Filetype PDF | Posted on 26 Jan 2023 | 2 years ago
Authentication
digital resources
208x Filetype PDF File size 0.16 MB Source: math.berkeley.edu
MATH 53 DISCUSSION SECTION PROBLEMS – 3/31 – SOLUTIONS
JAMESROWAN
1. Triple integrals
(1) True/false practice:
(a) There is no way to interpret any triple integral using only three-dimensional geometry.
False. An important application of triple integrals to three-dimensional geometry is in finding
the volumes for regions, since the volume of the region E is given by RRR 1dV .
RRR E
(2) (textbook 15.6.9) Evaluate the triple integral E ydV, where E = {(x,y,z)|0 ≤ x ≤ 3,0 ≤ y ≤
x,x−y≤z≤x+y}.
Translating the description of the region as a set of points into bounds for a triple integral, we
have
ZZZ ydV =Z 3Z xZ x+yydzdydx
E 0 0 x−y
=Z 3Z xy·2ydydx
0 0
Z 3 2 3
= 3x dx
0
= 27 .
2
(3) (textbook 15.6.21) Use a triple integral to find the volume of the solid enclosed by the cylinder
y = x2 and the planes z = 0 and y +z = 1.
The volume of the region is the triple integral of 1, so, as in many triple integral problems, the
only difficulty in this problem is finding the bounds for the triple integral. We see that this region
lies between the graphs of z = 0 and z = 1 − y and inside a surface described in terms of x and y,
so setting up our integral with z as the innermost variable is a good strategy here. The “shadow”
of this region onto the xy-plane is the region between the parabola y = x2 and the line y = 1 (the
line where the two planes z = 0 and y + z = 1 intersect), and in this region the graph of z = 1 − y
1
2 JAMESROWAN
lies above the graph of z = 0, so we can set up and evaluate our triple integral as follows:
V =Z 1 Z 1Z 1−y1dzdydx
−1 x2 0
=Z 1 Z 1(1−y)dydx
−1 x2
Z 1 2
y 1
= y − dx
2 2
−1 x
=Z 1 1 −x2+ x4dx
−1 2 2
3 5
1 x x 1
= x− +
2 3 10
−1
= 8
15
See the Zoom whiteboard images for today for a sketch of this region, which looks vaguely like a
doorstop.
(4) (textbook 15.6.36) Write five other iterated integrals equal to R1R1Rz f(x,y,z)dxdzdy.
0 y 0
The other five iterated integrals are:
Z 1Z 1Z zf(x,y,z)dydzdx
0 x 0
Z 1Z 1Z 1 f(x,y,z)dzdydx
0 0 max(x,y)
Z 1Z 1Z 1 f(x,y,z)dzdxdy
0 0 max(x,y)
Z 1Z zZ zf(x,y,z)dxdydz
0 0 0
Z 1Z zZ zf(x,y,z)dydxdz.
0 0 0
There are a few approaches one can take here; sketching out the region in three-dimensional space
(it is a relatively simple polyhedron, so if you’re particularly comfortable with 3D visualization this
is possible for this problem), sketching the shadows of the region on the xy-, yz-, and xz-planes,
or working to manipulate the inequalities that the bounds correspond to. I’ll illustrate the third
strategy for a few orders.
The given bounds correspond to the inequalities
0 ≤ y ≤ 1
y ≤ z ≤ 1
0 ≤ x ≤ z;
note that the top inequality must have bounds that are constants, the second one can have bounds
depending on the first variable, and the third one can have bounds depending on the first and second
variables.
Torewrite the integral as an integral dydzdx, we can look at what the largest and smallest possible
values for x are in this region. We see from the third inequality that x can be as small as 0 and
(when z = 1, which is allowed by the second inequality) as large as 1, so we have 0 ≤ x ≤ 1. We see
from the third inequality that z must be bigger than x and from the first equation that z must be
smaller than or equal to 1, so we have x ≤ z ≤ 1 as the bounds on z depending only on x. From the
first inequality, we see that 0 ≤ y, and from the second inequality, the biggest y can be at any point
MATH 53 DISCUSSION SECTION PROBLEMS – 3/31 – SOLUTIONS 3
(x,z) is z, so we have 0 ≤ y ≤ z. The three sets of inequalities we have found correspond exactly to
the first new order above.
The third new order is a bit trickier. We see that the smallest and largest possible values for y
are 0 and 1, but that at any given value of y, it is possible to have any value of x between 0 and 1,
since the only variable giving an upper bound on x is z, but we can’t use z to bound x because z is
the innermost variable in our triple integral. For the innermost variable z, the third inequality tells
us that x ≤ z, but the second inequality tells us that y ≤ z. Since we need z to be bigger than both
x and y, we know that z is bigger than the maximum of x and y, which we denote as max(x,y).
Another way to give this order would be to split the triple integral into two triple integrals, one
over the region in the xy-plane where x ≥ y and another over the region in the xy-plane where x < y.
(5) (textbook 15.6.53) Find the average value of the function f(x,y,z) = xyz over the cube with side
length L > 0 lying in the first octant with one vertex at the origin and three edges along the positive
x−, y−, and z-axes.
Theaveragevalueofafunctionofthreevariablesf(x,y,z)overaregionE isgivenby 1 RRR f(x,y,z)dV.
Vol(E)
E
The volume of a cube with side length L > 0 is L3. Integrating the function f(x,y,z) = xyz over
this cubical region and splitting the integral into a product of three single integrals since all the
bounds are constants and the integrand is a product of single-variable functions of x, y, and z, we
have
Z LZ LZ Lxyzdzdydx=Z LxdxZ LydyZ Lzdz
0 0 0 0 0 0
L2 L2 L2
= 2 · 2 · 2
L6
= 8 .
3
Dividing this triple integral by the volume L3, we have an average value of L for f(x,y,z) = xyz
8
over this cube.
2. Triple integrals in cylindrical coordinates
(6) True/false practice:
(a) When we defined cylindrical coordinates, the choice of the xy-plane as the plane we expressed
in polar coordinates was arbitrary; we could also have set up a polar coordinate system (r,θ,y),
with (r,θ) describing the xz-plane in polar coordinates.
True. For regions where you have rotational symmetry around the x or y-axes, cylindrical coor-
dinates can be set up with (r,θ) describing the shadow on the yz-plane or xz-plane, respectively.
R R√ 2 R
2 4−y 2
(7) (textbook 15.7.29) Evaluate −2 √ √ 2 2 xzdzdxdy by changing to cylindrical coordi-
− 4−y2 x +y
nates.
The outer bounds in x and y are a circle of radius 2 centered at the origin in the xy-plane, and
2 2
the inner bounds in z depend only on x +y , so cylindrical coordinates are a good choice for this
p 2 2
problem. We see that the lower bound z = x +y is the bound z = r, and the upper bound
remains z = 2, while the outer bounds become 0 ≤ θ ≤ 2π and 0 ≤ r ≤ 2. Setting up our integral,
4 JAMESROWAN
taking care to remember that in cylindrical coordinates, dV = rdzdrdθ, we have
Z 2 Z √4−y2 Z 2 Z 2π Z 2Z 2
√ 2 √ 2 2xzdzdxdy= rcosθ·z·rdzdrdθ
−2 − 4−y x +y 0 0 r
=Z 2πcosθdθZ 2r2Z 2zdzdr
0 0 r
Z 2π Z 2 2
z 2
= cosθdθ r2 dr
2
0 0 r
=0·Z 2r214−r2dr
0 2
= 0 .
Wecould also notice that the region of integration in this problem is the region between the cone
z2 = x2 +y2 and the plane z = 2, and that this region is reflection-symmetrical over the yz-plane,
while the integrand xz is an odd function of x, so this integral must be 0 by symmetry.
(8) (textbook 15.7.21) Evaluate RRR x2dV, where E is the solid that lies within the cylinder x2+y2 =
E
2 2 2
1, above the plane z = 0, and below the cone z = 4x +4y .
Since we are working inside a cylinder centered along the z-axis, cylindrical coordinates are a
good choice for this problem. The cylinder x2 + y2 = 1 corresponds to 0 ≤ r ≤ 1, 0 ≤ θ ≤ 2π. The
surface z2 = 4x2 + 4y2 corresponds to the surface z = 2r in cylindrical coordinates, and the plane
z = 0 will be the lower bound for z. We have
ZZZ x2dV =Z 2πZ 1Z 2rr2cos2θrdzdrdθ
0 0 0
E Z Z Z
2π 1 2r
2 3
= cos θdθ r dzdr
0 0 Z 0
cos(2θ)+1 2π 1
= · r3 · 2rdr
Z 2 0 0
1
=π· 2r4dr
0
= 2π.
5
(9) (textbook 15.7.27) Find the mass and center of mass of the solid S bounded by the paraboloid
z = 4x2 +4y2 and the plane z = a, a > 0 if S has a constant density K.
2 2 2 2
The paraboloid z = 4x + 4y intersects the plane z = a when 4x + 4y = a. This is a circle
(in the plane z = a) with radius √a centered along the z-axis. The paraboloid z = 4x2 +4y2 is the
2
paraboloid z = 4r2 in cylindrical coordinates as well, so cylindrical coordinates are a good choice
here.
