MA1132 (Advanced Calculus) - Solutions to tutorial sheet 8
[March 19 – 20, 2013]
1. For R = [2, 3] × [−1, 1] (therectangle
{(x, y) ∈ R2 : 2 ≤ x ≤ 3, −1 ≤ y ≤ 1}) and
RR
f (x, y) = cos π2 x + sin π3 y , find R f (x, y) dx dy
Solution:
Z
ZZ
f (x, y) dx dy =
R
=
=
=
=
=
=
1
Z
3
π π cos
x + sin
y dx dy
2
3
y=−1
x=2
Z 1 π π 3
2
sin
x + x sin
y
dy
2
3
y=−1 π
x=2
Z 1
π 2
π 2
3π
sin
+ 3 sin
y −
sin π + 2 sin
y
dy
2
3
π
3
y=−1 π
Z 1
π 2
− + sin
y dy
π
3
y=−1
π 1
2
3
− y − cos
y
π
π
3
y=−1
π 2
2
3
π
3
− − cos −
− cos −
π π
3
π π
3
4
−
π
2. Find and graph the region R in R2 so that the iterated integral
Z Z √
ZZ
2
x=
9−y 2
f (x, y) dx dy =
y=0
x=0
f (x, y) dx dy
R
Solution:
p
R = {(x, y) ∈ R2 : 0 ≤ x ≤ 9 − y 2 , 0 ≤ y ≤ 2}
= {(x, y) ∈ R2 : x2 + y 2 ≤ 9, x ≥ 0, 0 ≤ y ≤ 2}
so that R is part of the first quadrant inside the disk of radius 3 around the origin, below
the line y = 2.
3
2
1
-3
-2
1
-1
2
3
-1
-2
-3
RR
3. Find R x3 + y 2 dx dy where R is the region in the plane bounded by the y-axis and the
lines y = 2x, y = x + 1. For full credit, compute the integral in two ways: first, perform
the integration in the order dxdy, then in the order dydx.
Solution: A picture of R is good to have
3
2
1
-1.5
-1.0
0.5
-0.5
-1
-2
-3
The lines cross where 2x = x + 1, or x = 1 and y = 2.
2
1.0
1.5
In this case it is easier to integrate with respect to y first (keeping x fixed).
ZZ
Z x=1 Z y=x+1
3
2
3
2
x + y dy dx
x + y dy dx =
x=0
R
Z
x=1
=
x=0
Z x=1
=
=
=
=
=
y=2x
y3
x y+
3
3
y=x+1
dx
y=2x
1
8
x3 (x + 1 − 2x) + (x + 1)3 − x3 dx
3
3
Zx=0
1
1
8
x3 − x4 + (x + 1)3 − x3 dx
3
3
0
4
1
x
x5
1
2 4
4
−
+ (x + 1) − x
4
5
12
3
0
1 1 16 2
1
− +
− −
4 5 12 3 12
19
30
Now, integrating first with respect to x means that we have to divide R in two regions:
R1 := {(x, y) ∈ R : 0 ≤ y ≤ 1, 0 ≤ x ≤ y/2}
and
R2 := {(x, y) ∈ R : 1 ≤ y ≤ 2, y − 1 ≤ x ≤ y/2}
ZZ
3
2
ZZ
3
x + y dx dy =
R
Z
y=1
y=0
=
=
=
x=0
y=1
R2
x3 + y 2 dx dy
!
x=y/2
x3 + y 2 dx
dy
x=y−1
x=y/2
x=y/2
Z y=2 4
x
x4
2
2
+y x
dy +
+y x
dy
4
4
y=0
y=1
x=0
x=y−1
Z y=1 4
Z y=2 4
y
y3
y
y 3 (y − 1)4
2
+
dy +
+
−
− y (y − 1) dy
64
2
64
2
4
y=0
y=1
y=1 5
y=2
5
y
y4
y
y 4 (y − 1)5 y 4 y 4
+
+
+
−
−
+
5 · 64
8 y=0
5 · 64
8
20
4
4 y=1
1
1
8 1 1
1
1 1
3
1
+2−
−4+ + − =
+ + =
+ =
10
20
3 4 3
20 3 4
10 3
19
30
Z
=
x + y dx dy +
!
Z x=y/2
Z y=2 Z
3
2
x + y dx dy +
R1
=
=
ZZ
2
y=1
Petronela Radu
3