Download 1. Homework Set 1

Math 10B - Calculus of Several Variables II
Graded Homework Problem Solutions
1.
Homework Set 1
Problem (5.1 #10). Calculate the given integral and indicate which region of R3 it gives the volume of:
Z 2Z 3
2 dxdy.
1
0
Solution.
Z
2
Z
3
2
Z
(2x)|31 dy =
2 dxdy =
0
1
0
2
Z
Z
(6 − 2)dy =
0
2
4 dy = 4y|20 = 8 − 0 = 8.
0
This is the volume of the box [1, 3] × [0, 2] × [0, 2].
√
Problem (5.2 # 12). Integrate the function f (x, y) = 3xy over the region bounded by y = 32x3 and y = x.
√
Solution. First we find the intersection points of y = 32x3 and y = x. Clearly (0, 0) is an intersection point. Since
32 = 25 we have, setting the functions equal to each other, that the x-value of the other intersection point is:
1
1
1
1
25 x3 = x 2 =⇒ 210 x6 = x =⇒ x5 = 10 =⇒ x = 2 =
2
2
4
and the y-value is
r
1
1
y=
=
4
2
so the other point of intersection is 14 , 12 .
Notice that if we plug in x =
and for y =
√
1
1
= 4 we see that for y = 32x3 :
16
2
3
1
1
1
1
5
= 25 12 = 7 =
y=2
24
2
2
128
x:
r
y=
so between x = 0 and x =
1
1
=
16
4
√
1
we see that y = x is the top function. Thus our integral looks like:
4
Z 14 Z √x
3xy dydx.
0
32x3
Now evaluating the integral:
Z
1
4
√
Z
√x
3 2 xy dx
2
0
32x3
Z 41 3 2
9
7
x − 2 3x dx
2
0
41
1 3
6
8 x − 2 3x 2
0
1
3
− 10
27
2
8
3
− 10
210
2
5
5
=
210
1024
x
Z
3xy dydx =
0
32x3
=
=
=
=
=
1
4
1
2
2.
Homework Set 2
Problem (5.3 # 12). Rewrite the following sum of iterated integrals as a single iterated integral by reversing the
order of integration, and evaluate:
Z 2 Z 2−x
Z 1Z x
sin x dydx.
sin x dydx +
0
0
0
1
Solution. Let’s begin by switching the order on the first integral:
Z 1Z x
sin x dydx.
0
0
First we draw the region of integration:
from here we can tell that the x starts on the left at x = y and goes to x = 1 and y ranges from 0 to 1 so the integral
becomes
Z 1Z x
Z 1Z 1
sin x dydx =
sin x dxdy.
0
0
0
y
Now for the second integral:
Z
2
Z
2−x
sin x dydx
1
0
solving y = 2 − x for x we get that x = 2 − y, and so we see that the integral has become:
Z 2 Z 2−x
Z 1 Z 2−y
sin x dydx =
sin x dxdy.
1
0
0
1
Using the properties of integrals we have that the sum becomes:
Z 1Z x
Z 2 Z 2−x
Z 1Z 1
Z
sin x dydx +
sin x dydx =
sin x dxdy +
0
0
1
0
0
Z
y
1
Z
=
1
Z
Z
1
Z
=
y
2−y
1
sin x dxdy
0
y
2−y
Z
sin x dxdy
0
1
2−y
sin x dx +
0
1
sin x dx dy
3
Now let’s compute the integral:
Z 1 Z 2−y
sin x dxdy
0
1
Z
=
y
0
Z
2−y
dy
(− cos x)
y
1
(− cos(2 − y) + cos y)dy
=
0
1
=
(sin(2 − y) + sin y)|0
=
(sin(2 − 1) + sin 1) − (sin(2 − 0) + sin 0)
=
sin 1 + sin 1 − sin 2 − 0
=
2 sin 1 − sin 2
4
3.
Homework Set 3
Problem (5.4 # 17). Integrate the function f (x, y, z) = x + y over the region W which is bounded by the cylinder
x2 + 3z 2 = 9, and the planes y = 0 and x + y = 3. (Note that only the set-up of this integral is required by instruction
of Professor Henriques.)
Solution. First, here is a picture of the region in question (the axes are oriented as indicated, with the arrows pointing
in the positive direction):
Now, to set up the integral, notice that it is easiest to integrate along the y-direction first, thus the bounds on y are:
0 ≤ y ≤ 3 − x.
This gives us a shadow in the xz-plane which is the ellipse x2 +3z 2 = 9, and so it remains to find the bounds on x and z.
Let’s integrate x first, so that our bounds on x are:
p
p
− 9 − 3z 2 ≤ x ≤ 9 − 3z 2
and hence the bounds on z are:
√
√
− 3 ≤ z ≤ 3.
Thus the integral we are looking for is:
√
Z
3
√
− 3
√
Z
9−3z 2
√
− 9−3z 2
Z
3−x
(x + y)dydxdz.
0
5
4.
Homework Set 4
Problem (5.5 # 25). Determine
ZZZ
(x2 + y 2 + 2z 2 )dV ,
W
where W is the solid cylinder defined by the inequalities x2 + y 2 ≤ 4 and −1 ≤ z ≤ 2.
Solution. The fact that W is a cylinder should tip you off to use cylindrical coordinates, so let’s do so. Using the
simple fact that x2 + y 2 = r2 and the shadow of W in the xy-plane is just a disk of radius 2, we have that (don’t
forget that dV = rdrdθdz):
ZZZ
Z 2π Z 2 Z 2
2
2
2
(r2 + 2z 2 )rdzdrdθ
(x + y + 2z )dV =
W
−1
0
0
2
2
r3 z + rz 3 drdθ
=
3
0
0
−1
Z 2π Z 2 16
2
2r3 + r − −r3 − r drdθ
=
3
3
0
0
Z 2π Z 2
=
(3r3 + 6r)drdθ
Z
2π
Z
2π
0
Z
0
=
Z
0
2π
=
2
2
3 4
2 r + 3r drdθ
4
0
(12 + 12)dθ = 48π
0
6
5. Homework Set 5
Z
Problem (6.1 #11). Determine the value of
x dy − y dx, where x(t) = (cos 3t, sin 3t), 0 ≤ t ≤ π.
x
Solution. First note that this is a vector line integral. So to evaluate this integral, we notice that dx becomes the
derivative of the the first component of x and dy the derivative of the second. Thus:
Z
Z π
x dy − y dx =
[cos 3t(3 cos 3t)dt − sin 3t(−3 sin 3t)dt]
x
Z0 π
3 cos2 3t + 3 sin2 3t dt
=
Z0 π
=
3 dt
0
=
3π
7
6.
I
Problem (6.2 #11). Evaluate
Homework Set 6
(x4 y 5 − 2y)dx + (3x + x5 y 4 )dy where C is the oriented curve:
C
Solution. Since the curve C and the region it bounds, call it D, satisfy the hypothesis of Green’s theorem, except
that C is oriented the wrong way, we have, by Green’s theorem:
I
ZZ
ZZ
(x4 y 5 − 2y)dx + (3x + x5 y 4 )dy = −
3 + 5x4 y 4 − 5x4 y 4 − 2 dA = −
5 dA = −5 · Area(D).
C
D
D
Since the area of the region is easily seen to be 9, we have that:
I
(x4 y 5 − 2y)dx + (3x + x5 y 4 )dy = −45.
C
8
7.
Homework Set 7
Problem (6.3 #21).
(a) Determine where the vector field
F=
x2 + 1
x + xy 2
,−
2
y
y3
is conservative.
(b) Determine a scalar potential for F.
(c) Find the work done by F in moving a particle along the parabolic curve y = 1 + x − x2 from (0, 1) to (1, 1).
Solution.
(a) Let F = (M, N ), then checking the partial derivatives:
−2x
−2x
My = 3
Nx = 3
y
y
we see that the vector field is conservative as long as y 6= 0.
(b) Since F is conservative, that means that F = ∇f = (fx , fy ) = (M, N ). So to find f let’s integrate N with
respect to y:
Z
1 x2 + 1
+ φ(x).
f = N dy =
2 y2
Differentiating with respect to x now:
fx =
so φ0 (x) = x hence φ(x) =
2x
x + xy 2
x
+ φ0 (x) = M =
= 2 +x
2
y
y2
y
1 2
x + C and so a scalar potential is:
2
1 x2 + 1 1 2
x2 + 1 + x2 y 2
+
+C
f=
x
+
C
=
2 y2
2
2y 2
for any C ∈ R.
(c) Let A = (0, 1) and B = (1, 1), and let P be the path. Since our vector field is conservative, we have that the
work done is:
Z
1+1+1
0+0+1
W =
F · ds = f (B) − f (A) =
+C −
+ C = 1.
2
2
P
9
8.
Homework Set 8
Problem (7.1 #19). Let D denote the unit disk in the st-plane. Let X : D → R3 be defined by (s + t, s − t, s). Find
the surface area of X(D).
Solution. First we find the tangent vectors:
Ts =
and
Tt =
∂X
= (1, 1, 1)
∂s
∂X
= (1, −1, 0),
∂t
so a normal vector is:
î
N = Ts × Tt = 1
1
ĵ
1
−1
k̂
1
0
= (0 + 1, −(0 − 1), −1 − 1) = (1, 1, −2).
Thus the surface area of X(D) is:
ZZ
kN k dA =
SA =
D
ZZ √
6 dA =
√
6Area(D).
D
Since the area of the unit disk is π, we have that the surface area is
√
SA = 6π.
10
9.
Homework Set 9
Problem (7.2 #1). Let X(s, t) = (s, s + t, t), 0 ≤ s ≤ 1, 0 ≤ t ≤ 2. Find
ZZ
(x2 + y 2 + z 2 ) dS.
X
Solution. First we find the normal vector field:
Ts = (1, 1, 0)
and
Tt = (0, 1, 1),
so the normal vector field is:
i
N = Ts × Tt = 1
0
2
2
2
Letting f (x, y, z) = x + y + z , we see that
j
1
1
k
0
1
= (1, −1, 1).
f (X(s, t)) = s2 + (s + t)2 + t2 = 2(s2 + st + t2 ).
We also have that
kN k =
p
√
12 + (−1)2 + 12 = 3,
thus
ZZ
Z
f ds
1
2
Z
=
X
0
=
√
2(s2 + st + t2 ) 3 dtds
0
√ Z
2 3
1
0
Z
2
(s2 + st + t2 ) dtds
0
2
1
1
s2 t + st2 + t3 ds
2
3 0
0
Z
1
√
8
= 2 3
ds
2s2 + 2s +
3
0
1
√
8 2 3
s + s2 + s = 2 3
3
3
0
√
2
8
= 2 3
+1+
3
3
√
26 3
=
3
=
√ Z
2 3
1
11
10.
Homework Set 10 (Not Graded)
π
Problem (7.3 #2). S is parametrized by X(s, t) = (s cos t, s sin t, t), 0 ≤ s ≤ 1, 0 ≤ t ≤ . Verify Stokes’ theorem
2
for the vector field
F = (z, x, y).
Solution. First off, we need to find the curl of F:
i
j
k
∂
∂
∂
∇ × F = ∂x
∂y
∂z
z
x
y
= (1 − 0, −(0 − 1), 1 − 0) = (1, 1, 1).
Next, we need the normal vector field:
i
N = Ts × Tt = cos t
−s sin t
Then we have that:
ZZ
Z
∇ × F · dS =
S
0
1
Z
j
k sin t 0 = (sin t, − cos t, s).
s cos t 1 π
2
Z
1
Z
(1, 1, 1) · (sin t, − cos t, s) dtds =
0
π
2
(sin t − cos t + s) dtds =
0
0
π
.
4
Now, since the parametrization X is one-one on D, ∂S = X(∂D). The boundary of the region has four parts:
x1 (s) = (s, 0, 0)
0≤s≤1
x2 (t) = (cos t, sin t, t) 0 ≤ t ≤ π2
x3 (s) = 0, 1 − s, π2 0≤s≤1
π
0 ≤ t ≤ π2
x4 (t) = 0, 0, 2 − t
R
R
You can check that xi F · ds = 0 for i = 1, 3, 4 and that x2 F · ds = π4 . Then
I
Z
4
X
π
π
F · ds = 0 + + 0 + 0 = .
F · ds =
=
4
4
∂S
xi
i=1
Problem (7.3 #3). Let S be the surface x =
p
16 − y 2 − z 2 . Verify Stokes’ theorem for the vector field:
F = (x, y, z).
Solution. Again, the first step will be to compute the curl of F:
i
j
k ∂
∂
∂ ∇ × F = ∂x ∂y
∂z = (0 − 0, −(0 − 0), 0 − 0) = (0, 0, 0).
x
y
z Thus it follows that
ZZ
∇ × F · dS = 0.
S
Now, this surface is the hemisphere of radius 4, where x ≥ 0, so the boundary is given by y 2 + z 2 = 16, x = 0.
Parametrizing this, we get
x(t) = (0, 4 cos t, 4 sin t), 0 ≤ t ≤ 2π.
Then
F(x(t)) = (0, 4 cos t, 4 sin t)
and
x0 (t) = (0, −4 sin t, 4 cos t)
hence
F(x(t)) · x0 (t) = 0
so
I
F · ds = 0.
∂S
12
Problem (7.3 #13).
(a) Show that the path x(t) = (cos t, sin t, sin 2t) lies on the surface z = 2xy.
(b) Evaluate
I
(y 3 + cos x)dx + (sin y + z 2 )dy + x dz,
C
where C is the closed curve parametrized and oriented by the path x in part (a).
Solution.
(a) To do this, simply plug the first, second, and third components of x into the equation for x, y, and z
respectively:
?
z = sin 2t = 2 cos t sin t = 2xy
The question is asking if the middle equality is true. It is, since it is precisely the double angle formula for
sin. Thus the curve does indeed lie on the surface.