . . .. ..... .. . . . . . ... . ... .... . .... . . ... ... ...... . . ............ ... . . . ... ... ..... . . .. .. ... ... ..... . . . . ... ...... . . .. .. ... ... ......... . . . . ... . . ... . . ........ ..... ... . . . ... ...... . .... .... ....... ... . . . . . . ... ...... . . .. .... .... .. ..... ... . . . . . . . . . ....... . . .......................................................................................................................................................................... . . .... ...... .... ... ..... ... ... .. ..

1

y =1−x

x=1

x

1

0

Figure 11.8 To evaluate V we first look at the inner integral. Keeping x constant, we see: Z

y=1−x

y=0



y3 xy dy = x 3 2

y=1−x = y=0

1 x(1 − x)3 . 3

Hence it follows that Z 1 1 x(1 − 3x + 3x − x ) dx = (x − 3x2 + 3x3 − x4 ) dx V = 3 0 x=0 0     1 3 1 1 3 1 1 1 1 2 1 x − x3 + x4 − x5 = −1+ − − (0) = . = 3 2 4 5 3 2 4 5 60 0 Z

x=1

1 1 x(1 − x)3 dx = 3 3

Z

1

2

3

In order to reverse the order of integration you first have to find the appropriate limits. Simply reversing everything and writing V as Z y=1−x Z x=1 xy 2 dx dy y=0

x=0

clearly gives a nonsensical result. The correct approach is to realise that you need now to integrate with respect to x first and that this means keeping y fixed. This time, follow the line y = constant on Figure 11.9 from left to right and read off the limits in x for this y, noticing that the straight line y = 1 − x is the same as the line x = 1 − y. This procedure gives 0 ≤ x ≤ 1 − y. Now move the line from the bottom of the triangle to the top and notice that in the process it scans the whole triangle when y runs from 0 to 1. y

... .... ....... .... ........ ... .. .... .. ..... ........... ... ...... ... ..... ..... .... ..... ...... ... .... .... .... .... ...... ... ....... .... .... ....... ......... ... ... .... .. .... .................................................................. ................................................................. . .. . . . ... ...... .... ...... ..... .... . . . ... .... .. .... .. .... .... .... ...... ... . . .... ...... . ... .... .... .... .. ... ..... ... ... .... .... . . . . ........ . . . . . .... ... ........................................................................................................................................................................... . .... ... ... ... .... ... ...

1

x=1−y

0

x

1

Figure 11.9 The reversed order integral can thus be written Z

y=1

Z

x=1−y

V = y=0

x=0

–7–

xy 2 dx dy

and this evaluates as x=1−y Z 1 Z 1 Z y=1  1 2 2 1 2 2 x y (1 − y) dy = y dy = V = 2 y=0 0 2 0 x=0  1 Z 1 1 1 3 1 4 1 5 1 y − y + y (y 2 − 2y 3 + y 4 ) dy = = = 2 0 2 3 2 5 0 Example 3

Evaluate the integral

Z

y=1

Z

1 2 y (1 − 2y + y 2 ) dy 2    1 1 1 1 1 − + − (0) = . 2 3 2 5 60

x=y 2

V =

(x + y) dx dy, y=−1

x=0

and check the result by changing the order of integration. The integral represents the volume under the surface z = x + y which lies above the region of the xy-plane bounded by the lines y = −1, y = 1, x = 0 and x = y 2 , (the region shaded in Figure 11.10). y

... ..... .......... .. ................................................................................... . .... ... .... ....... ...... ... ........ .. ... ..... .... ....... 2 .... ........... ..... ... ............... . . ... ...... . ........ ... ..... .... . . . . . . . . . ........................... ...................... ... ....... ... ......... ... ........ .... .. .. ..... . ................................................................................................................................................................................. .... . ... ........ . ... .. .. .... ....... .. ..... .... ........ ......... ... ..... ....... . ... ..... ........ . .. .... ............ ....... ... ... ......... .... .... .... .... ...... ... .... .............. .. ..... .... .... . .. ........................................................................................... ... ... ....

1

x=y

1

0

x

−1

Figure 11.10 The inner integral may be evaluated as follows:  x=y2 Z x=y2 1 1 2 x + xy (x + y) dx = = y4 + y3 , 2 2 x=0 x=0 and hence 1      5  Z y=1  y4 1 1 −1 1 1 4 y 1 3 y +y + + − + = . dy = = V = 2 10 4 10 4 10 4 5 y=−1 −1 To change the order of integration draw lines parallel to the y-axis. The resulting strip splits into two parts and the y-integration must now also take place in two parts. y

. ....... .......... ............................................................................ . .. .. . ... ..... ...... ... .... ........ ... ..... .... .. .. .... ....... .... ....................... ... .... ... ............ . ..... . . ... ... ....... ................. ... ..... ... .. ..... ...... ..... . . . . .... ......... ........ .......... ...... ... .. . . ................................................................................................................................................................................... .... . .. ..... ........... .. .... ... ......... .. ... .. ...... ........ .... ...... .......... ... ..... ....... . .... ............ ....... ... .. ....... ...... ......... .... .. .. ......... ... ..... ... .. .. .... .............. ...... .. .... .. ..... ......................................................................................... ... ... ...

1

y=

0

√ x

1 √ y=− x

−1

Figure 11.11 –8–

x

It is apparent from Figure 11.11 that the double integral can also be expressed ) Z y=1 Z x=1 (Z y=−√x (x + y) dy + (x + y) dy dx, V = √ x=0

y= x

y=−1

where √the two parts correspond to the √ upper and lower areas: the lower area extends from y = −1 to y = − x and the upper area from y = x to y = 1. The two inner integrals can be evaluated to give Z

√ y=− x

y=−1

Z

   y=−√x   1 1 1 2 1 3 = −x3/2 + x − , (x + y) dy = xy + y = −x3/2 + x − −x + 2 2 2 2 2 y=−1 

y=1

1 (x + y) dy = xy + y 2 √ 2 y= x

y=1 √ y= x

    1 1 1 1 3/2 − x + x = −x3/2 + x + . = x+ 2 2 2 2

Hence, adding these results it follows that Z

x=1



V = x=0

3/2

−2x

1   1 4 5/2 −4 2 +1= . + 2x dx = − x + x = 5 5 5 0

In any given example, the shape of the boundary and the functions in the integrand will determine which order of integration is the simpler one to use. In some cases the functions in the integrand may mean that integration one way round is not possible. Z ***Do Exercise C

x=1

√ y= x

Evaluate the double integral x=0

***Do Exercise D

Z

(x2 + y 2 ) dy dx .

y=x

Let I denote the double integral ZZ x dx dy , ∆ y

where ∆ is the inside of the triangle bounded by the lines x = 1, y = 2 and y = x. Express I in two 3 1 different ways, and evaluate each expression in turn to show that I = − ln 2. 4 2 Z ***Do Exercise E

Evaluate 0

1

Z

1 √ x

3

ey dy dx .

5. When the region of integration is made up of arcs of circles and straight lines through the origin it is usually more convenient to use polar coordinates. With a polar representation each point in the plane is identified by its distance from the origin r (by definition r > 0) and from the angle that this position vector makes with the polar axis (usually coincident with the x-axis). The conversion formulae are x = r cos θ,

y = r sin θ.

To evaluate a double integral in polar coordinates you must 1. Express the region of integration in polar coordinates. 2. Express the integrand as a function of r and θ. 3. Write the area element in r and θ. –9–

An element of area in polar coordinates is obtained by considering the region bounded by arcs of radii r and r + δr centred on O and by radial lines making angles θ and θ + δθ with the x-axis. .. ...... .......... ... .... ...... .. .... .. .... ... .... ... ..... ...... .. ..... ... ... .... .... .... ..... . . ... . . . .. . . ... .... .. .. .. ... ..... ... .. . . . . .. ... ..... .. ... ... .. ... ... ... .... .......... ... ... .... .............. . ... . . . . . . . . ... . . . . . . . ... .. .. ....... ... ... ... .. ......... ... ............. ... .... ...... .... ... ...... ... . . . . . . ... . . . .. ..... . . . . . ... . . . ....... .... . . . . . . ... . . . . ... .. ... ... ... ............ .... .... ... .... ...... ... ... ....... .... ............ ... . . . . ... .... .......... ... .................. .... ... ... ................ ................................................................................................................................................................................................................... ......

y

rδθ

δr

δθ

θ

0

x

Figure 11.12 The shaded region is approximately a rectangle if δr and δθ are small enough and its area is therefore approximately δS = (r δθ) δr. Hence, in the limit as the area becomes smaller, the expression dS = dx dy used in earlier sections is replaced by dS = r dr dθ, and the double integral of a function g(r, θ) over a region R bounded by a closed curve lying in the plane can therefore be written ZZ g(r, θ) r dr dθ . R

The method is illustrated in the following Example. Example 4

Evaluate

ZZ I= R

y2 dx dy , x2

where R is the part of the annulus a2 ≤ x2 + y 2 ≤ b2 lying in the first quadrant and below the line y = x. The curves x2 + y 2 = a2 and x2 + y 2 = b2 are arcs of circles of radii a and b respectively centred at the origin. Hence the region of integration is that shown in Figure 11.13. ... ...... ......... .... .. ....... ... ....... ... ... ... .... .... ... ... . .. ... . .. ... ... .... ... ... ... ...... . .... . ... .... .. . . .. ... . ... .. ... . . . ... . ... ... .. . . . ... . ... ... .. . . . ... ... . ... .. . . ... . ... . ... ... ... . ... . . ... ... ... . ... . ... ... .... . ... . . ... ... ... ... . . . .. ... .. ... ... . . ... ... ... . ... . . .. ... ....... . ... .. . .. .. ... .. . ... . . . ... .. ... .. . ... . . . ... .. .. ... ...... . ... .. .. .. .. .......... . ............................................................................................................................................................................................................ .

y

y=x

R

π/4

0

a

b

x

Figure 11.13 The region R is specified in polar coordinates by 0 ≤ θ ≤ π/4 and a ≤ r ≤ b. The integrand becomes r2 sin2 θ y2 = tan2 θ, = x2 r2 cos2 θ – 10 –

and the area element is dS = dx dy = r dr dθ. Therefore, the value of the integral is found by r=b  r2 2 tan θ tan θ rdr dθ = dθ I= 2 r=a θ=0 r=a θ=0   2  Z θ=π/4  2 Z θ=π/4  b − a2 b − a2 dθ = = tan2 θ (sec2 θ − 1) dθ, using 1 + tan2 θ = sec2 θ, 2 2 θ=0 θ=0    2  2 i (4 − π)(b2 − a2 ) π b − a2 h b − a2 π/4 [tan θ − θ]0 = 1− − (0) = . = 2 2 4 8 Z

θ=π/4 Z r=b

***Do Exercise F

Z

2

θ=π/4

Calculate the area of the cardioid r = a(1 − cos θ)

0 ≤ θ ≤ 2π, using the formula

ZZ r dr dθ .

area = R

6. There are no conceptual differences between the evaluation of integrals in two-dimensions and integrals in three or more dimensions. The integration method involves writing the multiple integral as a succession of single integrals whose integration ranges generally depend on the other integration variables. Work through the calculation of a triple integral in the staightforward Example below. ZZZ Example 5 0 ≤ z ≤ c.

Evaluate B

(xy 2 + z 3 ) dx dy dz , where B is the rectangular box 0 ≤ x ≤ a, 0 ≤ y ≤ b,

To evaluate the integral you must follow a three step procedure: 1. Fix the value of one of the variables, for example z, within the domain 0 ≤ z ≤ c. In geometrical terms this choice is equivalent to slicing the box with a plane perpendicular to the z-axis. Since the region of integration in this Example is a box the slice obtained is a rectangle. .. ........ ......... ... ... ... ................................................................................................................. . . .... .... . .... .. . . . ..... ... ... ... . . . . .. ..... .... . ...... . . ... .... . ..... .... . . . . . . . . . ... . . ... . ... . . . . . . . . . ... . . . . ... .... . . . . . . . . ... . . . . ... .... . . . . . . . . . ... . . . .... .... . . . . . . . . ... . . ... . ... . . . . . . . . . . ... . . . .... .... . . . . . . . . ... . . .. ... . . . . . . . . . . ... . . . .............................................................................................................. ... .. .. .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ... . . . . . ... .... . ........... . .... ... . . . . . . . . . . ... . . . . . ............ .... ... .... ... . . . . . . . . . . . ... . . . . ... ... .......... . ....... . . . . . . . . . . . . ....... . . . ... .... . . . . . . . .. ... ... . ...... ... ...... ............................................................................................................................................................................................................... ..... . . .. .............. .... ..... ..... ...... ...... .... . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . ... ..... .. ............ .......... ...... ... ..... ..... ........... . ...... ..... ...... .... ............. .......... ...... ..... ...... . ........... .. ..... .. .......... ............... .... .......... ...... ..... . . . . . . . . . . . . ........... .. .. .. .. .. ..... ................................................................................................................................... ..... . ... ... ...... ..... ... .......... ... .......... ... ....... ... ....... ... .. ................................................................................................................. . ....... ............. ........

z

c

b

a x

y

Figure 11.14

2. Find the inequalities satisfied by a second variable, for example y, compatible with the chosen value of z. Since the slice obtained in the previous step is rectangular the limiting values of y are independent of the variable z: 0 ≤ y ≤ b. 3. Find the inequalities satisfied by the final variable, compatible with the values of y and z. Since the slice is rectangular the limiting values of x are independent of both y and z: 0 ≤ x ≤ a.

– 11 –

The calculation proceeds as follows (clearly the order of integration could be changed) x=a x2 y 2 3 + xz (xy + z ) dx dy dz = (xy + z ) dx dy dz = dy dz I= 2 B 0 0 0 0 0 x=0  y=b  Z c 2 3 Z c 2 3 Z cZ b 2 2 a y a y a b 3 3 3 + az dy dz = + ayz + abz dz dz = = 2 6 6 0 0 0 0 y=0 z=c  2 3 abz 4 a2 b3 c abc4 a b z + + . = = 6 4 z=0 6 4 ZZZ

2

Z

3

c

Z

b

Z

a

2

Z

3

c

Z

b



Note that the region of integration in the above Example is a volume, but it is not a volume which is being calculated. One interpretation of the answer is the mass of a rectangular solid box when the density of the solid is xy 2 + z 3 . Z ***Do Exercise G

3

Z

2

Z

Evaluate

1

x dx dy dz. 0

0

0

7. This module has concentrated on the techniques of evaluating double and triple integrals. Before finishing we list some of the many geometrical and physical applications of these integrals, a few of which have been discussed above. (a) The area A of a region R in the xy-plane is given by ZZ dx dy . A= R

(b) The volume V beneath the surface z = f (x, y) (> 0) and above a region R in the xy-plane is ZZ f (x, y) dx dy . V = R

(c) Let f (x, y) be the density (= mass per unit area) of a thin lamina of area R in the xy-plane. Then the total mass M of the lamina is ZZ M= f (x, y) dx dy , R

the centre of mass of the lamina has coordinates (x, y), where ZZ ZZ 1 1 x= x f (x, y) dx dy, y= y f (x, y) dx dy, M M R R the moments of inertia Ix and Iy of the lamina about the x- and y-axes respectively are ZZ ZZ 2 Ix = y f (x, y) dx dy, Iy = x2 f (x, y) dx dy, R

R

and the polar moment of inertia I0 of the lamina about the origin is ZZ (x2 + y 2 )f (x, y) dx dy. I0 = Ix + Iy = R

(d) The volume V of a three-dimensional region R is ZZZ V = dx dy dz. R

– 12 –

(e) Let f (x, y, z) be the density (= mass per unit volume) of a solid body of volume V in three dimensional space. Then the total mass M of the solid is ZZZ f (x, y, z) dx dy dz, M= V

and the centre of mass of the solid has coordinates (x, y, z), where x=

1 M

ZZZ x f (x, y, z) dx dy dz, V

y=

1 M

ZZZ y f (x, y, z) dx dy dz, V

z=

1 M

ZZZ z f (x, y, z) dx dy dz. V

Corresponding expressions for the moments of inertia of the solid could also be written down.

B:

Work Scheme based on STROUD (FIFTH EDITION)

S. contains more material for this Module than there exists in J. Turn to S. p.787 and work through Programme 23 frames 1-46. These frames include all the basic work on double and triple integrals, but do not consider changing the order of integration. Hence, after completing the stated frames in S. you should go through Work Scheme A presented in detail earlier in this Module, concentrating on the new material.

– 13 –

Specimen Test 11 1.

Evaluate the double integrals Z y=2 Z x=3 (i) xy 2 dx dy, y=1

2.

Z

1

Z

1

(ii)

y dy dx.

x=2

x2

0

A region of integration R is shaded on the diagram below y

... ...... .......... .. ... .... .. ...... .... ... .......... ... 2............................. .. ....... ...... . . . . . .... . . ..... .... ... .. ..... ....... ... ........ ..... ............. ... .... . . . . ... . . . . ... ..... ........ ...... ... ..... ...... ... ........ ..... ...... ........ ...... ..... . .. . ................................................................................................................................................................. . ....

1

x=y

y=x

0

x

1

For points (x, y) within R write down the inequalities which (i) show the limits between which y varies when x is fixed, and indicate the corresponding overall limits between which x varies; (ii) show the limits between which x varies when y is fixed, and indicate the corresponding overall limits between which y varies.

Z 3.

A double integral is defined by

x=1

Z

y=1−x2

xy dy dx .

I= y=0

x=0

(i) Sketch the region of integration. (ii) Evaluate the integral. (iii) Re-write the integral so that the integration with respect to x may be performed first. (iv) Evaluate the integral by first integrating with respect to x.

4. (i) State the expression for an element of area in terms of plane polar coordinates (r, θ). (ii) Evaluate the double integral

Z

θ=π

Z

r=a

sin θ r dr dθ. θ=0

r=0

– 14 –