integrals doubled

7/22/2019 integrals doubled

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4. DOUBLE INTEGRALS

4.1 Review of the Definite Integral

The area under the curve y = f(x) between x = a and x = b is given by

b

a dx)x(f. This is

illustrated by the figure below


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4.2 Iterated IntegralsThe following theorem tells us how to compute a double integral over a rectangle

Iff(x,y) is continuous onR = [a,b]x[c,d]then

==b

a

d

c

d

c

b

aR

dydx)y,x(fdxdy)y,x(fdA)y,x(f

Notice that the inner differential matches up with the limits on the inner integral

and similarly for the outer differential and limits. In other words if the innerdifferential dythen the limits on the inner integral must be ylimit of integration

and if the outer differential dxthen the limits on the outer integral must bexlimit

of integration.

How to compute the iterated integral


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dxdy)y,x(fdA)y,x(f

b

a

d

cR

= ?

We will compute the double integral by first computing

d

c

dy)y,x(f

By holding x constant and integrating with respect to y from y = ctoy = dand

then with respect toxfromx = atox = b.

Similarly, the iterated integral

dydx)y,x(fdA)y,x(f

d

c

b

aR

=

Means that we first integrate with respect to x(holding y fixed) from a tob and

then we integrate the resulting function of y with respect to y fromy = ctoy = d.

Notice that both the equations we workfrom the inside out

Example 1

Compute each of the following double integral

a) =R

2 ]2,1[x]4,2[R,dAxy6

b)

=

R

2 ]3,0[x]4,5[R,dA)y4x2(

c)

++1

0

1

2

22 dxdy))ysin()xcos(yx(

d) +2

1

1

0

2 dydx

)y3x2(

1

Example 2

Find the volume of the solid that lies under the plane 3x + 2y + z =12 and above

the rectangle { }3y2,1x0:)y,x(R = .

Exercises 4.1

1. Evaluate the iterated integral.


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2. Calculate the double integrala) { } =

R

432 1y0,3x0:)y,x(R,dA)y5yx6(

b) { } =+R

2

2

3y3,1x0:)y,x(R,dA1x

xy

c) =+R

]3

,0[x]6

,0[R,dA)yx(sinx

3. Find the volume of solid S that is bounded by the elliptic paraboloid

2 16zy2x 22 =++ , the plane x = 2 and y = 2, and the three coordinate planes.

Answer:

2. a) 21/2 b) 9 ln 2 c)122

13

4.3

Double Integrals over General regions

In the previous section we looked at double integrals over rectangular region. But

most of the regions are not rectangular so we need to look at the following doubleintegral,

Ad)y,x(fR

where R is any region.

(a)

(b)

(c)

(d)

(e)

(f)


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There two types of regions.

Type I;

A plane region R is said to be of type I if it lies between the graph of twocontinuous functions of x, that is

)}x(gy)x(g,bxa:)y,x{(R 21 =

where )x(g1 and )x(g2 are continuous on [a,b].

The double integral for both of these cases are defined in terms of iteratedintegrals as follows.

=R

b

a

)x(g

)x(g

2

1

dxdy)y,x(fdA)y,x(f

Type II;

A plane region R is said to be of type II if it lies between the graph of two

continuous functions of x, that is

)}()(,:),{( 21 yhxyhdycyxR =

where )(1 yh and )(2 yh are continuous on [c,d].

The double integral for both of these cases are defined in terms of iterated

integrals as follows.

=R

d

c

)y(h

)y(h

2

1

dydx)y,x(fdA)y,x(f

R

a b

)x(gy 2=

)x(gy 1=

x

y

Type I region

c

d

R

x

y

)y(hx 2= )y(hx 1=

Type II region


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Here are some properties of double integral that we should go over before weactually do some examples:

1) +=+R R R

dA)y,x(gdA)y,x(fdA)y,x(g)y,x(f

2) =R R

dA)y,x(fcdA)y,x(cf where c is any constant

3) If the region R can be split into two separate region R1and R2then the

integral can be written as

+=R R R1 2

dA)y,x(fdA)y,x(fdA)y,x(f

Example 3

Evaluate each of the following integrals over the given region R.

a) { } =R

3y/x yxy,2y1:)y,x(R,dAe

b) R

3 dA)yxy4( , R is the region bounded by xy = and y = x3

c) R

dA)y4x6( , R is the triangle with vertices (0,3), (1,1) and (5,3)

We can integrate the integrals in other order (i.e x followed by y or y followed by

x), although often one order will be easier than the other. In fact there will times

when it will not even be possible to do the integral in one order while it will be

possible to do the integral in the other order.

Example 4

Evaluate the following integrals by first reversing the order of integration.

a) 3

0

9

x

y3

2

3

dxdyex

b) +8

0

2

y

4

3

dxdy1x

Example 5

Find the volume of the solid that lies below the surface given by 2xy6z += and

lies above the region in the xy-plane bounded by 2xy = and 2x8y =


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Example 6

Find the volume of the solid enclosed by the planes ,10y2x4z =++ x3y = ,

0x,0z ==

Exercises 4.2

1. Evaluate the iterated integral

a) +1

0

x

0

2

dxdy)y2x( b) 1

0

e

y

y

dydxx

c) 2/

0

cos

0

sin ddre

d)

1

0

x2

x

2 dxdy)yx(

2. Evaluate the double integral

a) { } =R

23 xyx,2x0:)y,x(R,dAyx

b) { } =+R

2 xy0,1x0:)y,x(R,dA

1x

y2

c) { } =R

3y/x yxy,2y1:)y,x(R,dAe

d) R

,dAycosx R is bounded by 1x,xy,0y 2 ===

e) R

3 ,dAy R is triangular region with vertices (0,2), (1,1) and

(3,2)

f) R

,dA)yx2( R is bounded the circle with centre the origin and

radius 2

3. Find the volume of the given solid

a) Under the plane x + 2y z = 0 and above the region bounded by y

= x and y = x4

b) Under the surface z = xy and above the triangle with vertices (1,1),

(4,1) and (1,2)

c) Enclosed by the paraboloid z = x2+ 3y

2and the planes x =0 y = 1,

y = x , z = 0

d) bounded by the cylinder x2+ y

2= 1 and the plane y = z, x = 0 z = 0

in the first octant.


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4. Evaluate the integral by reversing the order of integraton

a) 1

0

x

3

y3

dydxe2

b) +1

0

1

y

3 dydx1x

c) 3

0

9

y

2 dydx)xcos(y2

d) 1

0

33

1

x

dxdy)y(sinx2

5. Find the volume of the solid in the first octant bounded by the cylinder z =9 y

2and the plane x = 2

Answer:

1) a) 9/20 b)

45

32e

9

4 2/3

c) e 1

2) a) 256/21 b) ln 2 c) e2e2

1 4

d) (1 cosx)/2 e) 147/20 f) 0

3) a) 7/18 b) 31/8 d) 1/3

4) a) 6/)1e( 9 c) 81sin4

1 5. 36

4.4 Double Integrals in Polar Coordinates

Suppose we want to evaluate the double integral R

dA)y,x(f , where R is a

region that is a disk, ring, or a portion of a disk or ring. In these case usingRectangular coordinate could be somewhat cumbersome. So we could convert the

double integral formula into one involving polar coordinates.

Recall from figure below that the polar coordinates ),r( of a point related to the

rectangular coordinates (x,y) by the equations

222 yxr += , cosrx = , sinry =

and dA can be written as dA = ddrr


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Now we need to find formula for double integral when we use polar coordinates

to define the region R.

The region R in the figure 2 will be defined by

, and )(hr)(h 21

We are now ready to write down a formula for double integral for double integralin polar form coordinates

=R

)(h

)(h

2

1

rdrd)isnr,cosr(dA)y,x(f

Example 7

Evaluate the following integrals by converting them into polar coordinates

a) R

dAxy2 , R is the portion of the region between the circles of radius 2

and radius 5 centred at the origin that lies in the first quadrant.

x

y

r

(r,)=(x,y)

x

y

Figure 2

Figure 1

dA

rd

dr

dr


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b) +

R

yx dAe22

, R is the unit circle centred at the origin.

Example 8

Find the volume of the solid bounded by the plane z = 0and the paraboloid22 yx1z =

Example 9

Evaluate +R

2 dA)y4x3( , where R is the region in the upper half plane bounded

by the circles 1yx 22 =+ and 4yx 22 =+

Exercises 4.3

1. Evaluate the given integral by changing to polar coordinate

a) R

dAxy , where R is the disk with centre the origin and radius 3

b) R

22 dA)yx(cos , where R is the region that lies above the x-axis within

the circle 9yx 22 =+

c)

R

yx dAe22

, where R is the region bounded by the semicircle

2y4x = and the y-axis

d) Rx

dAye , where R is the region in the first quadrant enclosed by the

circle 25yx 22 =+

2. Evaluate the iterated integral by converting to polar coordinate

a)

+1

0

x1

0

yx

2

22

dydxe b)

2

0

y4

y4

22

2

2

dydxyx

Answer: 1. a) 0 b) 9sin2

c) )e1(2

4

2. a) )1e(4

b)3

4


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4.5 Application of Double Integrals

Area of bounded region in the plane

Definition: The area of a closed, bounded plane region R is A = R

dA

Here we integrate the constant function f(x,y) = 1 over R

Example 1

Find the area of the region R bounded by the line y = x and the parabola y = x2in

the first quadrant.

Example 2

Find the area of the region R bounded by the line y = x + 2 and the parabola y =

x2in the first quadrant.

Surface Area

Figure 1 shows a surface S which projects onto the region R of the x-y-plane. If Sis given by the continuously differentiable function z = f(x,y) then the surface area

of S is given by

dAy

z

x

zS

R

+

+

= 1

22

x

z

y

R

S

Figure 1


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Example 1

Find the surface area of the part of the parabolic cylinder 9yz 22 =+ which lies

over rectangle R = { }3y3,2x0:)y,x(

Example 2

Find the surface area of the portion of the plane 2x + 2y + z = 8 in the first octant

Example 3

Find the surface area of the portion of the cone 222 y4x4z += that is above the

region in the first quadrant bounded by the line y = x and the parabola y = x2.

Example 4

Find the surface area of the portion of the surface y2x2z += that is above the

triangular region with vertices (0,0), (0,1) and (1,1).

Moment and Centre of Mass

Suppose the lamina occupies a region D and has density function )y,x( . We

define the moment of a particle about an x axis as the product of its mass and itsdistance from the axis. We divide D into small rectangle as in figure 2. The mass

of Rijis approximate A)y,x( ijij , so we can approximate the moment of Rijwith respect to the x-axis by

ijijij yA)y,x(

So the moment of the entire lamina about the x- axis:

dA)y,x(yA)y,x(ylimM

D

*ij

*ij

m

i

n

j

ij*

n,mx

1 1 ==

= =

Figure 2


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Similarly, the moment about the y-axis is

dA)y,x(xA)y,x(xlimM

D

*

ij

*

ij

m

i

n

j

ij*

n,my

1 1 ==

= =

As before, we define the center of mass

y,x so that m

x = My and m

y = Mx.

The physical significance is that the lamina behaves as if its entire massconcentrated at its center of mass. Thus, the lamina balances horizontally when

supported at its center of mass (see figure 3).

The coordinate

y,x of center of mass of a lamina occupying the region D and

having density function )y,x( are

x = =

D

ydA)y,x(x

mm

M 1

y = =

D

x dAyxymm

M ),(1

where the mass m is given by

=D

dA)y,x(m

Example 1:

Find the mass and the center of mass of a triangle lamina with vertices (0,0), (1,0)

and (0,2) if the density function is )y,x( = 1 + 3x + y.

Solution:

The mass of the lamina is

y

x

(1,2)

(1,0)

D

(3/8,11/16)

y = 2 - 2x

Figure 3


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++==R

x

dxdyyxdAyxm

1

0

22

0

)31(),(

= dx2

yxy3y

x22

0

1

0

2

++

=

==

1

0

1

0

32

3

8

3

xx4dx)x1(4

The moment about y-axis is

++==R

1

0

x22

0

2

y dxdy)xyx3x(dA)y,x(xM

= dx2

y

xyx3xy

x22

0

1

0

22

++

= 14

x

2

x4dx)xx(4

1

0

1

0

423 =

=

The moment about x-axis is

++==R

1

0

x22

0

2

x dxdy)yxy3y(dA)y,x(yM

= dx3

y

2

xy3

2

yx22

0

1

0

322

++

= dx3

)x22(

2

)x22(x3

2

)x22(1

0

322

+

+

=16

11

The centre of mass is at the point

m

M,

m

Mxy =

6

11,

8

3

When the density of lamina is 1 the center of mass is called the centroid of the

object.

Example 2

Find the center of mass of the lamina in the shape of region bounded by the

graphs of y = x2and y = 4 having mass density given by )y,x( = 1 + 2y + 6x

2


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Example 3

Find the centroid of the region in the first quadrant bounded by the x axis, the

parabola y2= 2x and the line x + y = 4.

Exercises 4.4

1. Use double integral to find the volume of the solid in the first octant that is

bounded by tha paraboloid 22 yxz += and the cylinder 4yx 22 =+ and the

coordinate plane

3

16:Ans

2. Find the volume of the solid whose base is the triangle in the x-y plane bounded

by the x-axis and the line y = x and x = 1 and whose top lies in the plane f(x,y) =

3 - x - yAns: (1)

3. Find the area of the region R enclosed by the parabola y = x2and the line

y = x + 2

Ans :

2

9

4. Use double integral to find the area of the region R enclosed between the curves

xy 2 = andn 3y x = 4

Ans :(11.68)

5. Use the double integral to find the area of the region R in the first quadrantbounded by 2xy = and 0xand4y ==

Ans:(16/3)

6. Use the double integral to find the area of the region R bounded by the parabolas2x2y = and 2x1y +=

Ans:(4/3)

7. Find the area of the portion of surface 2x + 2y + z = 8 in the first octant.

Ans:(24)

8. Find the surface area of the portion of the surface 2x4z = that lies above the

rectangle R in the x-y plane whose coordinate satisfies 1x0 and 4y0

Ans :

3

4

9. Find the area of the surface 22 y4x4z += that lies above the region R in the

first quadrant bounded by the parabola y = x2and the line y = x


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Ans :

6

5

10. Find the surface area of the portion of the surface z = 2x + 3y + 1 that lies above

the rectangle { }4y2,4x1),y,x(R =

Ans : ( )146 11. Use double integral to find the volume of the solid lying under the plane1y5x2z ++= and above the rectangle { }4y1,0x1),y,x(

Ans:(75/2)

12. Find the area of he portion of the surface 2yx2z += that is above the triangular

region with vertices (0,0), (0,1) and (1,1)

Ans:(1.318)

13. Find the mass and center of mass of the lamina occupied the region R in the first

quadrant bounded by the parabola 2xy = and the line y = x with density function

xy)y,x( =

Ans:(1/6, (16/7,3/4))

14. Find the center of mass of the lamina that occupies the region R which is bounded

by the parabola 2x9y = and the x-axis with density function y)y,x( =

15. Find the center of mass of a thin plate of density 3= bounded by the line y = x

and parabola y = x2in the first quadrant.

Ans :

5

2,

2

1

16. The point ( )y,x is the centroid of th eregion in the first quadrant that is bounded

above by the line y = x and below by the parabola 2xy = . Find ( )y,x

Ans :

5

2,

2

1

17. Find the center of mass of the triangular lamina with vertices (0,0), (0,1) and (1,0)

and the density function xy)y,x( =

Ans :

5

2,

5

2

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