Course: Accelerated Engineering Calculus II Instructor: Michael Medvinsky Average value: b
1
f ( x ) dx b − a ∫a
1
Given f(x,y) on rectangle R with area A(R) then fave =
f ( x, y ) ΔA A ( R ) ∫∫
R
Recall: Given f(x) on an interval [a,b] then fave =
Ex 2.
Let z = x 2 − y be a shape of a “sand dune” in a box with a base of a 2x2
units size. Answer the following question using the course approximation of the
volume of the sand from previous exercise. Could the entire dune be packed if a
box has a height of 1 unit?
In previous question we found that the volume of z = x 2 − y on 2x2 square is approximately one, V
1
while height of the box should of average value, which is given by fave =
≈ . Therefore we A ( R) 4
will have more then enough place for our sand in that box. Properties of Double Integrals: 1) ∫∫ f ( x, y ) + g ( x, y ) ΔA = ∫∫ f ( x, y ) ΔA + ∫∫ g ( x, y ) ΔA
R
R
R
2) ∫∫ cf ( x, y ) ΔA = c ∫∫ f ( x, y ) ΔA
R
R
3) f ( x, y ) ≥ g ( x, y ) ⇒ ∫∫ f ( x, y ) ΔA ≥ ∫∫ g ( x, y ) ΔA
R
R
5.2 Iterated Integrals (12.2)
It won‘t be easy to evaluate double integral using the definition with the limits. In Calculus I The Fundamental Theorem of Calculus provided a more convenient way. In this section we will learn an easy method to solve double integrals. d
Suppose f ( x, y ) defined on rectangle R = [ a,b ] × [ c,d ] and let g ( x ) = ∫ f ( x, y ) dy where the c
d
expression ∫ f ( x, y ) dy is understood as a partial integration with respect to y, i.e. the variable x c
considered a constant for the process of integration. Next we integrate g to get b
b d
⎧
⎫
g
x
dx
=
f
x,
y
dy
(
)
(
)
⎨
⎬ dx ∫a
∫a ⎪⎩ ∫c
⎪⎭
The last integral is called an iterated integral; we often omit the brackets and recognize 2 of them: b d
b
d b
d
⎧d
⎫
⎧b
⎫
1) ∫ ∫ f ( x, y ) dy dx = ∫ ⎨ ∫ f ( x, y ) dy ⎬ dx
2) ∫ ∫ f ( x, y ) dx dy = ∫ ⎨ ∫ f ( x, y ) dx ⎬ dy ⎪⎭
⎪⎭
a c
a ⎪
c a
c ⎪
⎩c
⎩a
61 Course: Accelerated Engineering Calculus II Instructor: Michael Medvinsky 1
Ex 3.
y2 x
1 x
1 x
e1 − e−1
ye
dA
=
ye
dy
dx
=
e
dx
=
e
dx
=
e
=
∫∫R
∫ ∫
∫2
2 ∫0
2 −1
2
x=−1 y=0
−1
0
Ex 4.
1 4
⎛ y3 ⎞
x
x5
1
y
x
dy
dx
=
⋅
x
dx
=
dx
=
=
∫x=0 y=0∫
∫0 ⎜⎝ 3 ⎟⎠
∫0 3
15 0 15
o
1
1
1
1
1
x
1
1
2
2
2
1 2⎞
⎛
2
3
∫x=0 y=0∫ x + 3y + 2xy dy dx = ∫0 ⎜⎝ xy + y + 2 xy ⎟⎠ 0 dx =
1
= ∫ 2x + 2 3 +
0
2
Ex 6.
1
x
x
1
Ex 5.
1
x
1
1
1 2
x2 dx = 2 ∫ 2x + 4 dx = 2 ( x 2 + 4x )0 = 10
2
0
1
1
2
∫ ∫
x + 3y 2 + 2xy dx dy =
y=0 x=0
⎛1 2
2
2 ⎞
∫y=0 ⎜⎝ 2 x + 3y x + x y⎟⎠ 0 dy =
2
2
=
1
1 ⎞
⎛1
2
3
3 2
∫y=0 2 + 3y + y dy = ⎜⎝ 2 y + y + 2 y⎟⎠ 0 = ( y + y )0 = 2 + 8 = 10
Thm (Fubini’s): If f(x,y) is continuous on rectangle R = [ a,b ] × [ c,d ] (or at least bounded with discontinuities on a finite number of smooth curves) then b d
d b
a c
c a
∫∫ f ( x, y ) dA = ∫ ∫ f ( x, y ) dy dx = ∫ ∫ f ( x, y ) dy dx R
d
We won’t prove this theorem, but to make some sense, recall that g ( x0 ) = ∫ f ( x0 , y ) dy c
represents area under curve described by f ( x0 , y ) and therefore the volume is given by ∫∫
R
b
b d
a
a c
f ( x, y ) dA = V = ∫ g ( x ) dx = ∫ ∫ f ( x, y ) dy dx . Similarly for d b
∫ ∫ f ( x, y ) dx dy , just x and y change c a
roles. Thm: Let f(x,y) = g(x)h(y) then b d
b d
b
d
a c
a c
a
c
∫ ∫ f ( x, y ) dy dx = ∫ ∫ g ( x ) h ( y ) dy dx = ∫ g ( x ) dx ⋅ ∫ h ( y ) dy
d b
d b
b
d
∫ ∫ f ( x, y ) dx dy = ∫ ∫ g ( x ) h ( y ) dx dy = ∫ g ( x ) dx ⋅ ∫ h ( y ) dy
c a
1
Ex 7.
1
c a
1
⎛y
2
⎞
∫ ∫ yx dy dx = ∫ ⎜⎝ 2 x ⎟⎠
x=0 y=0
x=0
1
dx =
0
a
2
∫
1
1
⎛ y2 ⎞ ⎛ y2 ⎞
1
yx
dy
dx
=
x
dx
⋅
y
dy
=
∫x=0 y=0∫
∫x=0 y=0∫
⎜⎝ 2 ⎟⎠ ⋅ ⎜⎝ 2 ⎟⎠ = 4
0
0
1
1
1
1
c
1
⎛1 x ⎞
1
1
x dx = ⎜
=
⎟
2
⎝ 2 2 ⎠0 4
x=0
1
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