EXAM 2 SOLUTIONS
√
1. If z = er cos θ, r = st, θ = s2 + t2 , find the partial derivatives
chain rule. Write your answers entirely in terms of s and t.
dz
ds
and
dz
dt
using the
Solution. This is a routine application of the chain rule.
dz
ds
=
dz dr
dr ds
dz dθ
dθ ds
+
r
= ter cos(θ) − se√ssin(θ)
2 +t2
√
= test cos( s2 + t2 ) −
dz
dt
=
dz dr
dr dt
+
√
sest sin( s2 +t2
√
s2 +t2
dz dθ
dθ dt
r
− te√ssin(θ)
2 +t2
= ser cos(θ)
√
= sest cos( s2 + t2 ) −
√
test sin( s2 +t2
√
s2 +t2
2. Use Lagrange Multipliers to find the minimum and maximum values of f (x) = x2 y
subject to the constraint g(x, y) = x2 + 2y 2 = 6.
Solution. We first need to find the gradients of the two functions and set them equal
to each other (with a factor of λ):
∇f =< 2xy, x2 >
∇g =< 2x, 4y >
2xy = 2xλ
x2 = 4yλ
The first
x = 0 in the constraint yields
√ λ equation requires either y = λ or x = 20. First,
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y = ± 3. The other option, y = λ, leads to x = 4y in the second equation. This
plugged into the constraint yields 6y 2 = 6, or y = ±1. In turn, x = ±2. In summary,
the points to be tested in f (x, y) are the following:
f (±2, 1) = 4
f (±2, −1) = −4
√
f (0, ± 3) = 0
Thus the maximum is 4 and the minimum is −4.
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Note. The most common mistake on this problem was forgetting to consider x = 0
when looking at the first λ equation. This is necessary to do, as it represents a legitimate solution to the system of equations! Those points did not turn out to be
min/maxes this time, but in general they might, so I took off points for not doing this
step.
Note. It is not necessary to perform the calculations from 12.8 in this problem to
find local min/maxes. The problem only asked to find the global max and min, not
other extreme values. The constraint represents a curve within 2 dimensions. The 12.8
extreme value methods are only needed for a 2-D region in 2 dimensions or a 3-D space
in 3 dimensions.
3. Evaluate the triple integral below.
π
Z
Z √1−y2
1
Z
y sin xdzdydx
0
0
0
Solution.
π
Z
0
Z
π
1
Z
2
[yz sin x]0sqrt1−y dydx
0
1
Z
0
sqrt1 − y 2 y sin xdydx
0
Set u = 1 − y 2 , du = −2ydy to integrate with respect to y.
12
−
23
1
3
Z
0
π
Z
π
[(1 − y 2 )3/2 sin x]10 dx
0
1
1
2
sin xdx = [− cos x]π0 = (1 + 1) =
3
3
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Note. This is a very typical kind of integral you would be expected to evaluate in
this course. It requires basic knowledge of what an integral with respect to 1 variable
is, plus a basic u-substitution which should be easy by now from Calc 1 and 2. If you
got this problem wrong (besides a silly mistake or missing a constant or something)
you MUST come see me if you are serious about this class. You cannot pass this class
without knowing basic Calc 1 concepts!
4. Evaluate the following integral by first sketching the region of integration implied by
the limits, then reversing the order of integration.
2
1
Z
1
Z
√
0
y
2
yex
dxdy
x3
The outside integral is with respect to y, which goes from 0 to 1. The inner integral
√
for x has two bounds, the first from left to right which is y(i.e.y = x2 ) and then the
line x = 1. So the region is this:
Changing order of integration, the outside integral is with respect to x, which ranges
from 0 to 1. Then y ranges from a lower bound of y = 0 to y = x2 . So the new integral
is:
1
Z
x2
Z
0
0
Z
1
0
1
2
2
yex
dydx
x3
2
y 2 ex x2
] dx
[
2x3 0
Z
1
2
xex dx
0
Use a u-substitution u = x2 , du = 2xdx.
1 x2 1 1
[e ]0 = (e − 1)
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Note. The change of variables is ”necessary” to calculate the integral. Without a
change of variables, the dx integral would require lots of integration by parts.
Note. This is another very important concept going forward, as I have stressed in
class over and over. If you do not understand how I reversed the order of integration,
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you MUST see me if you are serious about this class. You cannot simply switch the
integrals and switch the differentials. Doing this gives you a variable on the outer
integral limit! This is never a good thing since it means your final answer will have
variables in it, not just a number. Also, you cannot reverse the order of integration by
just solving all the limits for the opposite variable. I saw a lot of people calculate the
inner integral from x2 to 1. An accurate diagram is necessary to actually see which
functions should be in which limits.
5. Find the center of mass of the portion of the sphere x2 + y 2 + z 2 = 4 which is in the
first octant (x > 0, y > 0, z > 0) if it has density function δ(x, y, z) = 1 + x2 + y 2 + z 2
Solution. First it is necessary to find the mass of the sphere piece. The mass can be
found by integrating the density function over the sphere piece,
Z Z Z
(1 + x2 + y 2 + z 2 )dV
S
Both the region of integration and the density function suggested switching to spherical
coordinates. Since the integral is only over the first octant of the sphere, the limits
are 0 ≤ θ ≤ π/2 and 0 ≤ φ ≤ π/2. The sphere’s radius is 2. Remember to include the
standard spherical coordinates differential.
π/2
Z
0
Z
0
π/2
Z
π/2
Z
(1 + ρ2 )ρ2 sin(φ)dρdθdφ
0
π/2
Z
0
2
[(ρ3 /3 + ρ5 /5) sin(φ)]20 dθdφ
0
π/2
Z
π/2
Z
(8/3 + 32/5) sin(φ)dθdφ
0
0
π/2
Z
(π/2)(8/3 + 32/5) sin(φ)dφ
0
(π/2)(8/3 + 32/5)(1) ≈ 14.24
Now to calculate one of the center of mass coordinates, say x̄, calculate the same
integral except with an extra variable multiplied at the beginning, then divide by mass.
So, for example, the x̄ integral will have x multiplied by the density function. Since
we’re doing the calculation in spherical coordinates, we must use x = ρ sin(φ) cos(θ).
Z
0
π/2
Z
0
π/2
Z
2
(1 + ρ2 )ρ3 sin2 (φ) cos(θ)dρdθdφ
0
4
Z
π/2
0
Z
π/2
[(ρ4 /4 + ρ6 /6 sin2 (φ) cos(θ)]20 dθdφ
0
Z
0
π/2
π/2
Z
(4 + 32/3) sin2 (φ) cos(θ)dθdφ
0
Z
π/2
(4 + 32/3) sin2 (φ)dφ
0
(4 + 32/3)[φ/2 −
1
π
π/2
sin(2φ)]0 = (4 + 32/3) ≈ 11.52
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x̄ = 11.52/14.24 = 0.809
The other two coordinates actually turn out to be the same. The reason for this is
the original situation was entirely symmetric with respect to x, y, and z. Actually
calculating the components using integrals will support this.
Note. Some people tried to convert to cylindrical coordinates instead. Ths is a valid
thing to do, but it doesn’t really make the integral any easier to calculate so is not
recommended. The fact that x2 + y 2 + z 2 shows up twice in the problem is a solid tell
that spherical coordinates are the best choice.
Note. A very common mistake was setting up the integral in rectangular coordinates
and integrating from 0 to 2 in all three variables. This demonstrates to me a lack of
understanding in what the limits of integration mean. Integrating from 0 to 2 would
mean calculating the mass of a cube with 2 units per side, not a sphere of radius 2!
Note. Some people switched to spherical coordinates and attempted to find ρ̄ and θ̄
instead of x̄. Although this was a clever idea, it doesn’t quite work and gives the wrong
answer for reasons that are hard to describe. To put it simply, the mass distribution
in x, y, z is a different question than the mass distribution along a radial vector. You
can check for yourself
√ by calculating ρ̄ to be about 1.6, while the ρ value of the real
center of mass is 3(0.809) ≈ 1.41.
Bonus Problem (10 Pts.)
Use
transformation
x = u2 , y = v 2 , z = w2 to calculate the volume inclosed by the surface
√ the
√
√
x + y + z = 1 and the planes x = 0, y = 0, and z = 0.
Solution. First it is necessary to calculate the Jacobian for this transformation.
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dx dx dx 2u 0 0 dv
dw du
dy dy dy = 0 2v 0 = 8uvw
du
dv
dw dz dz dz 0 0 2w du
dv
dw
The region of integration changes to the region bounded by the planes u = 0, v = 0, w = 0,
and u + v + w = 1. The original desired integral was a volume integral, i.e. integrating the
constant function 1. This function is still 1 in uvw coordinates. Thus,
Z 1 Z 1−u Z 1−u−v
8
uvwdwdvdu
0
Z
0
1
Z
4
0
Z
1
Z
4
1−u
uv(1 − u − v)2 dvdu
0
1−u
Z
uv + u3 v + uv 3 − 2u2 v − 2uv 2 + 2u2 v 2 dvdu
4
0
0
0
1
u(1 − u)2 /2 + u3 (1 − u)2 /2 + u(1 − u)4 /4 − u2 (1 − u)2 − 2u(1 − u)3 /3 + 2u2 (1 − u)3 /3du
0
Z
4
1
u/12 − u2 /3 + u3 /2 − u4 /3 + u5 /12du
0
4(1/24 − 1/9 + 1/8 − 1/15 + 1/72) = 1/90
Note. The simplification of the mess with all the (1 − u)2 terms can be made a little
easier by first factoring out the biggest thing you can from all of the terms, which is u(1−u)2 .
It actually turns out that the entire expression can be written as 1/12(u)(1 − u)4 . This can
be integrated by parts without having to multiply it all out.
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