Triple integrals
Multiple Integrals
Advanced Calculus Lecture 3 Dr. Lahcen Laayouni Department of Mathematics and Statistics McGill University
January 11, 2007
Dr. Lahcen Laayouni
Advanced Calculus
Multiple Integrals
Triple integrals Change of variables Cylindrical coordinate
Triple integrals
Triple integrals Triple integrals are just like double integrals, but with another integration to perform. Let f (x, y, z) be a bounded function over a rectangular parallelepiped R , a ≤ x ≤ b ,
PARRALELEPIPED
c≤y ≤d , p≤z≤q .
then the triple integral of f over R ZZZ f (x, y, z)dV or R
ZZZ
f (x, y, z)dxdydz , R
can be defined as the limit of Riemann sums corresponding to partitions of R into sub-parallelepiped by planes parallel to each of the coordinate planes. Dr. Lahcen Laayouni
Advanced Calculus
Multiple Integrals
Triple integrals Change of variables Cylindrical coordinate
Triple integrals
Proposition Suppose that f (x, y, z) is a continuous function on the rectangular parallelepiped R , a ≤ x ≤ b , c ≤ y ≤ d , p ≤ z ≤ q , then ZZZ Z b Z d Z q f (x, y, z)dV = dx dy f (x, y, z)dz . R
a
c
p
Remark There are six different ways of evaluating the triple integral (1), depending upon which variable is chosen first. Example Find the integral of f (x, y, z) = x 4 + 2yz over the parallelepiped R given by 0 ≤ x ≤ 1 , −1 ≤ y ≤ 2 , −2 ≤ z ≤ 1 Dr. Lahcen Laayouni
Advanced Calculus
(1)
Multiple Integrals
Triple integrals Change of variables Cylindrical coordinate
Triple integrals
Solution Z 1 Z I = dx 0
= =
Z Z
1
dx 0 1 0
Z
2
dy −1 2 −1
(9x 4 −
Proposition
Z
1
(x 4 + 2yz)dz =
1
y =2 3 3x 4 y − y 2 2
dx
0
−2
(3x 4 − 3y)dy = 9 )dx = 2
Z
Z
1
dx 0
Z
2
−1
z=1 dy (x 4 z + yz 2 )
z=−2
y =−1
9 5 9 1 27 x − x =− . 5 2 10 0
Let f (x, y, z) be a continuous function on the region R . If R can be described by a ≤ x ≤ b , u(x) ≤ y ≤ v(x) , ψ(x, y) ≤ z ≤ φ(x, y) , then ZZZ Z b Z v (x) Z φ(x,y ) f (x, y, z)dV = dx dy f (x, y, z)dz . (2) R
a
Dr. Lahcen Laayouni
u(x)
ψ(x,y )
Advanced Calculus
Multiple Integrals
Triple integrals Change of variables Cylindrical coordinate
Triple integrals
Example Evaluate I =
ZZZ
xdV , where R is the region in the first octant R
bounded by the plane x + y + z = 1 . Solution The region can be described by the inequalities 0 ≤ x ≤ 1 , 0 ≤ y ≤ 1 − x ,
and 0 ≤ z ≤ 1 − x − y ,we obtain the iterate integral I=
Z
0
Dr. Lahcen Laayouni
1
dx
Z
1−x
dy 0
Advanced Calculus
Z
1−x−y
xdz 0
Multiple Integrals
Triple integrals Change of variables Cylindrical coordinate
Triple integrals
Solution (cont.) Thus I =
Z
=
Z
= = =
Z
1
1−x
dx
Z
1−x
dx
Z
0
0
1 0
0
1 0
1 2
Z
1 2
Z
z=1−x−y
dy (xz)|z=0
x(1 − x − y)dy
y =1−x y 2 dx x y − xy − 2 y =0 1
0 1
0
x(1 − x)2 dx
(x − 2x 2 + x 3 )dx =
Dr. Lahcen Laayouni
1 . 24
Advanced Calculus
Multiple Integrals
Triple integrals Change of variables Cylindrical coordinate
Triple integrals
Change of variables Consider a one-to-one transformation from S a uvw -space to D a xyz -space x = x(u, v, w ) y = y(u, v, w ) z = z(u, v, w ) where x, y , and z have continuous first partial derivatives with respect to u, v , and w . The volume element is given by ∂(x, y, z) dudvdw . dV = dxdydz = ∂(u, v, w )
If g(u, v, w ) = f (x(u, v, w ), y(u, v, w ), z(u, v , w )) , then ZZZ ZZZ ∂(x, y, z) dudvdw . f (x, y, z)dxdydz = g(u, v, w ) ∂(u, v, w ) D S Dr. Lahcen Laayouni
Advanced Calculus
Multiple Integrals
Triple integrals Change of variables Cylindrical coordinate
Triple integrals
Example Find the volume of the region R given by the inequalities 0 ≤ z ≤ 4, 0 ≤ y + z ≤ 3, 0 ≤ x + y + z ≤ 5. Solution The region R is a parallelepiped, using the change of variables u = x + y + z, v = y + z, w = z , it corresponds to the rectangular parallelepiped defined by 0 ≤ u ≤ 5, 0 ≤ v ≤ 3, 0 ≤ w ≤ 4. The required volume is ZZZ Z V = dxdydz =
5
3
4
∂(x, y, z) du dv ∂(u, v, w ) dw . D 0 0 0 Solving for x, y, z in terms of u, v, w , to obtain x = u − v,
Z
Z
y = v − w,
Dr. Lahcen Laayouni
z = w,
Advanced Calculus
Multiple Integrals
Triple integrals Change of variables Cylindrical coordinate
Triple integrals
Solution (cont.) The corresponding Jacobian is ∂x ∂y ∂u ∂u ∂x ∂y ∂(x, y, z) = ∂(u, v, w ) ∂v ∂v ∂x ∂y ∂w ∂w So Z 5 Z V = du 0
∂z ∂u ∂z ∂v ∂z ∂w 3
dv 0
Z
1 0 0 = −1 1 0 = 1. 0 −1 1 4
1dw = 60 . 0
Example Find the volume of the solid inside the ellipsoid E given by x 2 y 2 z2 + 2 + 2 ≤1, a2 b c Dr. Lahcen Laayouni
where a, b, c > 0 . Advanced Calculus
Multiple Integrals
Triple integrals Change of variables Cylindrical coordinate
Triple integrals
Solution Using the change of variables x = au , y = bv , and z = cw , the ellipsoid becomes the ball B given by u2 + v 2 + w 2 ≤ 1 .
The Jacobian of the transformation is a 0 0 ∂(x, y, z) = 0 b 0 = abc. ∂(u, v, w ) 0 0 c So the volume of the ellipsoid is given by ZZZ ZZZ V = dxdydz = abcdudvdw E
B
= abc(Volume of B) = Dr. Lahcen Laayouni
4 πabc cubic units. 3
Advanced Calculus
Multiple Integrals
Triple integrals Change of variables Cylindrical coordinate
Double Integrals
Cylindrical coordinates The cylindrical coordinates of any triple (x, y, z) are x = r cos θ,
y = r sin θ,
z = z.
The volume element is given by dV = r drd θdz . Example Evaluate
ZZZ
(x 2 + y 2 )dV over the first octant region bounded by the
D
cylinders x 2 + y 2 = 1 and x 2 + y 2 = 4 and the planes z = 0, z = 1, x = 0 , and x = y . Dr. Lahcen Laayouni
Advanced Calculus
Multiple Integrals
Triple integrals Change of variables Cylindrical coordinate
Double Integrals
Solution In cylindrical coordinates the region is bounded by r = 1, r = 2 , θ = π4 , θ = π2 , z = 0, and z = 1 . Since x 2 + y 2 = r 2 , then ZZZ Z 1 Z π/2 Z 2 2 2 (x + y )dV = dz dθ r 2 rdr D
0
= (1 − 0)
π/4
π
2
1
−
π 24
14 − 4 4
4
=
15π . 16
Example Find the volume of the region R bounded by the hyperboloid x2
+
y2
−
z2
= 1 and the planes
2.0
1.5
z 1.0 −2 −1
0.5
z = 0, z = 2 .
0 1
0.0 −2
−1
2 0 y
Dr. Lahcen Laayouni
Advanced Calculus
1
2
x
Multiple Integrals
Triple integrals Change of variables Cylindrical coordinate
Double Integrals
Solution In cylindrical coordinates the region is given by 0 ≤ θ ≤ 2π , √ 0 ≤ z ≤ 2 , and 0 ≤ r ≤ 1 + z 2 .Thus ZZZ Z Z Z √ 2π
V
=
dV =
0
R
=
=
Z
2π
dθ 0
1 2
Z
2
dz
0
Z
0
2π
dθ
1+z 2
2
dθ
dz
0
r = 2
r2
rdr
√
r =0
0
1+z 2
1 = 2
Z
2π
dθ 0
Z
2
0
z=2 Z z 3 7 2π 14π z+ = dθ = . 3 z=0 3 0 3
Dr. Lahcen Laayouni
Advanced Calculus
(1 + z 2 )dz
Advanced Calculus Lecture 3 Dr. Lahcen Laayouni Department of Mathematics and Statistics McGill University
January 11, 2007
Dr. Lahcen Laayouni
Advanced Calculus
Multiple Integrals
Triple integrals Change of variables Cylindrical coordinate
Triple integrals
Triple integrals Triple integrals are just like double integrals, but with another integration to perform. Let f (x, y, z) be a bounded function over a rectangular parallelepiped R , a ≤ x ≤ b ,
PARRALELEPIPED
c≤y ≤d , p≤z≤q .
then the triple integral of f over R ZZZ f (x, y, z)dV or R
ZZZ
f (x, y, z)dxdydz , R
can be defined as the limit of Riemann sums corresponding to partitions of R into sub-parallelepiped by planes parallel to each of the coordinate planes. Dr. Lahcen Laayouni
Advanced Calculus
Multiple Integrals
Triple integrals Change of variables Cylindrical coordinate
Triple integrals
Proposition Suppose that f (x, y, z) is a continuous function on the rectangular parallelepiped R , a ≤ x ≤ b , c ≤ y ≤ d , p ≤ z ≤ q , then ZZZ Z b Z d Z q f (x, y, z)dV = dx dy f (x, y, z)dz . R
a
c
p
Remark There are six different ways of evaluating the triple integral (1), depending upon which variable is chosen first. Example Find the integral of f (x, y, z) = x 4 + 2yz over the parallelepiped R given by 0 ≤ x ≤ 1 , −1 ≤ y ≤ 2 , −2 ≤ z ≤ 1 Dr. Lahcen Laayouni
Advanced Calculus
(1)
Multiple Integrals
Triple integrals Change of variables Cylindrical coordinate
Triple integrals
Solution Z 1 Z I = dx 0
= =
Z Z
1
dx 0 1 0
Z
2
dy −1 2 −1
(9x 4 −
Proposition
Z
1
(x 4 + 2yz)dz =
1
y =2 3 3x 4 y − y 2 2
dx
0
−2
(3x 4 − 3y)dy = 9 )dx = 2
Z
Z
1
dx 0
Z
2
−1
z=1 dy (x 4 z + yz 2 )
z=−2
y =−1
9 5 9 1 27 x − x =− . 5 2 10 0
Let f (x, y, z) be a continuous function on the region R . If R can be described by a ≤ x ≤ b , u(x) ≤ y ≤ v(x) , ψ(x, y) ≤ z ≤ φ(x, y) , then ZZZ Z b Z v (x) Z φ(x,y ) f (x, y, z)dV = dx dy f (x, y, z)dz . (2) R
a
Dr. Lahcen Laayouni
u(x)
ψ(x,y )
Advanced Calculus
Multiple Integrals
Triple integrals Change of variables Cylindrical coordinate
Triple integrals
Example Evaluate I =
ZZZ
xdV , where R is the region in the first octant R
bounded by the plane x + y + z = 1 . Solution The region can be described by the inequalities 0 ≤ x ≤ 1 , 0 ≤ y ≤ 1 − x ,
and 0 ≤ z ≤ 1 − x − y ,we obtain the iterate integral I=
Z
0
Dr. Lahcen Laayouni
1
dx
Z
1−x
dy 0
Advanced Calculus
Z
1−x−y
xdz 0
Multiple Integrals
Triple integrals Change of variables Cylindrical coordinate
Triple integrals
Solution (cont.) Thus I =
Z
=
Z
= = =
Z
1
1−x
dx
Z
1−x
dx
Z
0
0
1 0
0
1 0
1 2
Z
1 2
Z
z=1−x−y
dy (xz)|z=0
x(1 − x − y)dy
y =1−x y 2 dx x y − xy − 2 y =0 1
0 1
0
x(1 − x)2 dx
(x − 2x 2 + x 3 )dx =
Dr. Lahcen Laayouni
1 . 24
Advanced Calculus
Multiple Integrals
Triple integrals Change of variables Cylindrical coordinate
Triple integrals
Change of variables Consider a one-to-one transformation from S a uvw -space to D a xyz -space x = x(u, v, w ) y = y(u, v, w ) z = z(u, v, w ) where x, y , and z have continuous first partial derivatives with respect to u, v , and w . The volume element is given by ∂(x, y, z) dudvdw . dV = dxdydz = ∂(u, v, w )
If g(u, v, w ) = f (x(u, v, w ), y(u, v, w ), z(u, v , w )) , then ZZZ ZZZ ∂(x, y, z) dudvdw . f (x, y, z)dxdydz = g(u, v, w ) ∂(u, v, w ) D S Dr. Lahcen Laayouni
Advanced Calculus
Multiple Integrals
Triple integrals Change of variables Cylindrical coordinate
Triple integrals
Example Find the volume of the region R given by the inequalities 0 ≤ z ≤ 4, 0 ≤ y + z ≤ 3, 0 ≤ x + y + z ≤ 5. Solution The region R is a parallelepiped, using the change of variables u = x + y + z, v = y + z, w = z , it corresponds to the rectangular parallelepiped defined by 0 ≤ u ≤ 5, 0 ≤ v ≤ 3, 0 ≤ w ≤ 4. The required volume is ZZZ Z V = dxdydz =
5
3
4
∂(x, y, z) du dv ∂(u, v, w ) dw . D 0 0 0 Solving for x, y, z in terms of u, v, w , to obtain x = u − v,
Z
Z
y = v − w,
Dr. Lahcen Laayouni
z = w,
Advanced Calculus
Multiple Integrals
Triple integrals Change of variables Cylindrical coordinate
Triple integrals
Solution (cont.) The corresponding Jacobian is ∂x ∂y ∂u ∂u ∂x ∂y ∂(x, y, z) = ∂(u, v, w ) ∂v ∂v ∂x ∂y ∂w ∂w So Z 5 Z V = du 0
∂z ∂u ∂z ∂v ∂z ∂w 3
dv 0
Z
1 0 0 = −1 1 0 = 1. 0 −1 1 4
1dw = 60 . 0
Example Find the volume of the solid inside the ellipsoid E given by x 2 y 2 z2 + 2 + 2 ≤1, a2 b c Dr. Lahcen Laayouni
where a, b, c > 0 . Advanced Calculus
Multiple Integrals
Triple integrals Change of variables Cylindrical coordinate
Triple integrals
Solution Using the change of variables x = au , y = bv , and z = cw , the ellipsoid becomes the ball B given by u2 + v 2 + w 2 ≤ 1 .
The Jacobian of the transformation is a 0 0 ∂(x, y, z) = 0 b 0 = abc. ∂(u, v, w ) 0 0 c So the volume of the ellipsoid is given by ZZZ ZZZ V = dxdydz = abcdudvdw E
B
= abc(Volume of B) = Dr. Lahcen Laayouni
4 πabc cubic units. 3
Advanced Calculus
Multiple Integrals
Triple integrals Change of variables Cylindrical coordinate
Double Integrals
Cylindrical coordinates The cylindrical coordinates of any triple (x, y, z) are x = r cos θ,
y = r sin θ,
z = z.
The volume element is given by dV = r drd θdz . Example Evaluate
ZZZ
(x 2 + y 2 )dV over the first octant region bounded by the
D
cylinders x 2 + y 2 = 1 and x 2 + y 2 = 4 and the planes z = 0, z = 1, x = 0 , and x = y . Dr. Lahcen Laayouni
Advanced Calculus
Multiple Integrals
Triple integrals Change of variables Cylindrical coordinate
Double Integrals
Solution In cylindrical coordinates the region is bounded by r = 1, r = 2 , θ = π4 , θ = π2 , z = 0, and z = 1 . Since x 2 + y 2 = r 2 , then ZZZ Z 1 Z π/2 Z 2 2 2 (x + y )dV = dz dθ r 2 rdr D
0
= (1 − 0)
π/4
π
2
1
−
π 24
14 − 4 4
4
=
15π . 16
Example Find the volume of the region R bounded by the hyperboloid x2
+
y2
−
z2
= 1 and the planes
2.0
1.5
z 1.0 −2 −1
0.5
z = 0, z = 2 .
0 1
0.0 −2
−1
2 0 y
Dr. Lahcen Laayouni
Advanced Calculus
1
2
x
Multiple Integrals
Triple integrals Change of variables Cylindrical coordinate
Double Integrals
Solution In cylindrical coordinates the region is given by 0 ≤ θ ≤ 2π , √ 0 ≤ z ≤ 2 , and 0 ≤ r ≤ 1 + z 2 .Thus ZZZ Z Z Z √ 2π
V
=
dV =
0
R
=
=
Z
2π
dθ 0
1 2
Z
2
dz
0
Z
0
2π
dθ
1+z 2
2
dθ
dz
0
r = 2
r2
rdr
√
r =0
0
1+z 2
1 = 2
Z
2π
dθ 0
Z
2
0
z=2 Z z 3 7 2π 14π z+ = dθ = . 3 z=0 3 0 3
Dr. Lahcen Laayouni
Advanced Calculus
(1 + z 2 )dz
