MAT 213 Exam 3
Z 4Z 3
√
1. Evaluate
3x y dx dy
1
SOLUTIONS
1
Solution.
4Z 3
Z
1
1
3 2 √ 3
3x y dx dy =
x y dy
1 2
1
Z 4
Z 4
27 3 √
√
12 y dy
−
y dy =
=
2
2
1
1
4
= 8y 3/2 = 8(43/2 − 13/2 ) = 8(7) = 56
√
Z
4
1
2. Sketch the region D = {(x, y) : 0 ≤ x ≤ 3, 0 ≤ y ≤ 3x −
x2 ,
ZZ
and then write
D
iterated integral, but do not evaluate it.
1
dA as an
x+1
Solution.
Z
3 Z 3x−x2
0
0
3
3x − x2
dx
x+1
0
Z 3
4
=
4−x−
dx
x+1
0
3
1 2
9
=
4x − x − 4 ln (x + 1) = 12 − − 4 ln 4
2
2
0
Z
1
dy dx =
x+1
3. Use calculus to find the centroid of the part of the unit circle shaded in the figure below. (Note that
symmetry tells us that the x- and y-coordinates are equal, so we only need to use calculus to find
one of the coordinates.)
Solution.
xcm
RR
x dA
= RRD
D 1 dA
R π/2 R 1
=
r cos θ r dr dθ
π/4
1
r3 /3 cos θ 0 dθ
π/4
0
0
R π/2
=
=
So the centroid is located at the point
Z
1Z 1
Z
1Z 1
0
1/3
R π/2
0
cos θ dθ
4
=
π/4
3π
4 4
,
.
3π 3π
sin x
dx dy, sketch the region of integration, and write an equivalent iterated
x
0
y
integral in terms of dy dx. Do not evaluate it.
4. For the integral
Solution.
0
y
sin x
dx dy =
x
Z
1Z x
0
Z
1
=
0
Z
=
0
1
sin x
dy dx
x
0
sin x x
y dx
x
0
Z 1
sin x
(x) dx =
sin x dx = − cos 1 + 1
x
0
1
ZZ
p
1 + x2 + y 2 dA where D = {(x, y) : 1 ≤ x2 + y 2 ≤
5. (a) Use polar coordinates to set up
D
4 and x, y ≥ 0} as an iterated integral. Do not evaluate it.
Solution. We can rewrite D = {(r, θ) : 1 ≤ r ≤ π/2}, so the polar integral is
Z
2 Z π/2 p
1+
1
π
r dθ dr =
2
r2
0
Z
2p
1 + r2 r dr
1
Using the substitution u = 1 + r2 , du = 2r dr, we get
Z
Z
π 5√ 1
π 2p
1 + r2 r dr =
u du
2 1
2 2
2
π 1 2 3/2 5
=
·
u
2 2 3
2
π 3/2
3/2
(5 − 2 )
=
6
(b) Set up an iterated triple integral in spherical coordinates for the integral of f (x, y, z) = x + y + z
over the region W = {(x, y, z) : x2 + y 2 + z 2 ≤ 9 and z ≥ 0}. Do not evaluate it.
Solution. We can rewrite W = {(ρ, φ, θ) : 0 ≤ ρ ≤ 9, 0 ≤ φ ≤ π/2, 0 ≤ θ ≤ 2π}, so the
spherical integral is
Z
0
2π
π/2 Z 3
Z
0
(ρ cos θ sin φ + ρ sin θ sin φ + ρ cos φ) ρ2 sin φ dρ dφ dθ
0
If you evaluated this one, then kudos to you! I am not going to take the space to do it here.
Z
6. Evaluate the line integral
F · ds for each of the following. Use the Fundamental Theorem of Line
C
Integrals, if possible.
(a) F(x, y) = hx2 , 2xyi and C is the path along the unit semicircle x2 + y 2 = 1 from (1, 0) to (−1, 0)
in the counterclockwise direction.
Solution. Using parameterization c(t) = hcos t, sin ti for 0 ≤ t ≤ π, so
Z
Z π
F · ds =
F(c(t)) · c0 (t)dt
C
0
Z π
=
hcos2 t, 2 sin t cos ti · h− sin t, cos ti dt
0
Z π
=
sin t cos2 t dt
0
Using substitution u = cos t, du = − sin t dt, this becomes
Z
1
−1
−1
u3 −u du = − = 2/3
3 1
2
(b) F(x, y) = h4x, −3i and C is the line segment from (0, −1) to (3, 4).
Solution. In this case F = ∇φ(x, y), where φ(x, y) = 2x2 − 3y, so be the fundamental theorem
of line integrals
Z
F · ds = φ(3, 4) − φ(0, −1) = (18 − 12) − (0 + 3) = 3
C
2
7. (a) Let C beZthe line segment parameterized by c(t) = h3t, 4t − 1i for 0 ≤ t ≤ 1. Evaluate the line
(x + y) ds.
integral
C
Solution. To do this we will need to know kc0 (t)k = h3, 4ik = 5.
1
Z
Z
(x + y)ds =
C
(3t + (4t − 1))kc0 (t)kdt
0
1
Z
(7t − 1) dt =
= 5
0
25
2
(b) The plane 2x + y + 3z = 3 is parameterized by Φ(u, v) = hu + v, u − 2v, 1 − ui. Set up and
evaluate an appropriate surface integral to find the surface area of S = Φ(D) where D = {(u, v) :
u2 + v 2 ≤ 1}.
Solution.
√ We use Tu = h1, 1, −1i and Tv = h1, −2, 0i to find n = Tu × Tv = h−2, −1, −3i, so
knk = 14, and
ZZ
ZZ
√
√
1dS =
knkdA = 14 · area of D = 14 · π
S
D
8. Use Green’s Theorem to rewrite the following line integral as an iterated double integral. (Do not
evaluate the integral.)
I
(x + 2) dx + (2xy − y 2 ) dy
C
√
where D is the region between y = 1 − x2 and y = 0 for −1 ≤ x ≤ 1, and C denotes the boundary
of this region traversed in a counterclockwise direction.
I
ZZ
2
Solution. By Green’s Theorem
(x + 2) dx + (2xy − y ) dy =
2y dA, and the domain can be
C
D
√
described as D = {(x, y) : −1 ≤ x ≤ 1, 0 ≤ y ≤ 1 − x2 }. So
Z
ZZ
1
√
Z
1−x2
2y dA =
2y dy dx
−1
Z 1
D
=
−1
Z 1
=
0
√ 2
2 1−x
y 0
dx
1 3 1
(1 − x ) dt = x − x = 4/3
3
−1
−1
2
3