MAT2615 Assignment 4
(COMPLETE ANSWERS)
2025 - DUE 2025
For assistance contact
Email:
, MAT2615 Assignment 4 (2025) – Worked Solutions
This document provides step-by-step solutions for all five questions, with brief figures
where helpful.
Question 1
Region V is bounded above by the hemisphere z = sqrt(1 − x^2 − y^2) (unit sphere)
and below by the cone z = sqrt(x^2 + y^2) − 1. The two surfaces meet along the circle
x^2 + y^2 = 1, z = 0.
(a) Volume of V
Using cylindrical coordinates (x = r cosθ, y = r sinθ, z = z):
V = ∫_{θ=0}^{2π} ∫_{r=0}^{1} ∫_{z=r−1}^{sqrt(1−r^2)} r dz dr dθ = 2π ∫_{0}^{1} r
[ sqrt(1−r^2) − r + 1 ] dr.
Compute the three elementary integrals: ∫_0^1 r sqrt(1−r^2) dr = 1/3, ∫_0^1 r^2 dr =
1/3, ∫_0^1 r dr = 1/2. Hence V = 2π(1/3 − 1/3 + 1/2) = π.
Answer: V = 3.14159 (exactly π).
Remark: A spherical-coordinate setup is possible but piecewise (the cone does not
pass through the origin). Cylindrical coordinates give a clean one-piece integral.
(b) Sketch S and the XY-projection
• The boundary curve is the circle x^2 + y^2 = 1 at z = 0. • XY-projection of S is the
unit disk; a helpful r–z cross-section is shown below.
(COMPLETE ANSWERS)
2025 - DUE 2025
For assistance contact
Email:
, MAT2615 Assignment 4 (2025) – Worked Solutions
This document provides step-by-step solutions for all five questions, with brief figures
where helpful.
Question 1
Region V is bounded above by the hemisphere z = sqrt(1 − x^2 − y^2) (unit sphere)
and below by the cone z = sqrt(x^2 + y^2) − 1. The two surfaces meet along the circle
x^2 + y^2 = 1, z = 0.
(a) Volume of V
Using cylindrical coordinates (x = r cosθ, y = r sinθ, z = z):
V = ∫_{θ=0}^{2π} ∫_{r=0}^{1} ∫_{z=r−1}^{sqrt(1−r^2)} r dz dr dθ = 2π ∫_{0}^{1} r
[ sqrt(1−r^2) − r + 1 ] dr.
Compute the three elementary integrals: ∫_0^1 r sqrt(1−r^2) dr = 1/3, ∫_0^1 r^2 dr =
1/3, ∫_0^1 r dr = 1/2. Hence V = 2π(1/3 − 1/3 + 1/2) = π.
Answer: V = 3.14159 (exactly π).
Remark: A spherical-coordinate setup is possible but piecewise (the cone does not
pass through the origin). Cylindrical coordinates give a clean one-piece integral.
(b) Sketch S and the XY-projection
• The boundary curve is the circle x^2 + y^2 = 1 at z = 0. • XY-projection of S is the
unit disk; a helpful r–z cross-section is shown below.
