Practice problems
1. Consider a right circular cone of uniform density. The height is H.
Let’s say the distance of the centroid to the base is d. What is the
value d/H?
We can create a coordinate system such that the base is in xy plane
and the z-axis is the axis. Assume the radius of the base is R. Then,
the region can be written in cylindrical coordinates as 0 ≤ r ≤ R, 0 ≤
θ <RR2π, 0 ≤ z ≤ H
by d = z̄ =
R (R − r). The centroid can be computedRRR
1
1
zδdV
.
In
our
case,
we
can
cancel
δ
and
have
z̄
=
zdV
m
V
2
2. Set up the integral without
√ evaluation. The volume inside (x − 1) +
2
2
y + z = 1, below z = 3r but above z = −r.
This problem is very tricky in cylindrical or Cartesian since we must
divide the region into several blocks if we use these
√ two. It is convenient
to use spherical. π/6 ≤ φ ≤ 3π/4 by z = 3r and z = −r. The
sphere is ρ = 2 sin φ cos θ. The range for θ is from −π/2 to π/2. This
is because the sphere is on the right of yz plane. At the starting
point ρ = 0 and cos θ = 0. This means it starts from −π/2. Lastly,
0 ≤ ρ ≤ 2 sin φ cos θ
3. Set up the integral for the moment of inertia about z axis inside both
ρ = 2 and
p r = 2 cos θ, outside r = 1 and above xy plane. The density
is δ = x2 + y 2 + z 2 .
This problem is good for√cylindrical coordinates. −π/3 ≤ θ ≤ π/3 and
1 ≤ r ≤ 2 cos θ, 0 ≤ z ≤ 4 − r2
4. Find the centroid of the ice-cream cone enclosed by x2 + y 2 + z 2 =
1, z ≥ 0 and z = r−1, z ≤ 0. Suppose the density is δ = 1 (Hint: break
the region into two parts. One for cylindrical and one for spherical.)
RRR
x̄ = ȳ = 0. z̄ = V1
T zdV . The region can be written as the union
of two 0 ≤ φ ≤ π/2, 0 ≤ θ < 2π, 0 ≤ ρ ≤ 1 and 0 ≤ r ≤ 1, 0 ≤ θ <
2π, r − 1 ≤ z ≤ 0
2
2
2
5. Find the total mass
√ of the ice-cream cone inside x + y + (z − 1) = 1
and above z = 3r, assuming the density is δ = 1.
This is the example in the book. ρ = 2 cos φ. 0 ≤ φ ≤ π/6, 0 ≤ θ <
2π, 0 ≤ ρ ≤ 2 cos φ
1
6. Set up the integral for the volume outside r = 1 inside ρ = 2 in both
cylindrical and spherical coordinates.
The tricky part is the spherical. In spherical, the two intersects at
2 sin φ = 1 or π = π/6, 5π/6. Hence, in spherical, 0 ≤ θ < 2π, π/6 ≤
φ ≤ 5π/6. For the ρ part, r = 1 is ρ sin φ = 1. Hence, 1/ sin φ ≤ ρ ≤ 2
7. (The practice problems for 14.8 will be together with the one for surface integrals.)
**********************************************************
1. Line integrals
(a) Parametrization
• Parametrize the curve y = x2 r = ht, t2 i
• Parametrize x2 + 4y 2 = 1 r(t) = hcos t, 12 sin ti, 0 ≤ t < 2π
• Parametrize the boundary of the region bounded by x-axis,
√
y = x and x = 1. C = C1 +C2 +C3 . C1 : r(t)√= ht, 0i, 0 ≤
t ≤ 1. C2 : r(t) = h1, ti, 0 ≤ t ≤ 1. C3 : r = ht, ti, t : 1 → 0
• Parametrize the ellipse formed by the intersection of x2 +y 2 =
1 and x + z = 0.
r(t) = hcos t, sin t, − cos ti
(b) Usual line integrals (2 types)
• Consider the curve x2 /4 + y 2 = 1 with x ≥ 0, 0 ≤ y ≤ 1/2. If
the density (per unit length) is δ = y/x, compute the moment
of inertia
Iy .
R
R
Iy = C x2 δds = C xyds. r = h2 cos t, sin ti. 0 ≤ t ≤ π/6
• Compute the line integral of F = h3y, −2xi over the curve
y = x2 for 0 ≤ y ≤ 1 oriented from right to left.
R
• Let r = ht3 , t2 , ti, 0 ≤ t ≤ 1. Compute C F · T ds where
F = heyz , 0, yeyz i
(c) Conservative field.
R
• Compute the line integral C (x3 + y)dx + xdy where C is the
curve jointing (0, 0) and (1, 1). Justify your answer.
The field is conservative and is ∇(x4 /4 + xy)
9
3
100
• Let
R C be r(t) = hln(1 + t ), t + 1, t i, 0 ≤ t ≤ 1. Compute
C xdy + ydx + dz
R
The field is ∇(xy +z). Or you can notice that it is d(xy +z)
2
• Show that F = (3y 3 −10xz 2 )i+9xy 2 j−10x2 zk is irrotational
and thus conservative. Find a potential function φ.
R
• C F · T ds where F = hzexz + ex , 2yz, xexz + y 2 i. r =
2
3
het , et , t4 i. t ∈ [0, 1]
It is irrotational and thus conservative. φ = exz + y 2 z
3
• Let C be ~r(t) = hcos4 t, sin4 t, 7i, 0 ≤ t < 2π
R and F = hx −
3
3
z, y , y + z i. Compute the line integral C F · d~r. (Hint:
Split out a conservative field. The integral of the one you
split will be zero since it is on a closed curve.)
The field F = hx3 , y 3 , z 3 i is conservative.
Since the
R
R curve is
closed, weR only need to compute C h−z, 0, yi·dr = C −7dx+
ydz = −7 C dx = 0 because h1, 0, 0i is again conservative.
2. Green’s theorem
(a) Circulation and flux
3
• Compute the line integral of hx2 y + xex , xy − sin2 (ey )i over
the rectangle with vertices (0, 0), (2, 0), (0, 3) and (2, 3) oriented
counterclockwise.
H
H
3
P dx + Qdy. P = x2 y + xex , Q = xy − sin2 (ey ).
C F · dr =
Then, Qx − Py = y − x2 .
3Z 2
Z
0
(y − x2 )dxdy
0
H
• Compute the line integral C P dx + Qdy where P = xy, Q =
x2 and C is the loop of the curve in the first quadrant whose
polar equation is r = sin(2θ).
RR
We have Qx − Py = 2x − x = x. Hence, D xdA. In polar,
0 ≤ θ ≤ π/2, 0 ≤ r ≤ sin(2θ)
Z
π/2 Z sin(2θ)
r cos θrdrdθ
0
0
R π/2
The integral 0 sin3 (θ) cos4 (θ)dθ can be done by doing substitution u = cos θ
2
2
• Let F = xy
√ i + x yj. Let C be the union of x-axis(|x| ≤ 2)
and Hy = 4 − x2 , oriented counterclockwise. Compute the
flux C F · nds in two ways.
3
H
RR
The first way is RR
Green’s theorem F · nds = D ∇ · F dA.
Then, you have D (x2 + y 2 )dA which can be finished using
polar.
The second
way is to parametrize directly. nds = hdy, −dxi.
H
Then, P dy − Qdx. We parametrize the boundary with two
pieces.
• If v = hy 2 , xyi, Hand C is the ellipse x2 /9 + y 2 /4 = 1, compute
the circulation F · dr in two ways.
(b)
y
x
• Suppose v = h x2 +y
2 , x2 +y 2 i. Suppose C is a simple closed
curve
H in the plane, oriented counterclockwise. What values
can C v · nds be? (Hint: This field is divergence free.)
Since it is divergence free except at the origin, we discuss two
cases. If C doesn’t enclose the singular point, the answer is
zero by Green’s. If the curve encloses the origin, then we use
a small circle to isolate the origin. The flux then equals the
flux on the small circle. The answer is 2π. Then, the answer
could be 0 or 2π
• Suppose F = h x2−y
, x i. Suppose C is a simple closed
+y 2 x2 +y 2
curve
H in the plane, oriented counterclockwise. What values
can C F · T ds be? (Hint: This field is curl free or Py = Qx .)
(c)
• Compute the area bounded by y = x3 , y-axis and y = 1 using
both line integral and double integral.
• Find the area of the region enclosed by r(t) = hsin(2t), sin(t)i
above x-axis.
H
Rπ
For the loop, 0 ≤ t ≤ π. Then, A = − ydx = − 0 sin(t)2 cos(2t)dt.
u = cos t and cos(2t) = 2 cos2 t − 1 = 2u2 − 1. Answer is 4/3
• Find the area of the region enclosed by one arch of the cycloid
x(t) = a(t − sin(t)), y(t) = a(1 − cos(t)) and x-axis.
Example in class
3. Surface integrals (2 types)
(a) Parametrization, surface area element and the surface integral of
a function
• Parametrize the surface y = f (x, z). Use this to compute the
area of the plane y = 2x + 2z + 1 inside x2 + z 2 = 1.
Use x, z. r = hx, f (x, z), zi = hx, 2x + 2z + 1, zi, x2 + z 2 ≤ 1.
dS = |r z × r x |dzdx
4
• Consider the surface of revolution obtained by revolving x =
f (z) about z axis. Parametrize this surface.
Use z and θ as the parameters. x = f (z) cos θ, y = f (z) sin θ, z =
z
• Consider the fence S: x = 2 sin(t), y = 8 cos(3t),
0 ≤ t < 2π
RR
and 0 ≤ z ≤ 2. Set up the surface integral S 2yzdS.
Use t, z for the parameters. r = h2 sin t, 8 cos(3t), zi.
(b) Flux
• Compute the flux of G = h2x, x−y, y+zi through the surface
S, which is the portion of the plane 2x − 3y + 5z = 0 inside
x2 + y 2 = 1 oriented upward. Can we use Stokes’s theorem
here? Why?
The surface has a boundary and the field is not a curl(the
divergence is nonzero). We parametrize the surface r =
hx, y, 51 (3y − 2x)i, x2 + y 2 ≤ 1. Then, the flux is
ZZ
~ · (r x × r y )dxdy
Φ=
G
D
Note that r x × r y always points up for a function graph and
we are fine. D is the disk. For this problem, use polar. Don’t
forget to write z in terms of x, y
• Parametrize the upper hemi-sphere with radius 1. Then compute the flux of F = hx2 , 0, 0i across it.
r = hsin φ cos θ, sin φ sin θ, cos φi. 0 ≤ φ ≤ π/2, 0 ≤ θ <
2π, 0 ≤ ρ ≤ 1
RR
• S F · ndS where F = hy, −x, zi and S is the surface z = θ,
0 ≤ θ ≤ π and 1 ≤ x2 + y 2 ≤ 4.
Parametrize the surface. r = hr cos θ, r sin θ, θi, 1 ≤ r ≤
2, 0 ≤ θ ≤ π. Then, ndS = r r × r θ drdθ
• Compute the flux of F = h2, 2, 3i across the surface S: r(u, v) =
hu + v, u − v, uvi, 0 ≤ u, v ≤ 1
We compute directly that ndS = r u × r v dudv = . . .
4. Stokes Theorem
• Let F = zi+xj −yk. Let C be the intersection between x2 +y 2 =
1 and z +Hy = 3 oriented counterclockwise if viewed from above.
Compute C F · T ds in two ways.
The first way is direct computation by parametrization. r(t) =
hcos t, sin t, 3 − sin ti.
5
RR
For the second way, we can apply Stokes’s theorem. S ∇ × F ·
ndS. Then, we have to parametrize the surface r = hx, y, 3 − yi.
2
If the vector field is F = hz + x3 + sin(x), x + ey + ey , −y −
cos(8z) − z 100 i, can you reduce this to the problem here? Why?
2
Yes. The extra field G = hx3 + sin(x), ey + ey , − cos(8z) − z 100 i
is conservative. The integral on the closed curve is zero.
• Let F = h3y, −2x, x2 y 2 z 2 i. Let S be the surface x2 + y 2 + (z −
1)2 = 2 above xy plane, oriented upward. Compute
ZZ
(∇ × F ) · ndS
S
Apply Stokes and reduce it to the circulation over the curve x2 +
y 2 + (0 − 1)2 = 2. On this curve, the field is F = h3y, −2x, 0i.
The curve is r = hcos t, sin t, 0i. The computation is easy.
• Suppose F = hxy 2 z 2 + y, x2 yz 2 + z, x2 y 2 z + xi. C is the hexagon
with vertices (2, 0, 1), (1, 0, 2), (0, 1, 2), (0, 2, 1), (1, 2, 0) and (2, 1, 0),
which are all in the plane x + y + z = 3. Compute the circulation
of the field over this curve. (Hint: The best way is to split a
conservative field first. Anyway, using Stokes theorem directly
will give you the same answer.)
Apply Stokes. ∇ × F = ∇ × hy, z, xi because the extra field is
∇(x2 y 2 z 2 /2) which has zero curl. If you don’t notice this, it is
fine
RRsince it will be canceled as well. Then, the order is clockwise.
− D h−1, −1, −1i · h1, 1, 1idxdy = 3Area(D) = 9
~ = hx2 − 2y, 2ey − z, z 3 + 3xi along
• Compute the line integral of G
the curve r(t) = hcos t, sin t, cos t sin ti where t : 0 → 2π.
Note that the curve is closed since r(0) = r(2π). Then, we
can throw away a conservative field and have h−2y, −z, 3xi. At
this point, you can either use direct computation or use Stokes’s
theorem. If we apply Stokes’s, noticing that the curve is on z =
2
2
xy.
RR Hence, the surface is r = hx, y,2 xyi,2 x + y ≤ 1. We have
D (2 + 3x − y)dxdy. However, x + y ≤ 1 is the disk. The
integral of x, y are zero. Then, 2 ∗ Area(D) = 2π
5. Divergence theorem
(a) Computation and applications
6
RR
• Compute the flux S F · ndS where F = hy 3 + z 2 , xy −
xz 2 , xey i. S is the boundary of the solid x + y + z ≤ 1,
x, y, z ≥ 0, with n being RR
the outer normal.
By divergence, we have T (0 + x + 0)dV . This volume is
0 ≤ x ≤ 1, 0 ≤ y ≤ 1 − y, 0 ≤ z ≤ 1 − x − y. Set up the
volume integral and integrate.
8 9
• Let F = hx + ey z , y + 3y 2 + ln(x8 + 1000), z + cos(xy)i. Let
T be the upper hemi-ball with radius 1. Compute the flux
of this field out of T . RR
By divergence theorem, T (1 + 1 + 6y + 1)dV . The integral
of y is zero. Hence, 3V ol(T ) = 3 ∗ 23 π13 = 2π
• Suppose v is the gravitational field generated
by a cloud of
RR
mass. The physical law tells us that S v · ndS = −4πGm
where m is the total mass inside S. Suppose
RRRthe density of
the mass is δ. Then, the total mass is m =
T δdV . Using
1
the divergence theorem, show that δ = − 4πG
∇ · v. If the
gravitational field is given by v = h3ex + 4y 2 , 2y 2 + eyz , z 2 +
xyzi. Compute δ(0, 0, 1)/δ(0, 0, 0)
This is a direct application of the divergence theorem. It is
easy
(b) Surface independence
• Let F = − GM
r be the gravitational field generated by the
r3
Earth. S is any
RR surface that does not go across the Earth.
Find the flux S F · ndS
Two cases, the first case is when the surface doesn’t enclose
the origin and the answer is zero. The second case is when
the surface encloses the origin. We use a small sphere to
isolate the origin. Since the field is divergence free, we can
reduce the flux to be on the small sphere.
• Consider the surface z = (x2 RR
+ y 2 − 1)(x4 + y 4 + 1) for z ≤ 0,
oriented upward. Compute S F · ndS where F = hxy 3 +
y 2 z, x2 z 2 − x2 y, zx2 − zy 3 i.
Divergence free. We use surface independence. The surface
we pick then is the one in z = 0 plane. Hence, it is the region
inside x2 + y 2 − 1 = 0. We find that F · k = 0. The final
answer is therefore zero.
• let S be r(t, z) = h(1 − z)8 3 cos t, (1 − z)8 2 sin t, zi and 0 ≤
t < 2π, 0 ≤ z ≤ 1. The normal is upward. Compute the flux
of F~ = hy 2 z − z 2 , 4 − xy, 3 + xzi through this surface.
7
The surface is a surface with boundary r(t, z = 0) = h3 cos t, 2 sin t, 0i.
The field is divergence free. Hence, we can apply the surface
independence technique and have the flux to be equal to the
flux through theRRpart inside r(t, z = 0) in xy plane. Hence,
RR
D F · kdA =
D 3dA. The area can be computed using
whatever way you want. You can either use the line integral
or use double integrals by change of variable. If you notice
it is an ellipse, you can get the answer quickly as well.
8