Module MA2E02 (Frolov), Multivariable Calculus
Tutorial Sheet 8
Due: at the end of the tutorial session Tuesday/Thursday, 22/24 March 2016
Name and student number:
1. Consider the lamina that is the portion of the surface 6z − x2 − y 2 − 1 = 0 inside the surface
x2 + y 2 − 9 = 0. The density of the lamina is
δ(x, y, z) = δ0 (9 + x2 + y 2 )
where δ0 is a constant
(a) What is the surface 6z − x2 − y 2 − 1 = 0?
(b) What is the surface x2 + y 2 − 9 = 0?
(c) Sketch the projection of the lamina onto the xy-plane.
(d) Find the mass of the lamina.
Show the details of your work.
Solution:
(a) The surface 6z − x2 − y 2 − 1 = 0 is a paraboloid.
(b) The surface x2 + y 2 − 9 = 0 is a circular cylinder of radius 3.
(c) Below is the lamina σ, and its projection onto the xy-plane
3
2
1
1.5
1.0
0.5
0.0
-3
-2
1
-1
2
3
-1
2
-2
-2
0
0
2
-2
-3
The projection of the surface onto the xy-plane is the disk R: x2 + y 2 ≤ 9.
(d) The mass of the lamina is equal to the following surface integral
s
2 2
ZZ
ZZ
ZZ
∂z
∂z
1
2
2
M=
δ(x, y, z) dS = δ0
(9+x +y ) 1 +
+
dA = δ0
(9+x2 +y 2 )3/2 dA .
∂x
∂y
3
σ
R
R
The simplest way to compute the double integral over R is to use the polar coordinates
3
ZZ
Z 3
1
2π
2π 1
2
2 3/2
2 3/2
2 5/2 M =
δ0
(9 + x + y ) dA =
δ0
(9 + r ) rdr =
δ0 (9 + r ) 3
3
3 5
R
0
0
√
√
2π
162π
=
δ0 35 (4 2 − 1) =
δ0 (4 2 − 1) ≈ 474.01 δ0
15
5
1
2. Consider the surface σ which is the portion of the surface z =
z = 5/3, oriented by downward unit normals.
(a) Sketch the projection of the surface onto the xy-plane.
(b) Find the flux of the vector field F across σ.
x2 +y 2 +1
6
below the plane
F(x, y, z) = 3y j + 2k ;
Show the details of your work.
Solution:
(a) The surface σ is the same as the lamina in Problem 1. Thus, the projection of the
surface onto the xy-plane is the disk R: x2 + y 2 ≤ 9.
(b) Taking into account that through the surface z = f (x, y) oriented by down, the vector
field F(x, y, z) = M i + N j + P k has
ZZ
ZZ ∂f
∂f
+N
− P dA ,
M
flux =
F · n dS =
∂x
∂y
σ
R
we get by using the polar coordinates (M = 0 , N = 3y , P = 2)
ZZ
Z 2π Z 3
2
flux =
y − 2 dA =
(r2 sin2 φ − 2) rdr dφ
R
0
0
Z 2π
4
34
3
9
=
( sin2 φ − 9)dφ = π − 18π = π .
4
4
4
0
3.
(a) Express rectangular coordinates in terms of spherical coordinates
(b) Use triple integral and spherical coordinates to derive the formula for the volume of a
ball of radius R.
(c) Use the Divergence Theorem to find the flux of the vector field
F(x, y, z) = (2x + 15xy 2 + 5z 3 ) i − (y 2 + 3y + 3x2 ) j + (5 + 4z + 2yz − 5y 3 ) k
across the surface σ with outward orientation where σ is given by the equation
x2 + y 2 + z 2 = 4.
Show the details of your work.
Solution :
(a) We have (draw a picture)
x = r cos θ sin φ ,
y = r sin θ sin φ ,
z = r cos φ ,
0 ≤ θ ≤ 2π , 0 ≤ φ ≤ π
(b) We use the spherical coordinates to get (you must give more details here)
ZZZ
Z 2π Z π Z R
4
2
V =
dV =
r dr sin φ dφ dθ = πR3 .
3
V
0
0
0
(c) The surface is a sphere of radius 2. We have
ZZ
ZZZ
ZZZ
flux =
F · n dS =
div F dV =
(2 + 15y 2 − 2y − 3 + 4 + 2y) dV
V
Z ZσZ
ZZZ
ZVZ Z
ZZZ
4 3
2
2
=
(3 + 15y ) dV = 3
dV +
15y dV = 3 π 2 + 5
r2 dV
3
V
V
V
V
ZZZ
= 32π + 5
r2 dV ,
V
2
where we used that the volume of a ball of radius R is 43 πR3 and that
ZZZ
ZZZ
ZZZ
2
2
2
2
3y dV =
(x + y + z ) dV =
r2 dV
V
V
V
due to the symmetry between x, y, z coordinates for a sphere (rotational symmetry).
To compute the last integral we use again spherical coordinates
ZZZ
Z 2π Z π Z 2
4
128
4
2
r dr sin φ dφ dθ = π25 =
π.
r dV =
5
5
0
0
0
V
Thus the flux is
flux = 160π .
4. Let C be the triangle in the plane z = 12 y with vertices (2, 0, 0), (0, 2, 1) and (0, 0, 0) with a
counterclockwise orientation looking down the positive z-axis.
(a) Sketch the triangle and its projection
H onto the xy-plane.
(b) Use Stokes’ Theorem to evaluate C F · dr.
F(x, y, z) = (5x − 2z 2 ) i − (y − 4x) j + (2xz + 4y) k .
(c) Represent C as a union of three line segments C1 , C2 , C3 connecting the points (0, 0, 0)
and (2, 0, 0), (2, 0,H0) and (0, 2, 1), (0, 2, 1) and (0, 0, 0), respectively. Parameterize C1 , C2 and
C3 , and evaluate C F · dr.
Show the details of your work.
Solution :
(a) Below is the triangle, σ, and its projection, R, onto the xy-plane
2.0
1.5
1.0
2.0
0.5
1.0
1.5
0.0
0.0
1.0
0.5
0.5
0.5
1.0
1.5
2.0
0.0
0.5
1.0
(b) According to Stokes’ Theorem
I
ZZ
ZZ
∂z
∂z
F · dr =
(curl F) · n dS =
(curl F) · − i −
j + k dA ,
∂x
∂y
C
σ
R
3
1.5
2.0
where n is the normal vector to the surface z = 12 y looking up the positive z-axis.
To use Stokes’ Theorem we first compute
∂(2xz + 4y) ∂(4x − y)
∂(5x − 2z 2 ) ∂(2xz + 4y)
curl F =
−
i+
−
j
∂y
∂z
∂z
∂x
∂(4x − y) ∂(5x − 2z 2 )
+
−
k
∂x
∂y
= 4 i − 6z j + 4 k ,
and
−
∂z
∂z
1
i−
j + k = − j + k.
∂x
∂y
2
Thus we have
Z 2
I
ZZ
Z 2 Z 2−x
3
3
( (2 − x)2 + 4(2 − x)) dx
F · dr =
(3z + 4) dA =
( y + 4) dy dx =
2
4
0
C
0
0
Z 2R
3
1
( x2 + 4x) dx = 23 + 2 · 22 = 10 .
=
4
4
0
(c) The line segments C1 is parametrized as x = t, y = 0, z = 0, 0 ≤ t ≤ 2 and one gets
Z
Z 2
5
F · dr =
5t dt = 22 = 10 .
2
C1
0
The line segments C2 is parametrized as x = 2 − t, y = t, z = t/2, 0 ≤ t ≤ 2 and one gets
Z
Z
F · dr =
(5x − 2z 2 ) dx − (y − 4x) dy + (2xz + 4y)dz
C2
C
Z 21
Z 2
t(2 − t)
t2
+ 2t) dt =
(−2 + 3t) dt
(−5(2 − t) + − t + 4(2 − t) +
=
2
2
0
0
3
= −4 + 22 = 2 .
2
The line segments C3 is parametrized as x = 0, y = 2 − t, z = 1 − 2t , 0 ≤ t ≤ 2 and one gets
Z
Z
F · dr =
(5x − 2z 2 ) dx − (y − 4x) dy + (2xz + 4y)dz
C3
C
Z 21
Z 2
=
((2 − t) − 2(2 − t)) dt =
(−t) dt
0
0
1
= − 22 = −2 .
2
Summing up the contributions one gets
I
F · dr = 10
C
which agrees with the result obtained by using Stokes’ Theorem.
4