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Chapter 12 Multiple Integrals
Section 12.1 Double Integrals Over Rectangles
Examples :
- Evaluate the iterated integral a) (5 ) , {( , ) |0 1, 0 1} R
∫∫ −^^ x dA R^ =^ x y^ ≤^ x^ ≤^ ≤^ y ≤ and
b) (4 2 ) , [0,1] [0,1] R
∫∫ −^^ y dA R =^ ×
Solution: a)
1 1 2 1 0 0 0
(5 ) (5 ) [5 / 2] (1) 4.
R
∫∫ −^^ x dA^ =^ ∫ ∫ −^ x dydx^ =^ x^ −^ x =
b)
1 1 2 1 0 0 0
(4 2 ) (4 2 ) [4 ] 3
R
∫∫ −^^ y dA^ =^ ∫ ∫ −^ y dydx^ =^ y^ −^ y =
- If R = [0,1] × [0,1], show that 0 sin( ) 1 R
≤ ∫∫ x + y dA ≤
Solution: On R, 0 ≤ x + y ≤ 2 < π,sin( x + y ) ≥ 0. Thus we have
0 sin( ) 1 R R
≤ ∫∫ x + y dA ≤ ∫∫ dA =
Iterated Integrals
Fubini’s Theorem If f x y ( , )^ is differentiable on R = {( , x y ) | a ≤ x ≤ b c , ≤ y ≤ d }
then ( , )^ ( , )^ ( , )
b d d b
R a c c a
∫∫^ f x y dA^ =^ ∫∫ f x y dydx^ =∫∫ f x y dxdy
Examples Evaluate the integral in two different ways using Fubini’s theorem (1-4).
4 1 4 2 2 2 3 1 1 3 14 1 1 1
( x y ) dydx [ x y y / 3] yy ==− dx [2 x / 3 2 / 3 ] x 44 −
∫ ∫ +^ =^ ∫ +^ =^ +^ =
also
1 4 1 (^2 2 3 2 14 3 ) 1 1 1
( x y ) dxdy [ x / 3 xy ] xx == dy [21 y y ] (^) − 44 − −
∫ ∫ +^ =^ ∫ +^ =^ +^ =
1 2
0 1
xe x dydx ∫ ∫ y
2 1 2 1 0
∫ ∫^ (^ x^^ + y^ )− dxdy
1 1
0 0 2 2 1
xy (^) dydx ∫ ∫ x + y +
- Evaluate cos( 2 ) , {( , ) |0 , 0 / 2} R
∫∫^ x^ +^ y dA R^ =^ x y^ ≤^ x^ ≤^ π ≤^ y ≤π
- Evaluate
2 2
R^1
x (^) dA R x y x y y
∫∫ +
- Find the volume of the solid in the first octant bounded by the cylinder z = 9 − y^2 and the plane x = 2 Solution: On the xy plane z = 0 ⇒ y^2 = 9 ⇒ y = 3 in the first octant. Thus we have 2 3 2
0 0
V = (^) ∫ ∫ (9 − y ) dydx = 36
- Find the average value of f ( , x y ) = e y^ x + e y , R = [0, 4] × [0,1].
The average value is given by 1 ( , ) ave ( ) R f f x y dA A R
= (^) ∫∫ , where A(R) represents the area of the region R. Now
4 1
0 0
y y fave = (^) ∫ ∫ e x + e dydx = (Try to solve the integral)
- Evaluate
3 5
0 0 ∫ ∫ (5^ − x dxdy ).^ Ans: 37.
- From problem 9. draw the bounded region by z = 5 − x R , = {( , x y ) |0 ≤ x ≤ 5, 0 ≤ y ≤ 3}and evaluate 3 5
0 0 ∫ ∫ (5^ − x dxdy ) geometrically.
- Evaluate
3 4
1 2 ∫ ∫ (40^ −^2 xy dydx ).^ Ans: 112
- Evaluate 2 R ∫∫^ y xdA over the rectangle^ R^ =^ {( , x y^ ) |^^ −^3 ≤^ x^ ≤^ 2, 0^ ≤^ y ≤^ 1}. Ans: -5/
- Evaluate cos( ) cos (^2 ) ; [0,1/ 2] [0, ] R ∫∫^ x^ xy^^ π x dA R =^ × π Ans:^ 1/(3^ π^ ).
- Evaluate
- Sketch the region of integration and change the order of integration of 1 4
0 4
x
∫ ∫^ f^ x y dydx. Solution hint:
1 4 4 / 4
0 4 0 0
y
x
∫ ∫^ f^ x y dydx^ =∫ ∫ f^ x y dxdy
- Use symmetry to evaluate (2 3 4 ) D
∫∫ −^^ x^ + y dA , where D is the region bounded by the square with vertices ( 5, 0)± and (0, ±5) Solution hint: (^) ∫∫ − + =∫∫ = = = D D
( 2 3 x 4 y ) dA 2 dA 2 A ( D ) 2 ( 50 ) 100 using symmetry.
Section 12.3 Polar Coordinates
If f is continuous on a polar rectangle R given by 0 ≤ a ≤ r ≤ b , α ≤ θ≤ β where
0 ≤ β − α≤ 2 π then (^) ∫∫ =∫ ∫ R
b
a
f x ydA f r rdrd
β
α
Examples
- Evaluate (^) ∫∫ R
sin θ dA , where R is the region in the first quadrant that is outside the
circle r = 2 and inside the cardioid r = 2 ( 1 +cos θ)
Hint: (^) ∫∫ ∫ ∫
= =
/ 2
0
2 ( 1 cos )
2
sin sin 8 / 3
π θ
θ dA θ rdrd θ
R
- The sphere of radius a centered at origin is expressed in rectangular coordinates as x^2 + y^2 + z^2 = a^2 and hence its equation in cylindrical coordinates is r^2 + z^2 = a^2. Use this equation and a polar double integral to find the volume of the sphere. Hint: + = ⇒ =± − = ∫ ∫ − =
π
2
0
3 0
r^2 z^2 a^2 z a^2 r^2 , V 2^ a a^2 r^2 rdrd 4 / 3 a
- Use polar coordinate to evaluate (^) ∫ ∫ −
−
1
1
1
0
2 2 3 / 2
2 ( )
x x y dydx
Hint: draw a diagram and set (^) ∫ ∫ + =∫ ∫ = −
− π
0
1
0
(^14)
1
1
0
(^2 2 )^3 /^2 / 5
2 x y dydx r drd
x
- Find the area of the region given: a) between the circles of radius 1 and 3 in the first quadrant. Solution: (^) ∫∫ = ∫ ∫
/ 2
0
3
1
( , ) ( cos , sin )
π
f x y dA f r θ r θ rdrd θ
R
b) Use double integral to find the area of one loop of the rose r =cos 3 θ: Solution hint: (^) ∫∫ ∫ ∫ −
/ 6
/ 6
cos 3
0
π
π
θ
A dA rdrd θ π
D c) Use double integral to find the area of the region enclosed by
r = 4 + 3 cos θ, D = {( r , θ )| 0 ≤ θ≤ 2 π, 0 ≤ r ≤ 4 + 3 cos θ}
Solution hint:
2 4 3cos
0 0
D
A dA rdrd
π θ θ π
= (^) ∫∫ = (^) ∫ ∫ =
Section 12.5 Triple Integrals
Examples
- Evaluate xy z dv ∫∫∫ E 12 2 2 where E is defined as a region bounded by
− 1 ≤ x ≤ 2 , 0 ≤ y ≤ 3 , 0 ≤ z ≤ 2.
Hint: Draw he bounded region and set (^) ∫∫∫ ∫ ∫ ∫ −
2
1
3
0
2
0
12 xy^2 z^2 dv 12 xdx y^2 dy z^2 dz 648 E
- Evaluate zdv ∫∫∫ E where E is the wedge in the first octant that is cut from the
cylindrical solid y^2 + z^2 ≤ 1 by the planes y = x , x = 0
Hint: Draw diagram and set (^) ∫∫∫ =∫ ∫ ∫ =
1 −
0 0
1
0
y y^2
E
zdv zdzdxdy using type I region
- Evaluate zdv ∫∫∫ E where E is the solid tetrahedron bounded by the four planes
y = 0 , x = 0 , z = 0 , x + y + z = 1
Hint: (^) ∫∫∫ ∫ ∫ ∫
− −− = = =
1
0
1
0
1
0
x xy
E
V zdv zdzdydx
- Evaluate ydv ∫∫∫ E where E is the solid tetrahedron bounded by the four planes
y = 0 , x = 0 , z = 0 , 2 x + 2 y + z = 4 Answer: V = 4/
- Use triple integral to find volume of the tetrahedron bounded by the four planes 2 y = x , x = 0 , z = 0 , x + 2 y + z = 2 Answer: V = 1/
- Find the volume of the tetrahedron with vertices (0, 0, 0), (1, 2, 0), (0, 2, 2) and (0, 2, 0). Use Type I, Type II and Type III regions.
Hint: Find the plane through (0, 0, 0), (1, 2, 0), and (0, 2, 2) which is z = y − 2 x.
Type I: (^) ∫ ∫ ∫ ∫ ∫ ∫
− − = =
2
0
/ 2
0
2
0
1
0
2
2
2
0
y y x
x
y x V f x y zdzdydx f x y z dzdxdy , from xy – solid
Type II: (^) ∫ ∫ ∫ ∫ ∫ ∫
− − = =
2
0 0
1 / 2 ( )
0
2
0
2
0
1 / 2 ( )
0
y z y yz V f x y zdxdydz f x y z dxdzdy , from zy - solid
Solution hint: Type I, xy solid E ={( x , y , z )| 0 ≤ x ≤ 1 , 0 ≤ y ≤ x^2 , 0 ≤ z ≤ y }={( x , y , z )| 0 ≤ y ≤ 1 , y ≤ x ≤ 1 , 0 ≤ z ≤ y } Type II, zy solid E = {( x , y , z )| 0 ≤ y ≤ 1 , 0 ≤ z ≤ y , y ≤ x ≤ 1 }={( x , y , z )| 0 ≤ z ≤ 1 , z ≤ y ≤ 1 , y ≤ x ≤ 1 } Type III, xz solid E ={( x , y , z )| 0 ≤ x ≤ 1 , 0 ≤ z ≤ x^2 , z ≤ y ≤ x^2 }={( x , y , z )| 0 ≤ z ≤ 1 , z ≤ x ≤ 1 , z ≤ y ≤ x^2 }
- Find the average value of f ( , x y z , ) = x z^2 + y z^2 over the region enclosed by the paraboloid z = 1 − x^2^ − y^2 and the plane z = 0. Solution hint: f x y zdv V E
f E
ave =^ ( )∫∫∫ ( , , )
(^1) , where
∫− ∫ ∫
−
− −
− − = =
1
1
1
1
1
0
2
2
2 2 ( ) / 2
x
x
x y V E dzdydx π using polar coordinate. Now
∫∫∫ (^) − ∫ ∫ ∫
−
− −
− − = = + =
1
1
1
1
1
0
2 2
2
2
2 2 ( , , )^22 ( ) 1 / 12 ( )
1 x x
x y
E
fave (^) V E f x yzdv π x y dzdydx
Section 12.6-12.7 Triple Integrals in Cylindrical and Spherical
Coordinates
Cylindrical Coordinates:
∫∫∫ = ∫ ∫ ∫
β
α
θ
θ
θ
θ
θ θ
( )
()
(,)
(,)
2
1
2
1
h
h
u r
E u r
f x y zdv f r z rdzdrd
Spherical Coordinates:
∫∫∫ = ∫ ∫ ∫
d
c
b
E a
f x y zdv f d d d
β
α
( , , ) (ρ ,θ,φ)ρ^2 sinφ ρ θ φ
Examples
- Use triple integral in cylindrical coordinates to find the volume of the solid E that is bounded by the hemisphere z = 25 − x^2 − y^2 below bounded by the xy – plane and laterally by the cylinder x^2 + y^2 = 9
Hint:
2 2 2 2
2
3 9 25 2 3 25
3 9 0 0 0 0
x x^ y r
E (^) x
z dv dzdydx rdzdrd
π
− −^ − −
− (^) − − ∫∫∫ =^ ∫ ∫ ∫ =^ ∫ ∫ ∫ =
- Evaluate by cylindrical coordinate:
2 2 2
2
3 9 9 2 3 9 0
x x y
x
x dzdydx
− − −
− (^) − −
∫ ∫ ∫
Hint:
2 2 2
2
3 9 9 2 3 9 0
x x y
x
x dzdydx
− − −
− (^) − −
∫ ∫ ∫
2 3 9^2 3 2 0 0 0
cos 243 / 4
r r dzdrd
π θ θ π
− = (^) ∫ ∫ ∫ =
- Use spherical coordinate to evaluate
2 2 2
2
2 4 4 2 2 2 2 2 4 0
x^ x^ y
x
z x y z dzdydx
−^ −^ −
− (^) − −
∫ ∫ ∫^ +^ +
Hint:
2 2 2
2
2 4 4 2 2 2 2 2 4 0
x^ x^ y
x
z x y z dzdydx
−^ −^ −
− (^) − −
∫ ∫ ∫ +^ +
2 / 2 2 5 2 0 0 0
cos sin d d d 64 / 9
π π = (^) ∫ ∫ ∫ ρ φ φ ρ φ θ = π
- Evaluate ( 3 2 ) E
∫∫∫^ x^ + xy^ dv where E is the solid in the first octant that lies beneath the paraboloid z = 1 − x^2^ − y^2 Hint: E = {( , r θ, z ) |0 ≤ θ ≤ π/ 2, 0 ≤ r ≤ 1, 0 ≤ z ≤ 1 − r^2 }and / 2 1 1^2 3 2 4 0 0 0
( ) cos 2 / 35
r
E
x xy dv r dzdrd
π θ θ
− ∫∫∫ +^ =^ ∫ ∫ ∫ =
- Find the volume of the solid that lies within both the cylinder x^2 + y^2 = 1,and the sphere x^2 + y^2^ + z^2 = 4 Hint: E = {( , r θ, z ) |0 ≤ θ ≤ 2 π, 0 ≤ r ≤ 1, − 4 − r^2 ≤ z ≤ 4 − r^2 }and 2
2
2 1 4
(^0 0 )
r
r
V rdzdrd
π θ π
−
− −
= (^) ∫ ∫ ∫ = −
- Use spherical coordinates to evaluate ( 2 2 ) H
∫∫∫^ x^ + y^ dv where H is the hemispherical region that lies above the xy plane and below the sphere x^2^ + y^2 + z^2 = 1
Hint: 2 2 2 / 2 1 4 3 0 0 0
( ) sin 4 / H
x y dv d d d
π π ∫∫∫ +^ =^ ∫ ∫ ∫^ ρ^ φ^ ρ θ^ φ^ = π
- Evaluate 2 2 2 ( x^2^ y^2^ z^2 ) x y z e dxdydz
∞ ∞ ∞ − + +
−∞ −∞ −∞ ∫ ∫ ∫^ +^ + Hint: Use spherical coordinate and find ∞ ∞ ∞ x (^2) (^) y (^2) z e (^2) − ( x (^2) (^) + y (^2) (^) + z (^2) ) dxdydz
−∞ −∞ −∞ ∫ ∫ ∫ +^ +^2
(^23)
0 0 0
lim sin 2
R R^ e^ d^ d^ d
π π (^) ρ = (^) →∞∫ ∫ ∫ ρ − φ ρ θ φ = π
Section 12.8 Change of Variables in Multiple Integrals
Examples
Hint: Use the transformation (substitution) u = x / a v , = y / b w , = z / c ⇒ u^2 + v^2 + w^2 = 1 , which is a sphere of radius 1. Now ( , , ) 4 / 3 R S
∫∫∫^ dV^ =^ ∫∫∫ J u v w^ dudvdw^ =^ π abc. For volume of a sphere see example 2,
section 12.3 page # 685.
- Find the image of the set S which is a disk given by u^2 + v^2 ≤ 1 under the transformation x = au y , = bv.
- Evaluate ( 3 ) R
∫∫^ x^ − y dA , where R is the triangular region with vertices (0, 0), (2, 1) and (1, 2). Hint: The line thru (0, 0) and (2, 1) is y = 1/ 2 x , which is the image of v = The line thru (0, 0) and (1, 2) is y = 2 x , which is the image of u =0 and the line thru (1, 2) and (2, 1) is x + y = 3 , which is the image of u + v = 1
The Jacobian
x x J u v u^ v y y u v
= ∂^ ∂ = =
Now
1 1
0 0
u
R
x y dA u v dvdu
− ∫∫ −^ =^ ∫ ∫ −^ −^ = −
- Evaluate ( 2 2 ) R
∫∫^ x^ −^ xy^ + y^ dA , where R is the region bounded by the ellipse x^2^ − xy + y^2 = 2;under the substitution x = 2 u − 2 / 3 , v y = 2 u + 2 / 3 v
Solution: x^2^ − xy + y^2 = 2 ⇒ u^2^ + v^2 = 1 , The Jacobian ( , ) 4 / 3
x x J u v u^ v y y u v
= ∂^ ∂ =
2
2
1 1 2 2 2 2 (^1 )
u
R (^) u
x xy y dA u v dudv π
−
− (^) − −
∫∫ −^ +^ =^ ∫ ∫ +^ =
- Evaluate cos R
y x (^) dA y x
∫∫ (^) + , where R is the trapezoidal region with vertices (1, 0),
(2, 0), (0, 2), (0, 1). Hint: Use u = y − x v , = y + x , J u v ( , ) = −1/ 2 , cos 3 / 2 sin R
y x dA y x
∫∫ +
- Evaluate sin(9 2 4 2 ) R
∫∫^ x^ + y^ dA , where R is the region bounded by the ellipse 9 x^2 + 4 y^2 = 1. Use u = 3 , x v = 2 , y J u v ( , ) =1/ 6 2 2 2 2 / 2 1 2 0 0
sin(9 4 ) 1/ 6 sin( ) 1/ 6 sin R S
x y dA u v dudv r r drd
π ∫∫ +^ =^ ∫∫ +^ = ∫ ∫^ θ
