MA22S1: SOLUTIONS TO TUTORIAL 8
1. Find the average value of the function
f (x, y) = x2 y
on the rectangular region
R = [1, 3] × [−4, 4]
Solution: The average value of f on R is given by
1
average(f ) =
area(R)
ZZ
f (x, y) dA
R
We could compute the area of R using the formula
area(R) =
=
=
=
ZZ
Z
Z
Z
1
1 dA
R
3 Z 4
1 dy dx
−4
3
[y]y=4
y=−4 dx
1
3
8 dx
1
= [8x]31
= 16
But since R is a rectangular region it is easier to just multiply the
lengths of two adjacent sides:
area(R) = (3 − 1)(4 − (−4)) = 16
1
2
MA22S1: SOLUTIONS TO TUTORIAL 8
By Fubini’s theorem we have
ZZ
Z 3Z
f (x, y) dA =
R
1
=
Z
=
Z
=
Z
3
1
4
x2 y dy dx
−4
2 y=4
y
x2
dx
2
y=−4
3
8x2 − 8x2 dx
1
3
0 dx
1
= 0
Thus the average value of f (x, y) on the rectangle R is
average(f ) =
1
(0) = 0.
16
2. Evaluate the following integral,
Z
1
0
Z
1
2
ey dy dx
x
Solution: To evaluate the integral we will reverse the order of integration. Let D be the region of integration and suppose (x, y) ∈ D.
The x-limits of integration are 0 and 1. This means that x can take
any value between 0 and 1. The y-limits of integration are x and 1.
This means that if we fix a value for x then y can take any value
between x and 1. Thus the region of integration is the triangular
region bounded by the lines
x = 0, y = x, y = 1
See Figure 1. This is a region of Type I,
D = {(x, y) : x ∈ [0, 1], y ∈ [x, 1]}
MA22S1: SOLUTIONS TO TUTORIAL 8
3
y
y=x
1
D
1
x
Figure 1. Region of integration
We could also view D as a region of Type II. If a point (x, y) is in
D then y can take any value between 0 and 1. If we fix a value for y
then x can take any value between 0 and y. So to reverse the order
of integration the y-limits will be 0 and 1 and the x-limits will be 0
and y. Thus,
D = {(x, y) : y ∈ [0, 1], x ∈ [0, y]}
Z
1
0
Z
1
e
y2
dy dx =
Z
=
Z
x
1
0
1
1
=
=
"
y
2
ey dx dy
0
h
0
Z
=
Z
x ey
2
ix=y
dy
x=0
2
y ey dy
0
2
ey
2
#y=1
y=0
e−1
2
3. A thin sheet of metal encloses a region D in the first quadrant of the
xy-plane which is bounded by the circles x2 +y 2 = 1 and x2 +y 2 = 9.
4
MA22S1: SOLUTIONS TO TUTORIAL 8
The density of the metal sheet at each point is given by the function
δ(x, y) = 16 − x2 − y 2
Find the mass of the metal sheet.
Solution: The mass of the sheet is given by the formula
mass(D) =
ZZ
δ(x, y) dA
D
If we switch to polar coordinates (r, θ) where
x = r cos θ,
y = r sin θ
then
mass(D) =
Z
π
2
0
Z
3
δ(r cos θ, r sin θ) r dr dθ
1
We have
δ(r cos θ, r sin θ) = 16 − (r cos θ)2 − (r sin θ)2
= 16 − r 2
Thus
mass(D) =
Z
=
Z
=
Z
=
Z
π
2
0
π
2
0
π
2
Z
3
(16 − r 2 ) r dr dθ
1
2
r=3
r
r4
16
−
dθ
2
4 r=1
(72 −
0
π
2
44 dθ
0
θ= π
= 44 [θ]θ=02
= 22π
81
1
) − (8 − ) dθ
4
4
MA22S1: SOLUTIONS TO TUTORIAL 8
5
The extra r that appears when we switch to polar coordinates is
the Jacobian of the polar coordinate transformation T : (r, θ) →
(x, y):
∂(x, y) =
∂(r, θ)
∂x
∂r
∂x
∂θ
∂y
∂r
∂y
∂θ
cos θ −r sin θ
=
sin θ r cos θ
=r