Homework 7 Solution
1. Compute the surface integral ∬𝑆 𝑥 2 𝑧 𝑑𝑆 where S is that part of the cone 𝑧 = √𝑥 2 + 𝑦 2
that lies between the two planes 𝑧 = 1 and 𝑧 = 4.
The surface can be written in the form 𝑧 = √𝑥 2 + 𝑦 2 . So, the normal vector is
𝑥
𝑦
⃑ = ⟨−𝑓𝑥 , −𝑓𝑦 , 1⟩ = ⟨−
𝑁
,−
, 1⟩
√𝑥 2 + 𝑦 2 √𝑥 2 + 𝑦 2
2
2
⃑ ‖ = √ 2𝑥 2 + 2𝑦 2 + 1 = √2. We can write 𝑑𝑆 as
The magnitude of the normal vector is ‖𝑁
𝑥 +𝑦
𝑥 +𝑦
⃑ ‖𝑑𝐴 = √2𝑑𝐴
𝑑𝑆 = ‖𝑁
2
∬ 𝑥 𝑧 𝑑𝑆 = ∬ (𝑥
𝑆
2 √𝑥 2
z
+ 𝑦 2 ) √2 𝑑𝐴
𝑅
𝜃=2𝜋 𝑟=4
𝜃=2𝜋
∫ (𝑟 2 cos2 𝜃 𝑟)𝑟𝑑𝑟𝑑𝜃 = √2 ∫ [𝑟 4 cos 2 𝜃]𝑑𝜃
= √2 ∫
𝜃=0
𝑟=1
y
𝜃=0
𝜃=2𝜋
𝜃=2𝜋
𝑟=4
𝑟5
45 1
= √2 ∫ [ cos2 𝜃]
𝑑𝜃 = √2 ( − ) ∫ [cos 2 𝜃]𝑑𝜃
5
5 5
𝑟=1
𝜃=0
x
𝜃=0
𝜃=2𝜋
2𝜋
1023
1023
1
1023
= √2 (
) ∫ [1 + cos 2𝜃]𝑑𝜃 = √2 (
) [𝜃 + 𝑠𝑖𝑛2𝜃 ] = 2√2𝜋 (
).
10
10
2
10
0
𝜃=0
2. Compute the surface integral ∬𝑆 (𝑥 + 𝑦) 𝑑𝑆 where S is that part of the plane 2𝑥 + 3𝑦 +
𝑧 = 3 lying in the first octant.
The surface can be written in the form 𝑧 = 3 − 2𝑥 − 3𝑦. So, the normal
vector is
⃑ = ⟨−𝑓𝑥 , −𝑓𝑦 , 1⟩ = ⟨2, 3, 1⟩
𝑁
z
(0, 0,3)
⃑ ‖ = √14. We can write 𝑑𝑆 as
The magnitude of the normal vector is ‖𝑁
⃑ ‖𝑑𝐴 = √14𝑑𝐴
𝑑𝑆 = ‖𝑁
3−2𝑥
𝑥=3/2 𝑦= 3
∬(𝑥 + 𝑦)𝑑𝑆 = ∬(𝑥 + 𝑦)√14 𝑑𝐴 = √14 ∫
𝑆
𝑅
𝑥=
3
2
= √14 ∫ [𝑥𝑦 +
𝑥=0
𝑥=0
3−2𝑥
𝑦=
3
𝑦2
2
]
𝑦=0
𝑥=
∫
𝑦=0
3
2
1 𝑥 4𝑥 2
𝑑𝑥 = √14 ∫ [ + −
] 𝑑𝑥
2 3
9
𝑥=0
3/2
1
1
= √14 [2𝑥 − 𝑥 3 + 𝑥 4 ]
2
8
0
(𝑥 + 𝑦)𝑑𝑦𝑑𝑥
5
= √14.
8
-1-
x (1.5, 0, 0)
(0,1, 0) y
3. Find the surface area of the cone 𝑧 = √𝑥 2 + 𝑦 2 below the plane 𝑧 = 1.
The surface can be written in the form 𝑧 = √𝑥 2 + 𝑦 2 . So, the normal vector is
𝑥
𝑦
⃑ = ⟨−𝑓𝑥 , −𝑓𝑦 , 1⟩ = ⟨−
𝑁
,−
, 1⟩
√𝑥 2 + 𝑦 2 √𝑥 2 + 𝑦 2
2
2
⃑ ‖ = √ 2𝑥 2 + 2𝑦 2 + 1 = √2. We can write 𝑑𝑆 as
The magnitude of the normal vector is ‖𝑁
𝑥 +𝑦
𝑥 +𝑦
⃑ ‖𝑑𝐴 = √2 𝑑𝐴.
𝑑𝑆 = ‖𝑁
z
𝜃=2𝜋 𝑟=1
Surface Area = ∬ 1𝑑𝑆 = ∬ √2 𝑑𝐴 = ∫
𝑆
𝑅
𝜃=0
𝜃=2𝜋
𝜃=2𝜋
𝜃=0
𝜃=0
∫ √2 𝑟𝑑𝑟𝑑𝜃
z 1
𝑟=0
𝑟=1
1
1
= ∫ [ √2 𝑟]
𝑑𝜃 = √2 ∫ 𝑑𝜃 = √2𝜋
2
2
𝑟=0
y
x
4. Find the surface area of the part of the cylinder 𝑥 2 + 𝑧 2 = 4 that lies directly above the
rectangle 0 ≤ 𝑥 ≤ 1, 0 ≤ 𝑦 ≤ 4 in the 𝑥𝑦 – plane.
The surface can be written in the form 𝑧 = √4 − 𝑥 2 . So, the normal vector is
𝑥
⃑ = ⟨−𝑓𝑥 , −𝑓𝑦 , 1⟩ = ⟨
𝑁
, 0, 1⟩
√4 − 𝑥 2
2
⃑ ‖ = √ 𝑥 2 + 1 = 2 2 . We can write 𝑑𝑆 as
The magnitude of the normal vector is ‖𝑁
√4−𝑥
4−𝑥
⃑ ‖𝑑𝐴 =
𝑑𝑆 = ‖𝑁
Surface Area = ∬ 1 𝑑𝑆 = ∬
𝑆
𝑅
𝑥=1
=2 ∫ [
𝑥=0
2
√4 − 𝑥 2
2
√4 − 𝑥 2
𝑥=1 𝑦=4
𝑑𝐴 = 2 ∫
∫
𝑥=0 𝑦=0
𝑑𝐴
2
√4 − 𝑥 2
𝑥=1
𝑑𝑦𝑑𝑥 = 2 ∫ [
𝑥=0
2
√4 − 𝑥 2
𝑦=4
𝑦]
𝑑𝑥
𝑦=0
𝑥 1 8
] 𝑑𝑥 = 16 [sin−1 ( )] = 𝜋
2 0 3
√4 − 𝑥 2
8
5. Find the mass of a thin funnel in the shaped of a paraboloid 𝑧 = 𝑥 2 + 𝑦 2 , 1 ≤ 𝑧 ≤ 4 if its
density function 𝜌(𝑥, 𝑦, 𝑧) = 10 − 𝑧.
z
The surface can be written in the form 𝑧 = 𝑥 2 + 𝑦 2 . So, the normal
vector is
⃑ = ⟨−𝑓𝑥 , −𝑓𝑦 , 1⟩ = ⟨−2𝑥, − 2𝑦, 1⟩
𝑁
⃑ ‖ = √4𝑥 2 + 4𝑦 2 + 1. We can
The magnitude of the normal vector is ‖𝑁
write 𝑑𝑆 as
⃑ ‖𝑑𝐴 = √4𝑥 2 + 4𝑦 2 + 1𝑑𝐴
𝑑𝑆 = ‖𝑁
-2-
y
x
Mass = ∬ 𝜌(𝑥, 𝑦, 𝑧)𝑑𝑆 = ∬(10 − 𝑧)𝑑𝑆 = ∬(10 − (𝑥 2 + 𝑦 2 )) √4𝑥 2 + 4𝑦 2 + 1 𝑑𝐴
𝑆
𝑆
𝑅
𝜃=2𝜋 𝑟=2
∫ (10 − 𝑟 2 ) √4𝑟 2 + 1 𝑟 𝑑𝑟𝑑𝜃 = ∫
= ∫
𝜃=0
𝜃=2𝜋 𝑟=2
𝑟=1
𝜃=0
𝜃=2𝜋
∫ (10 − 𝑟 2 ) 𝑟√4𝑟 2 + 1 𝑑𝑟𝑑𝜃
𝑟=1
𝑟=2
3 𝑟=2
3
1
1
2 )(4𝑟 2
(10
= ∫ ([
−𝑟
+ 1)2 ]
+ ∫ ( 𝑟(4𝑟 2 + 1)2 ) 𝑑𝑟) 𝑑𝜃
12
6
𝑟=1
𝜃=0
𝑟=1
𝜃=2𝜋
3 𝑟=2
5 𝑟=2
1
1
2
2
2
= ∫ ([ (10 − 𝑟 )(4𝑟 + 1)2 ]
+
[(4𝑟 + 1)2 ] ) 𝑑𝜃
12
120
𝑟=1
𝑟=1
𝜃=0
𝜃=2𝜋
1309√17 95√5
1309√17 95√5
=(
−
) ∫ 1 𝑑𝜃 = (
−
) 𝜋.
120
24
60
12
𝜃=0
6. Evaluate the flux of the water through the paraboloid cylinder 𝑦 = 𝑥 2 , 0 ≤ 𝑥 ≤ 2, 0 ≤ 𝑧 ≤
3 if the velocity vector is 𝐹 = ⟨3𝑧 2 , 6, 6𝑥𝑧⟩.
First, observe that at any given point, the normal vectors for the paraboloid 𝑦 = 𝑥 2 are ± < 2𝑥, −1, 0 >.
For the normal vector to point upward, we need a positive 𝑦-component. In this case,
⃑ = < −𝑓𝑥 , 1, −𝑓𝑧 > = < −2𝑥, 1, 0 >
𝑁
z
is such a normal vector.
⃑ 𝑑𝐴
flux = ∬ 𝐹 ∙ 𝑁
𝑅
= ∬ < 3𝑧 2 , 6,6𝑥𝑧 > ∙ < −2𝑥, 1, 0 > 𝑑𝐴
𝑅
x
𝑧=3 𝑥=2
= ∬(−6𝑥𝑧 2 + 6)𝑑𝐴 = ∫
𝑅
y
∫ (−6𝑥𝑧 2 + 6) 𝑑𝑥𝑑𝑧 = −72.
𝑧=0 𝑥=0
7. Evaluate the flux of the water through the unit sphere lies in the first octant if the velocity
vector is 𝐹 = 𝑥 𝐣.
First, observe that at any given point, the normal vectors for the sphere in the first octant 𝑧 =
√1 − 𝑥 2 − 𝑦 2 are ± <
𝑥
√1−𝑥 2 −𝑦2
,
𝑦
√1−𝑥 2 −𝑦 2
z
, 1 >. For the normal
vector to point upward, we need a positive 𝑧-component. In this
case,
𝑥
𝑦
⃑ = < −𝑓𝑥 , −𝑓𝑦 , 1 > =<
𝑁
,
,1 >
√1 − 𝑥 2 − 𝑦 2 √1 − 𝑥 2 − 𝑦 2
x 2  y 2  z 2 1
is such a normal vector.
y
x
-3-
⃑ 𝑑𝐴 = ∬ < 0, 𝑥, 0 > ∙ <
flux = ∬ 𝐹 ∙ 𝑁
𝑅
𝑅
𝜃=
=∬
𝑥𝑦
√1 − 𝑥 2 − 𝑦 2
𝑅
𝜃=𝜋/2 𝑟=1
𝑑𝐴 = ∫
𝑟 3 cos 𝜃 sin 𝜃
√1 − 𝑟 2
, 1 > 𝑑𝐴
𝑑𝑟𝑑𝜃
1
𝑟=0
= ∫ cos 𝜃 sin 𝜃 ([−𝑟 2 (1 −
𝜃=0
1 𝑟=1
𝑟 2 )2 ]
𝑟=0
𝜋
2
𝑟=1
1
+ ∫ (2𝑟(1 − 𝑟 2 )2 ) 𝑑𝑟) 𝑑𝜃
𝑟=0
𝜃=
1 𝑟=1
= ∫ cos 𝜃 sin 𝜃 ([−𝑟 2 (1 − 𝑟 2 )2 ]
𝜃=0
=
√1 − 𝑥 2 − 𝑦 2 √1 − 𝑥 2 − 𝑦 2
𝜃=0 𝑟=0
𝜋
𝜃=
2
𝜃=
∫
𝑦
,
∫ 𝑟 2 cos 𝜃 sin 𝜃 𝑟 (1 − 𝑟 2 )−2 𝑑𝑟𝑑𝜃
= ∫
𝜃=0
𝜋
2 𝑟=1
𝑥
𝜋
2
3 𝑟=1
2
2
− [(1 − 𝑟 2 )2 ] ) 𝑑𝜃 =
∫ cos 𝜃 sin 𝜃 𝑑𝜃
3
3
𝑟=0
𝑟=0
𝜃=0
𝜋
1
1
(sin2 𝜃)02 = .
3
3
8. Use the divergence Theorem to find the flux of the vector 𝐹 = ⟨𝑥 3 , 𝑦 3 , 𝑧 3 ⟩ across the
surface of the hemisphere 𝑧 = √1 − 𝑥 2 − 𝑦 2 and 𝑧 = 0.
z
We show a sketch of the solid in opposite Figure. Notice that to
compute the flux directly, we must consider the two different portions
of 𝑆 (the surface of the paraboloid and its base in the 𝑥𝑦-plane)
separately. Alternatively, observe that the divergence of 𝐹 is given by
∇ · 𝐹 (𝑥, 𝑦, 𝑧) = ∇ · < 𝑥 3 , 𝑦 3 , 𝑧 3 > = 3𝑥 2 + 3𝑦 2 + 3𝑧 2 .
y
From the Divergence Theorem, we now have that the flux of 𝐹 over 𝑆
x
is given by
𝜃=2𝜋 𝜙=𝜋/2 𝜌=1
∯(𝐹 ∙ 𝑛̂) 𝑑𝑆 = ∭(3𝑥 2 + 3𝑦 2 + 3𝑧 2 ) 𝑑𝑉 = ∫
𝑆
𝐷
𝜃=0
∫
𝜃=0
𝜙=0
𝜋
𝜃=2𝜋 𝜙= 2
= ∫
𝜃=0
𝜙=0
𝜌=0
𝜋
𝜃=2𝜋 𝜙= 2
𝜃=2𝜋 𝜙=𝜋/2 𝜌=1
= ∫
∫ 3𝜌2 𝜌2 sin 𝜙 𝑑𝜌 𝑑𝜙 𝑑𝜃
∫
3 5 1
∫ [ 𝜌 ] sin 𝜙 𝑑𝜙 𝑑𝜃
5
0
4
∫ 3 𝜌 sin 𝜙 𝑑𝜌 𝑑𝜙 𝑑𝜃 = ∫
𝜌=0
𝜃=0
𝜃=2𝜋
𝜙=0
𝜋
𝜃=2𝜋
2
3
3
3
6
∫ sin 𝜙 𝑑𝜙 𝑑𝜃 = − ∫ [ cos 𝜙] 𝑑𝜃 = ∫
𝑑𝜃 = 𝜋.
5
5
5
5
0
𝜙=0
𝜃=0
𝜃=0
-4-
9. Use the divergence Theorem to find the flux of the vector 𝐹 = ⟨2𝑥, 3𝑦, 𝑧 2 ⟩ across the
surface of the unit cube has vertices
(0,0,0), (0,0,1), (0,1,0), (1,0,0), (1,1,0), (1,0,1), (0,1,1) and (1,1,1).
Rather than evaluate six surface integrals, we apply the divergence theorem.
Since div 𝐹 = ∇ · 𝐹 = 2 + 3 + 2𝑧, we have from
∯(𝐹 ∙ 𝑛̂) 𝑑𝑆 = ∭(5 + 2𝑧) 𝑑𝑉
𝑆
𝐷
1 1 1
1 1
1
𝑧=1
= ∫ ∫ ∫(5 + 2𝑧) 𝑑𝑥 𝑑𝑦 𝑑𝑧 = ∫ ∫(5 + 2𝑧) 𝑑𝑦 𝑑𝑧 = ∫(5 + 2𝑧) 𝑑𝑧 = (5𝑧 + 𝑧 2 )𝑧=0
= 6.
0 0 0
0 0
0
10.Use the divergence Theorem to find the flux of the vector 𝐹 = ⟨𝑥 3 + sin 𝑧 , 𝑥 2 𝑦 +
2
2
cos 𝑧 , 𝑒 𝑥 +𝑦 ⟩ across the surface of the paraboloid cylinder 𝑧 = 4 − 𝑥 2 , bounded by the
planes 𝑦 + 𝑧 = 5, 𝑥𝑦-plane and 𝑥𝑧-plane.
z  4x 2
z
∇ · 𝐹 (𝑥, 𝑦, 𝑧) = ∇ · < 𝑥 3 , 𝑦 3 , 𝑧 3 > = 3𝑥 2 + 𝑥 2 = 4𝑥 2 .
y z 5
From the Divergence Theorem, we now have that the flux of 𝐹 over
𝑆 is given by
y 0
∯(𝐹 ∙ 𝑛̂) 𝑑𝑆 = ∭(4𝑥 2 ) 𝑑𝑉
𝑆
𝐷
𝑅
z 0
𝑦=5−𝑧
=∬ ∫
𝑦=0
2
4𝑥 𝑑𝑦 𝑑𝐴 = ∬ (4𝑥
2
𝑦=5−𝑧
𝑦)𝑦=0
𝑑𝐴
𝑅
𝑥=2 𝑧=4−𝑥 2
= ∬ [4𝑥 2 (5 − 𝑧)] 𝑑𝐴 = ∫
𝑅
y
x
∫
[4𝑥 2 (5 − 𝑧)] 𝑑𝑧𝑑𝑥 =
𝑥=−2 𝑧=0
-5-
4608
.
35