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Math 55 chapter 9
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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