,MAT2615 Assignment 4 (COMPLETE ANSWERS) 2025 - DUE 2025; 100%
correct solutions and explanations
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]
Let's go step by step to solve this problem:
Given:
The region VVV is bounded above by the hemisphere: z=p1−x2−y2z = p_1
- x^2 - y^2z=p1−x2−y2
The region is bounded below by the cone: z=p(x2+y2)−1z = p(x^2 + y^2) -
1z=p(x2+y2)−1
(a) Compute the volume of VVV using spherical coordinates
We use spherical coordinates where:
x=rcosθ,y=rsinθ,z=ρcosϕx = r\cos\theta, \quad y = r\sin\theta, \quad z =
\rho\cos\phix=rcosθ,y=rsinθ,z=ρcosϕ
and the volume element is
dV=ρ2sinϕ dρ dϕ dθ.dV = \rho^2 \sin\phi \, d\rho \, d\phi \,
d\theta.dV=ρ2sinϕdρdϕdθ.
We need to determine the limits:
Azimuthal angle: 0≤θ≤2π0 \leq \theta \leq 2\pi0≤θ≤2π (full revolution
around zzz-axis)
Radial coordinate:
, o The cone is given by z=p(x2+y2)−1z = p(x^2 + y^2) -
1z=p(x2+y2)−1, which in spherical coordinates is:
ρcosϕ=pρ2sin2ϕ−1.\rho \cos\phi = p\rho^2\sin^2\phi -
1.ρcosϕ=pρ2sin2ϕ−1.
o The hemisphere is given by z=p1−x2−y2z = p_1 - x^2 - y^2z=p1
−x2−y2, which in spherical coordinates is:
ρcosϕ=p1−ρ2sin2ϕ.\rho \cos\phi = p_1 -
\rho^2\sin^2\phi.ρcosϕ=p1−ρ2sin2ϕ.
Polar angle ϕ\phiϕ: This is determined by the intersection of the
hemisphere and the cone.
We set up the volume integral in spherical coordinates:
V=∫02π∫0ϕmax∫ρminρmaxρ2sinϕ dρ dϕ dθ.V = \int_0^{2\pi}
\int_0^{\phi_{\max}} \int_{\rho_{\min}}^{\rho_{\max}} \rho^2 \sin\phi \, d\rho \,
d\phi \, d\theta.V=∫02π∫0ϕmax∫ρminρmaxρ2sinϕdρdϕdθ.
(b) Sketch SSS and its XY-projection
The surface consists of:
o A hemisphere z=p1−x2−y2z = p_1 - x^2 - y^2z=p1−x2−y2.
o A paraboloid z=p(x2+y2)−1z = p(x^2 + y^2) - 1z=p(x2+y2)−1.
The projection onto the XY-plane will be a circular region where these
surfaces intersect.
(c) Use a surface integral to evaluate the area of SSS
The surface area integral is given by:
A=∬SdS.A = \iint_S dS.A=∬SdS.
For each surface:
1. For the hemisphere: The surface element is:
dS=R2sinϕ dϕ dθ.dS = R^2 \sin\phi \, d\phi \, d\theta.dS=R2sinϕdϕdθ.
2. For the paraboloid: We use the normal vector and surface element
calculations.
The total surface area is:
correct solutions and explanations
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]
Let's go step by step to solve this problem:
Given:
The region VVV is bounded above by the hemisphere: z=p1−x2−y2z = p_1
- x^2 - y^2z=p1−x2−y2
The region is bounded below by the cone: z=p(x2+y2)−1z = p(x^2 + y^2) -
1z=p(x2+y2)−1
(a) Compute the volume of VVV using spherical coordinates
We use spherical coordinates where:
x=rcosθ,y=rsinθ,z=ρcosϕx = r\cos\theta, \quad y = r\sin\theta, \quad z =
\rho\cos\phix=rcosθ,y=rsinθ,z=ρcosϕ
and the volume element is
dV=ρ2sinϕ dρ dϕ dθ.dV = \rho^2 \sin\phi \, d\rho \, d\phi \,
d\theta.dV=ρ2sinϕdρdϕdθ.
We need to determine the limits:
Azimuthal angle: 0≤θ≤2π0 \leq \theta \leq 2\pi0≤θ≤2π (full revolution
around zzz-axis)
Radial coordinate:
, o The cone is given by z=p(x2+y2)−1z = p(x^2 + y^2) -
1z=p(x2+y2)−1, which in spherical coordinates is:
ρcosϕ=pρ2sin2ϕ−1.\rho \cos\phi = p\rho^2\sin^2\phi -
1.ρcosϕ=pρ2sin2ϕ−1.
o The hemisphere is given by z=p1−x2−y2z = p_1 - x^2 - y^2z=p1
−x2−y2, which in spherical coordinates is:
ρcosϕ=p1−ρ2sin2ϕ.\rho \cos\phi = p_1 -
\rho^2\sin^2\phi.ρcosϕ=p1−ρ2sin2ϕ.
Polar angle ϕ\phiϕ: This is determined by the intersection of the
hemisphere and the cone.
We set up the volume integral in spherical coordinates:
V=∫02π∫0ϕmax∫ρminρmaxρ2sinϕ dρ dϕ dθ.V = \int_0^{2\pi}
\int_0^{\phi_{\max}} \int_{\rho_{\min}}^{\rho_{\max}} \rho^2 \sin\phi \, d\rho \,
d\phi \, d\theta.V=∫02π∫0ϕmax∫ρminρmaxρ2sinϕdρdϕdθ.
(b) Sketch SSS and its XY-projection
The surface consists of:
o A hemisphere z=p1−x2−y2z = p_1 - x^2 - y^2z=p1−x2−y2.
o A paraboloid z=p(x2+y2)−1z = p(x^2 + y^2) - 1z=p(x2+y2)−1.
The projection onto the XY-plane will be a circular region where these
surfaces intersect.
(c) Use a surface integral to evaluate the area of SSS
The surface area integral is given by:
A=∬SdS.A = \iint_S dS.A=∬SdS.
For each surface:
1. For the hemisphere: The surface element is:
dS=R2sinϕ dϕ dθ.dS = R^2 \sin\phi \, d\phi \, d\theta.dS=R2sinϕdϕdθ.
2. For the paraboloid: We use the normal vector and surface element
calculations.
The total surface area is:
