Applications of Double Integrals

Math 55 - Elementary Analysis III

Institute of MathematicsUniversity of the Philippines

Diliman

Math 55 Applications of Double Integrals 1/ 21

Volume of a Solid

Let f(x, y) 0 for all (x, y) in a closed and bounded regionR R2.

The volume V of the solid under the surface z = f(x, y) abovethe region R is given by

V =

R

f(x, y) dA.

Math 55 Applications of Double Integrals 2/ 21

Volume of a Solid

Example

Determine the volume of the tetrahedron bounded by the planex+ y + z = 1 and the coordinate planes.

Solution. Let S be the solid. Note that S lies under the surfacez = 1 x y and above the triangular region R in the xy-plane.

V =

R

1 x y dA

=

10

1x0

1 x y dy dx

=

10

y xy y2

2

y=1xy=0

dx

=

10

[(1 x) x(1 x) (1 x)

2

2

]dx

=

10

(x2

2 x+ 1

2

)dx =

x3

6 x

2

2+x

2

10

=1

6

Math 55 Applications of Double Integrals 3/ 21

Volume of a Solid

Example

Find the volume of the solid enclosed by the paraboloidsz = x2 + y2 and z = 8 x2 y2.

Solution. The projection of the solid onto the xy-plane is theregion bounded by the intersection of the two paraboloids as shownbelow.

Solving for the intersection,

x2 + y2 = 8 x2 y22x2 + 2y2 = 8

x2 + y2 = 4

Notice that the region R is a disk of radius 2 centered at the origin.In polar form,

R = {(r, )|0 r 2, 0 2pi}Math 55 Applications of Double Integrals 4/ 21

Volume of a Solid

Example

Find the volume of the solid enclosed by the paraboloidsz = x2 + y2 and z = 8 x2 y2.

Solution(cont). Hence, the volume is

V =

R

(8 x2 y2) dA

R

(x2 + y2) dA

=

R

8 2(x2 + y2) dA

=

2pi0

20

(8 2r2)r drd = 2pi0

20

(8r 2r3) drd

=

2pi0

(4r2 r

4

2

) r=2r=0

d

=

2pi0

8 d = 8

2pi0

= 16pi

Math 55 Applications of Double Integrals 5/ 21

Area of a Plane Region

Suppose a solid lies above R under the plane z = 1.

The volume of the solid is V = AR height = AR, the area ofR. Hence,

The area of R is

AR =

R

dA.

Math 55 Applications of Double Integrals 6/ 21

Area of a Plane Region

Example

Use double integrals to find the area of the region R boundedby x = y2 1 and x+ y = 1.

Solution. The region R is a Type II region and hence, thearea of R is

x = y2 1x + y = 1x = 1 y

R

dA =

12

1yy21

dx dy

=

12x

x=1yx=y21

dy

=

12(1 y) (y2 1) dy

=

12

2 y y2 dy

= 2y y2

2 y

3

3

12

=9

2

Math 55 Applications of Double Integrals 7/ 21

Mass and Center of Mass

Consider a lamina (thin sheet of continuously distributed mass,e.g. paper or thin metal sheet) as a region R in the xy-planeand suppose its density at any point (x, y) is (x, y), continuousR.

The mass of the lamina is given by

M =

R

(x, y) dA.

Math 55 Applications of Double Integrals 8/ 21

Mass and Center of Mass

The moment of a particle about an axis is defined to be theproduct of its mass and its directed distance from the axis.

Therefore, the moment of a lamina about the x-axis is

Mx =

R

y(x, y) dA

and the moment of a lamina about the y-axis is

My =

R

x(x, y) dA.

Math 55 Applications of Double Integrals 9/ 21

Mass and Center of Mass

The center of mass is the point (x, y) such that

x =MyM

and y =MxM

The physical significance is that the lamina balanceshorizontally when supported at its center of mass.

Math 55 Applications of Double Integrals 10/ 21

Mass and Center of Mass

Example

Find the center of mass of the triangular lamina with vertices at(0, 0), (2, 0) and (2, 2) if the density is given by (x, y) = 6xy.

Solution. Let R be the lamina. The mass is given by

1 2

1

2

0

y = x

M =

R

(x, y) dA =

R

6xy dA

=

20

x0

6xy dy dx

=

20

3xy2y=xy=0

dx

=

20

3x3 dx =3x4

4

20

= 12

Math 55 Applications of Double Integrals 11/ 21

Mass and Center of Mass

Example

Find the center of mass of the triangular lamina with vertices at(0, 0), (2, 0) and (2, 2) if the density is given by (x, y) = 6xy.

Solution (contd). Now, we consider the moments:

Mx =

R

y(x, y) dA

=

R

6xy2 dA

=

20

x0

6xy2dydx

=

20

2xy3y=xy=0

dx

=

20

2x4 dx =2

5x520

=64

5

My =

R

x(x, y) dA

=

R

6x2y dA

=

20

x0

6x2ydydx

=

20

3x2y2y=xy=0

dx

=

20

3x4 dx =3

5x520

=96

5

Math 55 Applications of Double Integrals 12/ 21

Mass and Center of Mass

Example

Find the center of mass of the triangular lamina with vertices at(0, 0), (2, 0) and (2, 2) if the density is given by (x, y) = 6xy.

Solution (contd). So far we have M = 12, Mx =64

5and

My =96

5.

Finally, we have

x =MyM

=965

12

=8

5

y =MxM

=645

12

=16

15

Hence, the center of mass is at

(8

5,16

15

).

Math 55 Applications of Double Integrals 13/ 21

Surface Area

Let S be a surface parametrized by r(u, v), whose parameterdomain D is a rectangle.

Subdivide D into subrectangles Rij , with dimensions u andv. Choose (ui , v

j ) to be the lower left corner of Rij .

Each Rij corresponds to a patch Sij on S.

Math 55 Applications of Double Integrals 14/ 21

Surface Area

Let ru = ru(ui , vj ) and r

v = rv(u

i , v

j ) be the tangent vectors,

respectively at Pij , a point given by r(ui , v

j ).

Sij can be approximated by the parallelogram determined byuru and vrv

So the area of this parallelogram is

(uru) (vrv) = uvru rv

Math 55 Applications of Double Integrals 15/ 21

Surface Area

So the area of S is approximately

mi=1

nj=1

ru rvuv.

Hence, we have the following

Definition

If S is a smooth parametric surface S given by r(u, v), coveredjust once as (u, v) ranges over the parameter domain D, thenthe surface area of S is

A(S) =

D

ru rv dA

Math 55 Applications of Double Integrals 16/ 21

Surface Area

Example

Find the area of the portion of the surface defined by the vectorfunction r(u, v) = u+ v, uv, u v, for points (u, v) satisfyingu2 + v2 = 6.

Solution. Note that ru(u, v) = 1, v, 1 and rv(u, v) = 1, u,1.Then

ru rv = v u, 2, u v ru rv =

2u2 + 2v2 + 4.Also, since D is a circular disk in the uv-plane, we use polarcoordinates, where 0 r 6, 0 2pi. Hence,

A(S) =

D

ru rv dA =

D

2u2 + 2v2 + 4 dA

=

2pi0

60

2r2 + 4 r dr d =

2pi0

(2r2 + 4)32

6

r=6

r=0

d

=

2pi0

28

3d =

28

3

2pi0

=56pi

3

Math 55 Applications of Double Integrals 17/ 21

Surface Area

Suppose S is given b z = f(x, y). Then S can be parametrizedby r(x, y) = x, y, f(x, y) and hence

rx(x, y) = 1, 0, fx(x, y) , ry(x, y) = 0, 1, fy(x, y) .

It follows that

rx ry = fx(x, y),fy(x, y), 1and hence

rx ry =

[fx(x, y)]2 + [fy(x, y)]2 + 1

and we have

A(S) =

D

[fx(x, y)]2 + [fy(x, y)]2 + 1 dA

Math 55 Applications of Double Integrals 18/ 21

Surface Area

Example

Find the surface area of the portion of the cone z =x2 + y2

in the first octant between the cylinders x2 + y2 = 1 andx2 + y2 = 4

Solution. Let f(x, y) =x2 + y2.

A(S) =

R

[fx(x, y)]2 + [fy(x, y)]2 + 1 dA

=

R

( xx2 + y2

)2+

(y

x2 + y2

)2+ 1 dA

=

pi2

0

21

2 r dr d =

pi2

0

2r2

2

r=2r=1

d

=

pi2

0

32

2d =

32

2

pi20

=32pi

4

Math 55 Applications of Double Integrals 19/ 21

Exercises

1 Find the volume of the solid enclosed by the paraboloidz = 3x2 + y2 and the planes x = 0, y = 1, y = x and z = 0.

2 The density at any point on a semicircular lamina isdirectly proportional to the distance from the center of thecircle. Find the center of mass of the lamina.

3 Find the surface area of the helicoid (spiral ramp) given byr(u, v) = u cos v+ u sin v+ vk for 0 u 1, 0 v pi.

4 Find the surface area of the portion of the plane3x 2y z + 6 = 0 that lies above the region enclosed bythe parabola y = x2 4 and the line y = x+ 2.

5 Find the area of the portion of the sphere x2 + y2 + z2 = 4zthat lies inside the paraboloid z = x2 + y2.

Math 55 Applications of Double Integrals 20/ 21

References

1 Stewart, J., Calculus, Early Transcendentals, 6 ed., ThomsonBrooks/Cole, 2008

2 Leithold, L., The Calculus 7, Harper Collins College Div., 1995

3 Dawkins, P., Calculus 3, online notes available athttp://tutorial.math.lamar.edu/

Math 55 Applications of Double Integrals 21/ 21