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The University of Zambia
Department of Mathematics and Statistics
Academic Year 2022/
MAT 3110 Test 3
1st October, 2023 11.00 - 13.
Instructions:
- This paper has 2 pages and 4 questions. Answer all the questions.
- Duration is 2 hours. Total marks is 60 marks
Name: ......................................................................................
Computer Number: ..........................................................................
Department: .......................................................................................
Question: 1 2 3 4 Total
Points: 15 15 15 15 60
Score:
- (a) Evaluate the double integral [3]
Z 3
0
Z 0
− 2
x^2 y − 2 xy
dydx.
(b) Evaluate the double integral [5]
Z ∞
0
Z ∞
0
(1 + x^2 + y^2 )^2
dxdy.
(c) Suppose that the area of a region in the polar coordinate plane is given by the [7] double integral A =
Z 34 π π 4
Z (^) 2 sin θ
csc θ
r drdθ.
Sketch the region and find its area.
Total marks : 15
- (a) Evaluate the triple integral [3]
Z (^) e
1
Z (^) e
1
Z (^) e
1
xyz dx dy dz.
(b) Use an appropriate triple integral to determine the volume the polyhedron (a solid [7] with flat faces) show below with vertices (0, 0 , 0), (0, 0 , 1), (1. 2 , 0), (1, 0 , 0) and (0, 2 , 0).
x
y
z
( 0 , 0 , 0 )
( 1 , 0 , 0 )
( 0 , 2 , 0 )
( 1 , 2 , 0 )
( 0 , 0 , 1 )
(c) Evaluate the triple integral [5]
Z 2
0
Z (^4) −x 2
0
Z (^) x
0
2 sin (2z) 4 − z
dy dz dx.
Total marks : 15
- (a) Evaluate the cylindrical coordinate integral [5]
Z (^2) π
0
Z 1
0
Z √ 2 −r 2
r
r dz dr dθ.
(b) Give the limits of integration for evaluating the integral [3]
Z Z Z
E
f (r, θ, z) r dz dr dθ where E is the solid that is bounded below by the plane z = 0, on the side by the cylinder r = 4 cos θ, and on top by the paraboloid z = 3r^2.
(c) Use an appropriate triple integral to find the volume of the solid that is bounded [7] on the sides by the cylinders x^2 + y^2 = 1 and x^2 + y^2 = 4, bounded below by the cone z = −
p x^2 + y^2 , and bounded above by the cone z =
p x^2 + y^2.
Total marks : 15
- (a) Evaluate the spherical coordinate integral [5]
Z (^2) π
0
Z π 4
0
Z 2
0
ρ^2 sin ϕ dρ dϕ dθ
(b) Give the limits of integration for evaluating the integral [3] Z Z Z
E
f (ρ, θ, ϕ) ρ^2 sin ϕ dρ dϕ dθ.
where E is the solid with points satisfying the inequalities x^2 + y^2 + z^2 ≤ 4 and y ≥ 0. (c) Use a triple integral in spherical coordinates to find the volume of the solid bounded [7] below by z = 0, on the sides by the sphere ρ = 2, and above by the cone ϕ = π 3. Total marks : 15
End of Test!
Z 34 π π 4
Z (^) 2 sin θ
csc θ
r drdθ =
Z 34 π π 4
r^2 2
2 sin θ
csc θ
dθ✓
Z 34 π π 4
4 sin^2 θ 2
csc^2 θ 2
dθ✓
Z 34 π π 4 1 − cos(2θ) − csc^2 θ 2
dθ✓
θ − sin(2θ) 2
cot θ 2
^34 π
π 4
3 π 4
π 4
π 2
- (a) Z (^) e
1
Z (^) e
1
Z (^) e
1
xyz dx dy dz =
Z (^) e
1
x dx ×
Z (^) e
1
y dy ×
Z (^) e
1
z dz
= ln x e 1 × ln y e 1 × ln z e 1 ✓ = (1 − 0) × (1 − 0) × (1 − 0)✓ = 1 ✓
(b) The solid is bounded by the planes z = 0 at the bottom, and on the sides by the planes x + z = 1, y = 0, x = 0 and y + 2z = 2. Therefore,
Volume =
Z 2
0
Z 2 − 2 y
0
Z (^1) −z
0
dx dz dy✓✓✓
Z 2
0
Z 2 − 2 y
0
1 − z dz dy✓
Z 2
0
z − z^2 2
^2 − 2 y
0
dy
Z 2
0
y 2
(2 − y)^2 8 dy✓
y − y^2 4
(2 − y)^3 24
0
(c)
Z (^4)
0
Z √ 4 −z
0
Z (^) x
0
2 sin (2z) 4 − z dy dz dx =
Z 2
0
Z (^4) −x 2
0
Z (^) x
0
2 sin (2z) 4 − z dy dx dz✓✓
Z 2
0
Z (^4) −x 2
0
2 x sin (2z) 4 − z dx dz✓
Z 2
0
x^2 sin (2z) 4 − z
√ 4 −z
0
dz
Z 2
0
(4 − z) sin (2z) 4 − z dz✓
Z 2
0
sin (2z) dz
cos (2z) 2
0 = 1 − cos 8 2
- (a)
Z (^2) π
0
Z π 4
0
Z 2
0
ρ^2 sin ϕ dρ dϕ dθ =
Z (^2) π
0
dθ ×
Z π 4
0
sin ϕ dϕ ×
Z 2
0
ρ^2 dρ✓✓
= 2π × [− cos ϕ]
π 4 0 ×
ρ^3 3
0
= 2π ×
×
8 π 3
(b) Z (^) π
0
Z (^) π
0
Z 2
0
f (ρ, θ, ϕ) ρ^2 sin ϕ dρ dϕ dθ✓✓✓
(c)
Volume =
Z (^2) π
0
Z π 2 π 3
Z 2
0
ρ^2 sin ϕ dρ dϕ dθ✓✓✓
Z (^2) π
0
dθ ×
Z π 2 π 3 sin ϕ dϕ ×
Z 2
0
ρ^2 dρ✓✓
= 2π × [− cos ϕ]
π 2 π 3 ×
ρ^3 3
0
= 2π ×
×
8 π 3
