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Average Value of a Function over a Region

Prove the Mean Value Theorem for double integrals, find the average distance from points in a disk to the origin, and define the improper integral of a 2-variable function over the entire plane.

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Added on  2022-12-14

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Learn how to calculate the average value of a function over a given region using integration techniques. Understand the concept of average value and its application in various scenarios. Get step-by-step solutions and examples.

Average Value of a Function over a Region

Prove the Mean Value Theorem for double integrals, find the average distance from points in a disk to the origin, and define the improper integral of a 2-variable function over the entire plane.

   Added on 2022-12-14

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Proof
Let f (x, y) be a continuous real valued function D be a closed, connected
region, that has area A(D).
We first state the Extreme Value Theorem for function in two variables:
According to B, Hass, Heil and Weir (2018), “Let f (x, y) be a continuous
function. If D is a closed and bounded region then there exists (x1, y1),(x2, y2)
D such that f (x1, y1) is an absolute maximum of f and f (x2, y2) is an ab-
solute minimum of f .”
Thus, by Extreme Value Theorem
f (x1, y1) f (x, y) f (x2, y2) (1)
Integrating (1) over D
∫ ∫
D
f (x1, y1) dA
∫ ∫
D
f (x, y) dA
∫ ∫
D
f (x2, y2) dA
[
∫ ∫
D
dA = A(D)]
A(D)f (x1, y1)
∫ ∫
D
f (x, y) dA A(D)f (x2, y2)
[Divide throughout by A(D)]
1
Average Value of a Function over a Region_1
f (x1, y1) 1
A(D)
∫ ∫
D
f (x, y) dA f (x2, y2)
According to B, Hass, Heil and Weir (2018), “By Intermediate Value The-
orem, a continuous function f (x, y) on D, takes every value between it’s
maximum and minimum, therefore, there must a point (x0, y0) D, such
that”
f (x0, y0) = 1
A(D)
∫ ∫
D
f (x, y) dA
Multiplying throughout by A(D)
∫ ∫
D
f (x, y) dA = f (x0, y0) A(D)
which is the Mean Value Theorem in two variables. Hence proved.
Answer
A disk D centered at origin with radius a is shown in figure-1 below.
Figure 1: Disk at origin with radius=a
2
Average Value of a Function over a Region_2
Consider any point P (x, y) which lies inside D. Its distance from the origin
O is given by
d(x, y) = x2 + y2
Average value of a function f (x, y) over a bounded region R in the xy- plane
is calculated using formula
fave = 1
A(R)
∫ ∫
R
f (x, y) dA (1)
where, A(R) is the area of R.
Area of disk D is: A(D) = πr2 = πa2.
Therefore, by (1), average distance of points in D from origin O is
dave = 1
A(D)
∫ ∫
D
d(x, y) dA
= 1
π a2
∫ ∫
D
x2 + y2 dA
For convenience, switch to polar coordinates (r, θ).
x = r cos θ, y = r sin θ
Area element of a disk is given by: dA = rdrdθ.
In the region D, r varies from 0 to a while θ varies from 0 to 2π. Therefore,
dave = 1
π a2
∫ ∫
D
r2 rdrdθ
[Applying limits to the integral]
= 1
π a2
2π
0
a
0
r2 drdθ
= 1
π a2
2π
0
[r3
3
]a
0
dθ
3
Average Value of a Function over a Region_3
= 1
π a2
2π
0
a3
3 dθ
= a
3π
2π
0
dθ
= a
3π 2π
dave = 2a
3
Thus, the average value of distance of points from origin, in a disk centered
at origin with radius a is 2a
3 .
4
Average Value of a Function over a Region_4

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