,MAT2615 Assignment 4 (COMPLETE ANSWERS) 2025 –
DUE 2025..WE WISH YOU A GOOD LUCK
1. (Section 14.6, Chapter 17)
Let V be a region in R3 bounded above by the hemisphere z =
p1 − x2 − y2 and below by
the cone z = px2 + y2 − 1. Let S be the surface of V (consisting
of the hemisphere on top
and the paraboloid below).
(a) Compute the volume of V using spherical coordinates. (10)
(b) Sketch S and the XY-projection of S. (3)
(c) Use a surface integral to evaluate the area of S. (
[21]
(a) Compute the volume of VVV using spherical coordinates
𝑇ℎ𝑒 𝑟𝑒𝑔𝑖𝑜𝑛 𝑉𝑉𝑉 𝑖𝑠 𝑏𝑜𝑢𝑛𝑑𝑒𝑑 𝑎𝑏𝑜𝑣𝑒 𝑏𝑦 𝑡ℎ𝑒 ℎ𝑒𝑚𝑖𝑠𝑝ℎ𝑒𝑟𝑒 𝑧
= 1 − 𝑥2 − 𝑦2𝑧 = \𝑠𝑞𝑟𝑡{1 − 𝑥^2 − 𝑦^2}𝑧
= 1 − 𝑥2 − 𝑦2 𝑎𝑛𝑑 𝑏𝑒𝑙𝑜𝑤 𝑏𝑦 𝑡ℎ𝑒 𝑐𝑜𝑛𝑒 𝑧 = 𝑥2 + 𝑦2 − 1𝑧
= 𝑥^2 + 𝑦^2 − 1𝑧
= 𝑥2 + 𝑦2
− 1. 𝑊𝑒 𝑛𝑒𝑒𝑑 𝑡𝑜 𝑐𝑜𝑚𝑝𝑢𝑡𝑒 𝑡ℎ𝑒 𝑣𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑡ℎ𝑒 𝑟𝑒𝑔𝑖𝑜𝑛 𝑒𝑛𝑐𝑙𝑜𝑠𝑒𝑑 𝑏𝑦 𝑡ℎ
Step 1: Convert the equations to spherical coordinates.
In spherical coordinates:
𝑥 = 𝜌𝑠𝑖𝑛 𝜃𝑐𝑜𝑠 𝜙𝑥 = \𝑟ℎ𝑜 \𝑠𝑖𝑛\𝑡ℎ𝑒𝑡𝑎 \𝑐𝑜𝑠\𝑝ℎ𝑖𝑥 =
𝜌𝑠𝑖𝑛𝜃𝑐𝑜𝑠𝜙
𝑦 = 𝜌𝑠𝑖𝑛 𝜃𝑠𝑖𝑛 𝜙𝑦 = \𝑟ℎ𝑜 \𝑠𝑖𝑛\𝑡ℎ𝑒𝑡𝑎 \𝑠𝑖𝑛\𝑝ℎ𝑖𝑦 =
𝜌𝑠𝑖𝑛𝜃𝑠𝑖𝑛𝜙
, 𝑧 = 𝜌𝑐𝑜𝑠 𝜃𝑧 = \𝑟ℎ𝑜 \𝑐𝑜𝑠\𝑡ℎ𝑒𝑡𝑎𝑧 = 𝜌𝑐𝑜𝑠𝜃
𝑇ℎ𝑒 ℎ𝑒𝑚𝑖𝑠𝑝ℎ𝑒𝑟𝑒 𝑧 = 1 − 𝑥2 − 𝑦2𝑧
= \𝑠𝑞𝑟𝑡{1 − 𝑥^2 − 𝑦^2}𝑧
= 1 − 𝑥2 − 𝑦2 𝑏𝑒𝑐𝑜𝑚𝑒𝑠:
𝑧 = 1 − 𝜌2𝑠𝑖𝑛 2𝜃𝑧 = \𝑠𝑞𝑟𝑡{1 − \𝑟ℎ𝑜^2 \𝑠𝑖𝑛^2\𝑡ℎ𝑒𝑡𝑎}𝑧
= 1 − 𝜌2𝑠𝑖𝑛2𝜃
This is the upper boundary of the region. However, in spherical
coordinates, this is the upper surface of the region where ρ\rhoρ
will be constrained.
𝑇ℎ𝑒 𝑐𝑜𝑛𝑒 𝑧 = 𝑥2 + 𝑦2 − 1𝑧 = 𝑥^2 + 𝑦^2 − 1𝑧
= 𝑥2 + 𝑦2 − 1 𝑏𝑒𝑐𝑜𝑚𝑒𝑠:
𝑧 = 𝜌2𝑠𝑖𝑛 2𝜃 − 1𝑧 = \𝑟ℎ𝑜^2 \𝑠𝑖𝑛^2\𝑡ℎ𝑒𝑡𝑎 − 1𝑧
= 𝜌2𝑠𝑖𝑛2𝜃 − 1
This is the lower boundary of the region.
Step 2: Set up the volume integral.
To find the volume, we will integrate over the region bounded
by the hemisphere and the cone. The bounds for ρ\rhoρ,
θ\thetaθ, and ϕ\phiϕ must be determined from the geometry of
the problem.
ρ\rhoρ will range from 0 to 1 because the maximum value
of ρ\rhoρ is constrained by the hemisphere at z=1z = 1z=1.
θ\thetaθ will range from 0 to π/2\pi/2π/2, assuming the
symmetry is in the first quadrant (in x,y,zx, y, zx,y,z-
space).
ϕ\phiϕ will range from 000 to 2π2\pi2π.
DUE 2025..WE WISH YOU A GOOD LUCK
1. (Section 14.6, Chapter 17)
Let V be a region in R3 bounded above by the hemisphere z =
p1 − x2 − y2 and below by
the cone z = px2 + y2 − 1. Let S be the surface of V (consisting
of the hemisphere on top
and the paraboloid below).
(a) Compute the volume of V using spherical coordinates. (10)
(b) Sketch S and the XY-projection of S. (3)
(c) Use a surface integral to evaluate the area of S. (
[21]
(a) Compute the volume of VVV using spherical coordinates
𝑇ℎ𝑒 𝑟𝑒𝑔𝑖𝑜𝑛 𝑉𝑉𝑉 𝑖𝑠 𝑏𝑜𝑢𝑛𝑑𝑒𝑑 𝑎𝑏𝑜𝑣𝑒 𝑏𝑦 𝑡ℎ𝑒 ℎ𝑒𝑚𝑖𝑠𝑝ℎ𝑒𝑟𝑒 𝑧
= 1 − 𝑥2 − 𝑦2𝑧 = \𝑠𝑞𝑟𝑡{1 − 𝑥^2 − 𝑦^2}𝑧
= 1 − 𝑥2 − 𝑦2 𝑎𝑛𝑑 𝑏𝑒𝑙𝑜𝑤 𝑏𝑦 𝑡ℎ𝑒 𝑐𝑜𝑛𝑒 𝑧 = 𝑥2 + 𝑦2 − 1𝑧
= 𝑥^2 + 𝑦^2 − 1𝑧
= 𝑥2 + 𝑦2
− 1. 𝑊𝑒 𝑛𝑒𝑒𝑑 𝑡𝑜 𝑐𝑜𝑚𝑝𝑢𝑡𝑒 𝑡ℎ𝑒 𝑣𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑡ℎ𝑒 𝑟𝑒𝑔𝑖𝑜𝑛 𝑒𝑛𝑐𝑙𝑜𝑠𝑒𝑑 𝑏𝑦 𝑡ℎ
Step 1: Convert the equations to spherical coordinates.
In spherical coordinates:
𝑥 = 𝜌𝑠𝑖𝑛 𝜃𝑐𝑜𝑠 𝜙𝑥 = \𝑟ℎ𝑜 \𝑠𝑖𝑛\𝑡ℎ𝑒𝑡𝑎 \𝑐𝑜𝑠\𝑝ℎ𝑖𝑥 =
𝜌𝑠𝑖𝑛𝜃𝑐𝑜𝑠𝜙
𝑦 = 𝜌𝑠𝑖𝑛 𝜃𝑠𝑖𝑛 𝜙𝑦 = \𝑟ℎ𝑜 \𝑠𝑖𝑛\𝑡ℎ𝑒𝑡𝑎 \𝑠𝑖𝑛\𝑝ℎ𝑖𝑦 =
𝜌𝑠𝑖𝑛𝜃𝑠𝑖𝑛𝜙
, 𝑧 = 𝜌𝑐𝑜𝑠 𝜃𝑧 = \𝑟ℎ𝑜 \𝑐𝑜𝑠\𝑡ℎ𝑒𝑡𝑎𝑧 = 𝜌𝑐𝑜𝑠𝜃
𝑇ℎ𝑒 ℎ𝑒𝑚𝑖𝑠𝑝ℎ𝑒𝑟𝑒 𝑧 = 1 − 𝑥2 − 𝑦2𝑧
= \𝑠𝑞𝑟𝑡{1 − 𝑥^2 − 𝑦^2}𝑧
= 1 − 𝑥2 − 𝑦2 𝑏𝑒𝑐𝑜𝑚𝑒𝑠:
𝑧 = 1 − 𝜌2𝑠𝑖𝑛 2𝜃𝑧 = \𝑠𝑞𝑟𝑡{1 − \𝑟ℎ𝑜^2 \𝑠𝑖𝑛^2\𝑡ℎ𝑒𝑡𝑎}𝑧
= 1 − 𝜌2𝑠𝑖𝑛2𝜃
This is the upper boundary of the region. However, in spherical
coordinates, this is the upper surface of the region where ρ\rhoρ
will be constrained.
𝑇ℎ𝑒 𝑐𝑜𝑛𝑒 𝑧 = 𝑥2 + 𝑦2 − 1𝑧 = 𝑥^2 + 𝑦^2 − 1𝑧
= 𝑥2 + 𝑦2 − 1 𝑏𝑒𝑐𝑜𝑚𝑒𝑠:
𝑧 = 𝜌2𝑠𝑖𝑛 2𝜃 − 1𝑧 = \𝑟ℎ𝑜^2 \𝑠𝑖𝑛^2\𝑡ℎ𝑒𝑡𝑎 − 1𝑧
= 𝜌2𝑠𝑖𝑛2𝜃 − 1
This is the lower boundary of the region.
Step 2: Set up the volume integral.
To find the volume, we will integrate over the region bounded
by the hemisphere and the cone. The bounds for ρ\rhoρ,
θ\thetaθ, and ϕ\phiϕ must be determined from the geometry of
the problem.
ρ\rhoρ will range from 0 to 1 because the maximum value
of ρ\rhoρ is constrained by the hemisphere at z=1z = 1z=1.
θ\thetaθ will range from 0 to π/2\pi/2π/2, assuming the
symmetry is in the first quadrant (in x,y,zx, y, zx,y,z-
space).
ϕ\phiϕ will range from 000 to 2π2\pi2π.
