MATH 234 EXAM 1 REVIEW PROBLEMS
WITH SOLUTIONS
Problem 1. Consider the integral
ˆ
0
−1
ˆ
√
ˆ
1+x
√
− 1+x
f (x, y) dydx +
0
3
ˆ
1−x
f (x, y) dydx.
√
− 1+x
(1) Sketch the region of integration.
(2) Rewrite the integral as one double integral with the order of integration reversed.
Solution. This is the region bounded by the parabola x = y 2 − 1 and x = y − 1. These lines intersect at
(0, 1) and (3, −2). The integral is:
ˆ 1 ˆ y−1
f (x, y) dxdy .
y 2 −1
−2
Problem 2. Consider the following iterated integral
ˆ 2ˆ 4
I=
cos x2 dxdy.
0
2y
Sketch the region on which this integral is defined, and evaluate the integral.
Solution. To evaluate this integral, change the order of integration:
ˆ 4 ˆ x/2
I =
cos x2 dydx
0
ˆ
=
=
=
=
0
4
x
cos x2 dx
2
0
ˆ
1 4
2x cos x2 dx
4 0
ˆ
1 4 d
sin x2 dx
4 0 dx
1
sin (16)
(sin (16) − 0) =
.
4
4
Problem 3. Let D be the region bounded by y = x, y = 4, x = 0. Set up the iterated integral for both
orders of integration:
¨
y 2 exy dA,
D
and evaluate using the easier order.
Solution. The two integrals are:
ˆ 4ˆ 4
y 2 exy dydx,
0
ˆ
4
ˆ
y
y 2 exy dxdy.
x
0
1
0
WITH SOLUTIONS
2
It is easiest to integrate this in the second case:
ˆ
4
ˆ
ˆ
y
y 2 exy dxdy
0
4
y2
=
0
0
ˆ
4
1 xy
e
y
y
dy
0
2
yey − ydy
=
0
=
1 y2 y 2
e −
2
2
4
0
1 16
e − 17 .
2
=
Problem 4. Compute
ˆ
ˆ
0
2y
4 − x2
3/2
dxdy.
−1
−1/2
Solution. This integral is easier to do if we reverse the order of integration:
ˆ
0
ˆ
2y
4 − x2
−1/2
3/2
ˆ
dxdy
0
ˆ
0
4 − x2
=
−1
−1
ˆ 0
−1
dydx
x/2
−
=
3/2
3/2
x
4 − x2
dx
2
Now, apply u-substitution with u = 4 − x2 , du = −2xdx:
1
4
=
ˆ
4
u3/2 du
3
4
2 5/2
u
5
3
√ 1
32 − 9 3 .
10
1
4
=
=
Problem 5. Evaluate the following integral by converting to polar coordinates:
ˆ
1
ˆ
√
2−x2
p
x2 + y 2 dydx.
I=
x
0
Solution. In polar coordinates, this is
ˆ
I
π/2
ˆ
√
2
r2 drdθ
=
π/4
0
√2
r3
dθ
3 0
π/4
√
√ π 2 2
π 2
=
.
4
3
6
ˆ
=
=
π/2
Problem 6. Compute
ˆ
1
ˆ √1−y2
(1 − x) dxdy.
−1
0
WITH SOLUTIONS
Solution. In polar coordinates, this is
ˆ π/2 ˆ 1
(1 − r cos θ) rdrdθ
−π/2
ˆ
3
π/2
=
−π/2
0
ˆ
π/2
=
−π/2
r2
r3
−
cos θ
2
3
1
dθ
0
1 1
− cos θdθ
2 3
π 1
π/2
− (sin θ)−π/2
2
3
π 2
− .
2
3
=
=
Problem 7. Evaluate the following integral by converting to polar coordinates:
ˆ 3 ˆ √9−x2
sin x2 + y 2 dydx.
−3
0
Solution. In polar coordinates, this is
ˆ πˆ 3
sin r2 rdrdθ
0
ˆ
π
=
0
0
=
1
− cos r2
2
3
dθ
0
π
(1 − cos (9)) .
2
Problem 8. Find the center of mass of a lamina in the shape of an isosceles right triangle with equal sides
of length a, where the density at any point is proportional to the square of the distance from the vertex
opposite the hypotenuse.
Solution. The density at a point (x, y) is ρ (x, y) = x2 + y 2 . The hypotenuse is given by the equation
y = a − x. So, to compute the total mass, integrate
ˆ a ˆ a−x
M =
x2 + y 2 dydx
0
0
ˆ a
1
3
=
(a − x) x2 + (a − x) dx
3
0
ˆ a
ˆ
1 a 3
2
3
=
ax − x dx +
u du
3 0
0
a
x4
1 u4
ax3
−
+
=
3
4
3 4 0
1 1
1
a4
4
= a
− +
=
.
3 4 12
6
The x center of mass is
ˆ a ˆ a−x
1
x3 + xy 2 dydx =
M 0 0
=
1
M
ˆ
a
0
1
3
(a − x) x3 + x (a − x) dx
3
1 a5
2a
=
.
M 15
5
And the y center of mass is also 2a/5 (because the domain {x ≥ 0, y ≥ 0, x + y ≤ a} and the density x2 + y 2
are symmetric under the swap x ↔ y) .
Problem 9. Find the volume of the region inside the sphere x2 + y 2 + z 2 = 25 and outside the cylinder
x2 + y 2 = 16.
Solution. This integral is easiest to do in cylindrical
coordinates.
The curves are r2 + z 2 = 25 and r = 4.
√
√
2
2
The domain is 0 ≤ θ ≤ 2π, 4 ≤ r ≤ 5, and − 5 − r ≤ z ≤ 5 − r , so the integral is
ˆ 2π ˆ 5 ˆ √25−r2
ˆ 2π ˆ 5 p
rdzdrdθ
=
2r 25 − r2 dr.
√
0
4
− 25−r 2
0
4
WITH SOLUTIONS
4
Use u-substitution: u = 25 − r2 , then du = −2rdr, and this is
ˆ 2π ˆ 0
√
− ududθ
=
9
0
ˆ
9
2 3/2
u
dθ
3
0
0
4π 3/2 = 36π .
9
=
3
Problem 10. A swimming pool is circular with a 40-ft diameter. The depth is constant along the east-west
lines, and increases linearly from 2 ft at the south end to 7 ft at the north end. Find the volume of the pool.
2π
=
Solution. First, a function for the depth of the pool as a function of the position:
y + 20
y − 20
1
h (x, y) = 7
−2
=
(7y + 140 − 2y + 40)
40
40
40
1
9
5y + 180
= y+ .
=
40
8
2
Now, the volume of the pool will be
¨
ˆ 2π ˆ 20 9
1
h (x, y) dA =
r sin θ +
rdrdθ
8
2
D
0
0
20
ˆ 2π 9 2
1 3
r sin θ + r
dθ
=
24
4
0
0
ˆ 2π 3
9 · 202
20
=
sin θ +
dθ
24
4
0
9 · 24 · 52
26 · 53
2π
·
2π
+
(− cos θ)0
=
3 · 23
22
24 · 53
=
·π
3
2000
2000
=
·π =
π,
3
3
´ 2π
2π
since 0 sin θdθ = (− cos θ)0 = 1 − 1 = 0.
Problem 11. Express the following iterated integral
ˆ 1/2 ˆ √2x−x2 ˆ x+y+4
0
√
x2 + y 2 + z 2 dzdydx
−x2 −y 2 −2
3x
as an interated integral in cylindrical coordinates.
Solution. In cylindrical coordinates,
z = x + y + 4 is z = r√
(cos θ + sin θ) + 4, and z = −x2 − y 2 − 2 is
√
z = −r2 − 2. The line y = 3x is θ = π/3, and the curve y = 2x − x2 is
√
y 2 + x2
=
2x
r2
=
2r cos θ
r
=
2 cos θ.
θ ranges from θ = π/3 (the line y = 3x) to the line θ = π/2 (the line x = 0), so the integral is
ˆ π/2 ˆ 2 cos θ ˆ r(cos θ+sin θ)+4
r2 + z 2 rdzdrdθ .
π/3
−r 2 −2
0
Problem 12. Consider the following triple integral:
ˆ 2 ˆ √2x−x2 ˆ √4−x2 −y2
√
0
0
−
dzdydx.
4−x2 −y 2
WITH SOLUTIONS
5
(1) Describe the solid region of integration by giving the equations of the surfaces that bound the solid.
(2) Convert the triple integral into an iterated integral in cylindrical coordinates, but do not evaluate.
Solution. The bounds in z give x2 + y 2 + z 2 = 4, a sphere of radius 2 centered at the origin. In the xy-plane,
2
y 2 + x2 − 2x = 0 is the circle y 2 + (x − 1) = 1 of radius 1 centered at (1, 0). So, this integral calculates the
volume of the region inside cylinder with center axis (1, 0, z) , bounded above and below by the sphere of
radius 2 centered at the origin. √
In cylindrical coordinates, y = 2x − x2 is r = 2 cos θ, θ ∈ [−π/2, π/2], so this is
ˆ
π/2
−π/2
ˆ
2 cos θ
0
ˆ
√
4−r 2
√
− 4−r 2
rdzdrdθ .
˝
Problem 13. Express the triple integral D xdV as an iterated integral in spherical coordinates, where D
is the solid region bounded above by the sphere x2 + y 2 + z 2 = 2z and below by the plane z = 1. (Do not
evaluate).
2
2
2
Solution. The plane z = 1 is ρ cos
qφ = 1. The sphere x + y + z = 2z is ρ = 2 cos φ. ρ cos φ = 1 and
1
2,
ρ = 2 cos φ intersect when cos φ =
ˆ
2π
ˆ
i.e., at φ = π/4. So, this integral is
π/4
ˆ
2 cos φ
(ρ sin φ cos θ) ρ2 sin φdρdφdθ .
0
1/ cos φ
0
Problem 14. Suppose T is a transformation of the uv plane to the xy plane and we know that in the limit
∆u, ∆v → 0, both the rectangle with lower left corner (0, −2) width ∆u and height ∆v, and the rectangle
with lower left corner (5, 2) width ∆u and height ∆v map to patches of eight times their original areas.
(1) Which two of the following transformations
could be T ?
(a) x = u/3, y = v 12 + v 2
(b) x = u2 /3, y = v (2v − 1)
(c) x = v (v + 2), y = u/3
(d) x = 4v 3 , y = −u/6
(2) Could you differentiate between the two transformations from (1) if you knew how T changed the
area of the rectangle with lower left corner (u0 , v0 ), where v0 is equal to neither −2 or 2? Why/Why
not?
Solution.
(1) For this problem, we need to compute the Jacobian, as dxdy = J (u, v) dudv, so
∆x∆y
J (u, v) ≈
.
∆u∆v
Recall that J (u, v) =
∂x
∂u
∂y
∂u
∂x
∂v
∂y
∂v
, the absolute value of the determinant of the matrix of partial
derivatives.
1/3
0
(a) J =
= 13 12 + 3v 2 , so J (0, −2) = 8 and J (5, 2) = 8. Hence this is a possible
0
12 + 3v 2
T.
(b) J = (2u/3) (4v − 1); J (0, −2) = 0 6= 8, so this is not a possibility for T .
0
2v + 2
(c) J =
= − (2v + 2) /3; J (0, −2) = 2/3, J (5, 2) = 4, so this isn’t a possible T .
1/3
0
(d) J = 12v 2 /6 = 2v 2 ; J (0, −2) = 8 and J (5, 2) = 8, so this is a possible T .
Therefore, the possible choices are transformations (a) and (d).
(2) Yes, you could tell the difference between (a) and (d) if v0 6= 2 or −2. This is because 2 and -2 are
the two values where transformation (a) and (d) agree, i.e.,
1
12 + 3v 2
= 2v 2
3
v 2 = 4.
If we knew the value of J at some v0 6= 2, we could determine which of these transformations was
correct.
WITH SOLUTIONS
Problem 15. Compute the integral
6
¨
y 2 dA,
D
where D is the region in the first quadrant bounded by the curves xy = 1, y = x, y = 4x, and xy = 2.
Solution. Use the parametrization u = xy, v = xy , then the bounds on this region are u = 1 to u = 2 and
v = 1 to v = 4.
r
√
u
x=
,
y = uv,
v
so the Jacobian is
q
p
1
1
1
1
1
− 12 vu2
J = 2 puv
+
=
.
pu =
1
v
1
4v 4v
2v
u
2
Then, the integral is
ˆ
2
ˆ
4
(uv)
1
1
2
1
2v
v
ˆ
¨
ˆ
4
1
udvdu
1 2
1
2 2
u
9
=
3
.
4 1
4
dvdu =
=
Problem 16. Compute the integral
2
x + 2y
dA,
3/2
(3x − 2y)
where D is the region bounded by the curves x + 2y = 0, x + 2y = 2, 3x − 2y = 4, and 3x − 2y = 1.
D
Solution. Let u = x + 2y, v = 3x − 2y. Then,
dudv =
∂u
∂x
∂v
∂x
∂u
∂y
∂v
∂y
dxdy =
so
dxdy =
This integral is
ˆ
2
ˆ
0
4
1
u 1
dudv
3/2
8
v
=
=
1
3
2
dxdy = 8dxdy,
−2
1
dudv.
8
2 4
u2
−2v −1/2
2 0
1
1
1
1
(4)
.
=
8
2
4
1
8
Problem 17. Sketch the region of integration and evaluate the double integral
ˆ 4 ˆ y/2+1
(2x − y) sin (2x − y) dxdy.
y/2
0
Solution. The line x = y/2 + 1 is 2x − y = 1, and the line x = y/2 is 2x − y = 0. So, let u = 2x − y, v = y,
then
2 −1
dudv =
dxdy = 2dxdy,
0 1
so dxdy = 21 dudv. This integral is then
ˆ 4ˆ 1
0
0
1
u sin (u) dudv = 2
2
Apply integration by parts:
ˆ 1
ˆ
u sin (u) du.
0
ˆ
u sin udu =
0
=
1
(−u cos (u))0
1
1
cos (u) du
+
0
1
(− cos (1)) + (sin u)0 = sin (1) − cos (1) ,
WITH SOLUTIONS
7
so the final answer is
2 (sin (1) − cos (1)) .
Problem 18. Use a change of variables to express the following integral as an integral over the unit sphere:
˚ p
2
y2
z2
− x
2 + b 2 + c2
a
x2 + y 2 e
dV,
E
where
E=
y2
z2
x2
(x, y, z) : 2 + 2 + 2 ≤ 1 .
a
b
c
(Do not evaluate).
Solution. The unit sphere is ρ = 0 to ρ = 1, φ = 0 to π, θ = 0 to 2π. We do the conversion in two steps:
First, do substitution on x, y, and z:
x
dx = adx0 ,
x0 = ,
a
y
y0 = ,
dy = bdy 0 ,
b
z
dz = cdz 0 .
z0 = ,
c
o
n
2
2
2
This changes E to the region E 0 = (x0 , y 0 , z 0 ) : (x0 ) + (y 0 ) + (z 0 ) ≤ 1 , and the integrand to
2
2
2
p
p
+ yb2 + zc2
− x
02
02
02
a2
2
2
x +y e
= a2 x02 + b2 y 02 e−(x +y +z ) .
Note that now dxdydz = abcdx0 dy 0 dz 0 .
Second, convert to spherical coordinates in x0 , y 0 , z 0 :
x0 = ρ cos θ sin φ,
and the integral is
ˆ
0
2π
ˆ
0
π
ˆ
0
1
y 0 = ρ sin θ sin φ,
z 0 = ρ cos φ,
q
2
abc a2 ρ2 cos2 θ sin2 φ + b2 ρ2 sin2 θ sin2 φe−ρ ρ2 sin φdρdφdθ .