Name:
Practice Final
Chapter 15 - Integration
1. Compute the integral
ZZ
xy dxdy,
D
where D is described by 1 ≤ x2 + y 2 ≤ 4 and 0 ≤ y ≤ x.
Solution. The region D in polar coordinates is given by
1 ≤ r ≤ 2,
0≤θ≤
π
,
4
so the integral becomes
r=2
Z π
Z π
Z πZ 2
4
1 4 4
15 4
2
r cos(θ) sin(θ)r drdθ =
r cos(θ) sin(θ)
dθ =
cos(θ) sin(θ) dθ
4 0
4 0
0
1
r=1
π
15
cos(2θ) θ= 4
15
= .
−
=
4
4
16
θ=0
2. Evaluate the following double integral
Z 8Z
0
2
√
3
y
2
y 2 ex
dxdy
x8
by switching the order of integration.
Solution. The given boundaries of integration are
0 ≤ y ≤ 8,
√
3
y ≤ x ≤ 2,
which is a region of type II in the xy-plane. Expressing it as a type I region, we get
0 ≤ x ≤ 2,
0 ≤ y ≤ x3 .
The integral then becomes
Z 2 Z x3
0
0
2
y 2 ex
1
dydx =
8
x
3
Z
2
xe
x2
0
2
1 ex x=2 e4 1
dx =
=
− .
3 2 x=0
6
6
3. Find the total mass of the solid E which lies above the cone z 2 = 4x2 + 4y 2 with z ≥ 0 and
below the plane z = 4, where the density is given by ρ(x, y, z) = k, with k constant.
1
Recall that the total mass is
ZZZ
ρ(x, y, z) dV.
M=
E
Solution. We use cylindrical coordinates. The projection of the solid E onto the xy-plane is the
disc with center the origin and radius 2 (notice that intersecting the plane with the cone, we get
16 = 4x2 + 4y 2 ). In polar coordinates this region is described by
0 ≤ r ≤ 2,
0 ≤ θ ≤ 2π.
Above this projection, the z component goes from the surface of the cone z =
to the plane z = 4. The total mass of E is therefore given by
Z
2π
2Z 4
Z
2π
Z
2
Z
0
0
0
2r
r=2
64πk
16πk
2r3
2
(4r − 2r )drdθ = 2πk 2r −
=
=
.
3 r=0
12
3
2
r dzdrdθ = k
k
p
4x2 + 4y 2 = 2r up
0
4. Compute the triple integral
ZZZ
x2 dxdydz,
E
where E is the tetrahedron with vertices (0, 0, 0), (1, 0, 0), (0, 2, 0), (0, 0, 3).
Solution. We write the tetrahedron E as a region of type 1 in R3 . Its projection D is the triangle
in the xy-plane described by
0 ≤ x ≤ 1,
0 ≤ y ≤ 2 − 2x.
The tetrahedron E lies above D between the graph of the function u1 (x, y) = 0 and the plane
passing through the points (1, 0, 0), (0, 2, 0), (0, 0, 3). To write down the equation of this plane, we
choose two vectors on this plane h−1, 0, 3i, h−1, 2, 0i and compute their cross product, which is
h−6, −3, −2i. It follows that the equation of the plane is
6x + 3y + 2z = 6.
Solving for z, we see that the plane is the graph of the function
u2 (x, y) = 3 −
3y
− 3x.
2
The triple integral then becomes
Z
0
1 Z 2−2x Z 3−3y/2−3x
0
0
1 Z 2−2x 3x2 y
x dzdydx =
3x −
− 3x3
2
0
0
x=1
3x4 3x5
1
+
= .
= x3 −
2
5 x=0 10
Z
2
5. Calculate
√
Z
π
Z
1Z
2
1−y 2
y sin(x) dzdydx.
0
0
0
2
Z
dydx =
0
1
3x2 − 6x3 + 3x4 dx
Solution. The bounds are given to us; we simply compute. We have
Z πZ 1 p
Z π Z 1 Z √1−y2
y 1 − y 2 sin(x)dydx
y sin(x) dzdydx =
0
0
0
0
0
Z π
Z 1 p
2
=
sin(x) dx
y 1 − y dy
0
=
[− cos(x)]x=π
x=0
=
2
.
3
0
1
· − (1 − y 2 )3/2
3
y=1
y=0
6. Compute
ZZZ
x2 y 2 dxdydz,
E
where E is the region in xyz-space bounded by the paraboloid z = 1 − x2 − y 2 and the plane z = 0.
Solution. The projection of D onto the xy-plane is a circle of radius 1, centered at the origin.
Therefore, we may express our region E in cylindrical coordinates by
0≤r≤1
0 ≤ θ ≤ 2π
0 ≤ z ≤ 1 − r2
In cylindrical coordinates, the integral becomes
Z 1 Z 2π Z 1−r2
ZZZ
x2 y 2 dxdydz =
r4 cos(θ)2 sin(θ)2 r dzdθdr
0
0
0
E
Z 1 Z 2π
=
(r5 − r7 ) cos(θ)2 sin(θ)2 dθdr
0
0
Z 1
Z 2π
1 + cos(2θ) 1 − cos(2θ)
5
7
=
r − r dr
dθ
2
2
0
0
Z 2π
1 6 1 8 r=1
1 − cos(2θ)2
=
r − r
dθ
6
8
4
0
r=0
Z 2π
1
1 1 cos(4θ)
=
− −
dθ
24
4 8
8
0
1 θ sin(4θ) θ=2π
=
−
24 8
32
θ=0
π
=
.
96
Chapter 16 - Fundamental Theorem of Line Integrals, Green’s, Stokes’, Divergence
1. Compute the line integral of the vector field
F~ (x, y, z) = hcos(x), 2 + cos(y), ez i,
3
along the spiral curve C parametrized by ~r(t) = ht, cos(t), sin(t)i with 0 ≤ t ≤ 3π. (Hint: is F~
conservative?)
Solution. It is straightforward to check that ∇ × F~ = ~0, and since F~ is defined everywhere, it
is conservative F~ = ∇f . We can easily compute a potential function f (for example by partial
integration, or in this case just guessing)
f (x, y, z) = sin(x) + 2y + sin(y) + ez .
By the Fundamental Theorem for Line Integrals
Z
F~ ·d~r = f (~r(3π))−f (~r(0)) = f (3π, −1, 0)−f (0, 1, 0) = −2+sin(−1)+1−2−sin(1)−1 = −4−2 sin(1).
C
2. Compute the surface area of the surface S parametrized by
2
~r(u, v) = 3u + 2v, 4u + v, v 7/2 ,
7
with 0 ≤ u ≤ 1 and u1/4 ≤ v ≤ 1.
Solution. We compute
~ru = h3, 4, 0i,
~rv = h2, 1, v 5/2 i,
~ru × ~rv = h4v 5/2 , −3v 5/2 , −5i,
p
p
||~ru × ~rv || = 25v 5 + 25 = 5 v 5 + 1.
The surface area is then given by
Z
1Z 1
p
v 5 + 1 dvdu,
5
u1/4
0
which is impossible to compute, so we have to switch the order of integration. The given boundaries
of integration are
0 ≤ u ≤ 1,
u1/4 ≤ v ≤ 1,
which is a region of type I in the uv-plane. Expressing it as type II we get
0 ≤ v ≤ 1,
0 ≤ u ≤ v4.
The integral then becomes
Z
1 Z v4
5
0
0
Z
p
v 5 + 1 dudv = 5
0
1
v
4
p
√
5 + 1)3/2 v=1
2(v
4
2 2
v 5 + 1 dv = 5
=
− .
5·3
3
3
v=0
4
3. Find the flux of the vector field
F~ (x, y, z) = h5x3 + 12xy 2 , y 3 + ey sin(z), 5z 3 + ey cos(z)i,
through the surface S of the sphere with center the origin, radius 3 and outward orientation. (Hint:
Divergence Theorem).
Solution. The divergence of F~ is
∇ · F~ = 15x2 + 15y 2 + 15z 2 ,
and so by the Divergence Theorem, the flux equals the triple integral
ZZZ
15(x2 + y 2 + z 2 ) dxdydz,
E
where E is the ball with center the origin and radius 3. We use spherical coordinates and get
Z π Z 2π Z 3
ρ2 · ρ2 sin(φ) dρdθdφ = 3 · 35 · 2π · 2 = 2916π.
15
0
0
0
4. Use Stokes’ Theorem to find the flux
ZZ
~
(∇ × F~ ) · dS,
S
where
2
2
2
2
F~ (x, y, z) = hxz 3 ey + 2xyzex +z , x + z 2 ex +z , yex +z + zex i,
and the surface S is the upper hemisphere x2 + y 2 + z 2 = 4 with z ≥ 0 and upward orientation
(notice that the surface S is not closed).
Solution. We use Stokes’ Theorem to convert the flux integral to
ZZ
Z
~
~
(∇ × F ) · dS =
F~ · d~r,
S
C
where C is the circle x2 + y 2 = 4 in the xy-plane z = 0, with the counterclockwise orientation. We
can parametrize C by
~r(t) = h2 cos(t), 2 sin(t), 0i,
with 0 ≤ t ≤ 2π, so the line integral equals
Z 2π
F~ (~r(t)) · ~r 0 (t) dt.
0
We compute
2
F~ (~r(t)) = h0, 2 cos(t), 2 sin(t)e4 cos(t) i,
~r 0 (t) = h−2 sin(t), 2 cos(t), 0i,
F~ (~r(t)) · ~r 0 (t) = 4 cos(t)2 .
5
The original flux integral is then equal to
Z
4
2π
cos(t)2 dt = 4π.
0
5. Find the flux
ZZ
~
F~ · dS,
S
where
F~ (x, y, z) = hxz, x, yi
and S is the hemisphere x2 + y 2 + z 2 = 1, y ≥ 0, with the normal vector pointing outwards.
Solution. We may compute this directly. We parametrize the surface S by
~r(θ, φ) = hcos(θ) sin(φ), sin(θ) sin(φ), cos(φ)i,
where we take 0 ≤ θ ≤ π, 0 ≤ φ ≤ π in order to get the right half of the sphere. We’ve computed
in class that the normal vector pointing outwards is given by
~rφ × ~rθ = sin(φ)2 cos(θ)~i + sin(φ)2 sin(θ)~j + sin(φ) cos(φ)~k.
We have
F~ (~r(θ, φ)) = sin(φ) cos(φ) cos(θ)~i + sin(φ) cos(θ)~j + sin(φ) sin(θ)~k
F~ (~r(θ, φ)) · (~rφ × ~rθ ) = sin(φ)3 cos(φ) cos(θ)2 + sin(φ)3 sin(θ) cos(θ) + sin(φ)2 cos(φ) sin(θ)
Thus, the flux is
ZZ
Z πZ π
~ =
F~ · dS
F~ (~r(θ, φ)) · (~rφ × ~rθ ) dθdφ
S
0
0
Z πZ π
Z πZ π
3
2
=
sin(φ) cos(φ) cos(θ) dθdφ +
sin(φ)3 sin(θ) cos(θ)dθdφ
0
0
0
Z π0Z π
+
sin(φ)2 cos(φ) sin(θ)dθdφ
0
0
Z π
Z π
Z π
Z π
3
2
3
=
sin(φ) cos(φ)dφ
cos(θ) dθ +
sin(φ) dθ
sin(θ) cos(θ)dθ
0
0
0
0
Z π
Z π
2
+
sin(φ) cos(φ)dφ
sin(θ)dθ
0
0
= 0.
6. Find the work done by the vector field
F (x, y) = hex−1 , xyi
on a particle moving along the path ~r(t) = ht2 , t3 i, 0 ≤ t ≤ 1.
6
Solution. We evaluate directly. We have
~r 0 (t) = h2t, 3t2 i
2
F~ (~r(t)) = het −1 , t5 i
2
F~ (~r(t)) · ~r 0 (t) = 2tet −1 + 3t7
Hence
Z
F~ · d~r =
C
1
Z
F~ (~r(t)) · ~r 0 (t) dt
0
1
Z
2 −1
2tet
=
0
+ 3t7 dt
1
3
2
= et −1 + t8
8
11
=
− e−1 .
8
0
Complex Analysis
1. Evaluate in both polar and cartesian coordinates:
(a) (−3 + 3i)11
(b) (−1 +
√
3i)6
Solution. We write the base in polar form and use de Moivre’s Theorem.
11
√
3π
33π
33π
3π
11
11 11/2
(a) (−3 + 3i) = 3 2 cos
+ i sin
=3 2
cos
+ i sin
4
4
4
4
√
1
i
= 311 32 2 √ + √
= 311 32(1 + i).
2
2
6
√ 6
2π
2π
12π
12π
6
(b) (−1 + 3i) = 2 cos
+ i sin
= 2 cos
+ i sin
= 64.
3
3
3
3
2. Is the complex function
f (z) = ez
holomorphic or not?
Solution. We have z = x − iy, so f (z) = ez = ex e−iy = ex cos(y) − iex sin(y). We compute the
partial derivatives of the real and imaginary parts:
∂u
= ex cos(y)
∂x
∂u
= −ex sin(y)
∂y
∂v
= −ex cos(y)
∂y
∂v
−
= ex sin(y)
∂x
Since the Cauchy-Riemann equations aren’t satisfied, f (z) cannot be holomorphic.
7
3. At which points z is the complex function
f (z) =
z2
1
+1
holomorphic? Compute its complex derivative f 0 (z) at those points.
Solution. We know that the function z 2 + 1 is holomorphic (it is the sum of z 2 and 1, both of
which are holomorphic). The quotient
1
2
z +1
is holomorphic wherever the denominator does not vanish. The denominator vanishes precisely
when
z 2 + 1 = 0,
which means z = i or z = −i. At these two points the function f (z) is not defined. Everywhere
else, it is defined and it is holomorphic. Its complex derivative is computed with the standard
quotient rule
2z
f 0 (z) = −(z 2 + 1)−2 · 2z = − 2
.
(z + 1)2
4. Let C be the straight segment joining 2i to 1 + i. Compute
Z
z 2 dz.
C
Solution. We could do this directly, but we can also proceed as follows. The function z 2 has
antiderivative F (z) = z 3 /3, so the line integral equals
F (1 + i) − F (2i) =
(1 + i)3 (2i)3
2 10
−
= − + i.
3
3
3
3
5. Let C be the curve given by
2
z(t) = t sin(t)e1−t−t + ln(t(2π − t) + 1)i,
Evaluate
Z
0 ≤ t ≤ 2π.
2
cos(ez ) dz.
C
2
Solution. The curve C is closed (note that z(0) = 0 = z(2π)), and the function cos(ez ) is
analytic everywhere (but note that we can’t write down an antiderivative!). Therefore, by Cauchy’s
Theorem,
Z
2
cos(ez ) dz = 0.
C
6. Let C be the unit circle in the plane centered at the origin and oriented counterclockwise.
Evaluate each of the following integrals, giving a concise explanation if you use any Theorem.
Z
Z
Z
Z
Z
1
1
z+i
4
dz
(c)
dz
(d)
dz
(e)
(a)
ez dz
(b)
(3z − 1)−5 dz
i
z
z
+
2
z
−
C
C
C
C
C
2
8
Solution. Call D the unit disk enclosed by C.
4
(a) The function ez is holomorphic on D, so by Cauchy’s Theorem
Z
4
ez dz = 0.
C
(b) The point z0 = i/2 is inside D, so by Cauchy’s Integral Formula
Z
z+i
i
= −3π.
dz = 2πi i +
i
2
C z− 2
(c) The point z0 = 0 is inside D, so by Cauchy’s Integral Formula (with f (z) = 1)
Z
1
dz = 2πi.
z
C
1
(d) The function z+2
is holomorphic on the whole of D (since the denominator does not vanish on
D), therefore by Cauchy’s Theorem
Z
1
dz = 0.
z
+
2
C
(e) The function (3z − 1)−5 has antiderivative
−
1
(3z − 1)−4
12
which is defined on the curve C, which is closed, so
Z
(3z − 1)−5 dz = 0.
C
9