Q6. The point (0, 0, k) lies outside the sphere
x2 + y2 + z2 - 6z + 8 = 0 if
Answers
Answer:
Step-by-step explanation:
1. Compute the total mass of the solid which is inside the sphere x2 + y2 + z2 = a2 and
outside the sphere x2+y2+z2 = b2 if the density is given by ρ(x, y, z) = c
px2 + y2 + z2
.
Here a, b, c are positive constants and 0 < b < a.
Solution: The total mass is
m =
Z θ=2π
θ=0
Z φ=π
φ=0
Z ρ=a
ρ=b
c
ρ
ρ2 sin φdρdφdθ = 2πc
a2 − b2
2 (− cos φ)
φ=π
φ=0
= 2πc(a2 − b2
)
2. Compute the volume of the solid region which is inside the sphere x2 +y2 +z2 = 2 and
above the paraboloid z = x2 + y2.
Solution:
The intersection of the sphere and the paraboloid is the circle x2 + y2 = 1, z = 1.
Therefore
V =
Z θ=2π
θ=0
Z r=1
r=0
Z z=
√2−r2
z=r2
dzrdrdθ = 2π
Z r=1
r=0
r(
√
2 − r2 − r2
)dr
= −2π
3 (2 − r2
)
3/2
r=1
r=0
−π
2 = −2π
3 +
2π
3
2
√
2 − π
2
=
4
√
2
3 − 7
6
!
π
3. Find the volume inside the sphere ρ = a that lies between the cones φ = π
6
and φ = π
3
.
Solution:
V =
Z θ=2π
θ=0
Z φ=π/3
φ=π/6
Z ρ=a
ρ=0
ρ2 sin φdρdφdθ = 2π
a3
3 (− cos π/3 + cos π/6)
= 2π
a3
3
−1
2 +
√3
2
!
= π(
√3 − 1)
3
a3
4. Find the surface area of that part of the sphere z = pa2 − x2 − y2 which lies within
the cylinder x2 + y2 = ay. Here a is a positive constant.
Solution: The surface area of the graph of z = f(x, y) over a domain D in the x, y
plane is S =
Z Z
D
p1+(∂z/∂x)2 + (∂z/∂x)2dxdy. In this case D is the interior of the
circle x2 + y2 = ay, which in polar coordinates is 0 ≤ θ ≤ π, 0 ≤ r ≤ a sin θ. Therefore
S =
Z Z
x2+y2≤ay s
1 + x2 + y2
a2 − x2 − y2 dxdy =
Z θ=π
θ=0
dθ Z r=a sin θ
r=0
a
√a2 − r2
rdr
= a
Z θ=π
θ=0
−
√
a2 − r2
r=a sin θ
r=0
dθ = a
Z θ=π
θ=0
(−a| cos θ| + a)dθ = a2
(π − 2)
1
5. Find the surface area of that part of the hemisphere of radius √2 centered at the origin
that lies above the square −1 ≤ x ≤ 1, −1 ≤ y ≤ 1.
Solution: The surface area is what you get by subtracting 4 half caps from the area
of the hemisphere z = p2 − x2 − y2. By question 2 on quiz 5 each half cap has area
π
√2(√2 − 1), and therefore the surface area of the canopy is
2π(
√
2)2 − 4π
√
2(√
2 − 1) = 4π(
√
2 − 1)
6. Find the centroid of the region inside the cube 0 ≤ x, y, z ≤ a and below the plane
x + y + z = 2a.
Solution: The centroid of R is given by
x¯ =
RRR
R xdV
V , y¯ =
RRR
R ydV
V , z¯ =
RRR
R zdV
V where V is the volume of R
In this case the plane x+y+z = 2a cuts off 1/6 of the cube so V = 5a3/6. We evaluate
ZZZ
R
xdV by splitting the region R into 2 separate pieces, R1 and R2.
ZZZ
R1
xdV =
Z x=a
x=0
dx Z y=a−x
y=0
dy Z z=a
z=0
xdz =
Z x=a
x=0
ax(a − x)dx = a4
6
ZZZ
R2
xdV =
Z x=a
x=0
dx Z y=a
y=a−x
dy Z z=2a−x−y
z=0
xdz =
Z x=a
x=0
Z y=a
y=a−x
x(2a − x − y)dy
=
Z x=a
x=0
x(2a − x)x − x
2
(a2 − (a − x)
2
)
dx
=
Z x=a
x=0
ax2 − x3
2
dx = a4
3 − a4
8 = 5a4
24
Therefore ¯x = a4/6+5a4/24
5a3/6 = 9a
20, and by symmetry we know that ¯x = ¯y = ¯z = 9a
20.
7. Find the total mass of the region which is above the cone z = px2 + y2 and below the
sphere x2 + y2 + z2 = 1, if the density function is δ(x, y, z)=1+ κ
px2 + y2 + z2, for
a positive constant κ.
Solution: The cone and sphere intersect where φ = π/4 and therefore the total mass is
M =
Z θ=2π
θ=0
dθ Z φ=π/4
φ=0
dφ Z ρ=1
ρ=0
ρ2 sin φ(1 + κρ)dρ
=
Z θ=2π
θ=0
dθ Z φ=π/4
φ=0
dφ Z ρ=1
ρ=0
ρ2 sin φdρ + κ
Z θ=2π
θ=0
dθ Z φ=π/4
φ=0
dφ Z ρ=1
ρ=0
ρ3 sin φdρ
= 2π
3
Z π/4
0
sin φdφ +
κπ
2
Z π/4
0
sin φdφ = π(1 − 1/
√
2)(κ/2+2/3)
Answer:
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