1.
REVIEW 3: Key
Z 3Z 9
y sin(x2 ) dx dy, draw the correspondGiven the iterated integral
y2
0
ing region of integration on the xy-plane, reverse the order of integration, and
evaluate the integral.
Solution:
y
3
Z
3
y sin(x2 ) dx dy
y2
√
9 Z y= x
0
2.5
9
Z
Z
=
2
0
1.5
=
1
0.5
2
4
6
x
8
1
2
Z
sin(x2 ) y dy dx
y=0
9
x sin(x2 ) dx
0
= − 41 cos(x2 )|90
= 41 (1 − cos 81) ≈ 0.0558.
2. Find the volume under the graph of the function f (x, y) = 3xy + y 2 over
the region R defined by the inequalities x > 0, y > 0, and x + 2y 6 2.
Solution:
The region of integration is the triangle bounded by lines x = 0, y = 0, and
x + 2y = 2.
2
Z
Z
y=1−x/2
(3xy + y 2 ) dy dx
V =
0
y=0
2
Z
=
0
2
Z
=
2
=
0
y 3 y=1−x/2
dx
3 y=0
xy 2 +
2
3
x(1 − x/2)2 +
2
0
Z
3
(1 − x/2)3 dx
3
2
1
(4x3 − 15x2 + 12x + 4) dx = .
24
3
3. Find the mass of the pyramid bounded by the planes x = 0, y = 0,
z = 1, and 3x + 2y + z = 7 given the mass density function δ(x, y, z) = 2z.
Key:
Z 2 Z 3−3x/2 Z 7−3x−2y
2z dz dy dx = 30.
0
4.
0
1
Find the double integral
Z
3π/4
Z
4/ sin θ
r dr dθ
π/4
0
1
in polar coordinates and sketch the region of integration. Interpret the double
integral geometrically as an area or volume.
Key:
Z 3π/4
Z 3π/4 Z 4/ sin θ
8
r dr dθ =
2 dθ = 16.
sin
θ
π/4
0
π/4
The region is the triangle bounded by the lines y = x, y = −x, and y = 4.
The double integral is the area of the region; it is also the volume of the solid
between the planes z = 0 and z = 1 over the region.
5. Using cylindrical coordinates, find the mass of the upper half (above
the xy-plane) of the unit ball x2 + y 2 + z 2 6 1 given the mass density function
δ(x, y, z) = z. Find the same mass using spherical coordinates.
Key:
Z
2π
Z
1
√
Z
!
1−r 2
z dz
0
0
6.
0
π
r dr dθ = =
4
Z
0
2π
Z
0
π/2
Z
1
ρ cos ϕ ρ2 sin ϕ dρ dϕ dθ.
0
Given the parametric equations

√

x = 3 cos t


√
y = 3 sin t


 z = 4√t
of a curve C, where 0 6 t 6 π 2 , find the velocity and acceleration vectors as
they depend on t. Find also the speed and the length of curve C. Describe the
motion and its trajectory in words.
Key:
√
√ 2
dx dy dz
3
3
√
√
√
~v =
, ,
sin t,
cos t,
;
= −
dt dt dt
2 t
2 t
t
dvx dvy dvz
,
,
~a =
dt dt dt
√
√
√
√
3
3
3
3
1
√ sin t − cos t, − √ cos t − sin t, − √ ;
=
4t
4t
4t t
4t t
t t
5
k~v k = √ ;
2 t
Z π2
Z π2
5
√ dt = 5π.
length =
k~v k dt =
2 t
0
0
The motion starts at time t = 0 at point (3, 0, 0) and proceeds along the helix
5
trajectory with decreasing speed k~v k = √ , to arrive at point (−3, 0, 4π) at
2 t
time t = π 2 .
2