Engineering Math II – Spring 2015
Quiz #2
Name:
Section: 561 / 562 / 563
Directions. Read each problem carefully and work it out on a separate sheet of paper.
Put a box around the choice you believe best answers the question. Turn in your work
with this sheet of paper stapled on top.
Problem 1 (5 pts). The base of a solid is the parabolic region between the graphs of
y = 1 and y = x2 . Cross sections perpendicular to the y-axis are squares. Find the
volume of the solid.
(a)
4
3
(b) 2
(c) 1
(d)
1
3
(e) 3
Solution. The correct answer choice is (b). The base of the solid is the region between
the graphs of y = 1 and y = x2 , shown below.
The cross sections perpendicular to the y-axis are squares, whose side is just 2x. Since
the cross sections are perpendicular to the y-axis, we need to integrate over y, so the side
√
of each square is 2 y. Thus the area of each square is 4y, so that the volume of the solid
is given by
Z 1
1
2
4ydy = 2y = 2(1 − 0) = 2.
0
0
MATH 152:561-563 – Spring 2015
Quiz #2
2
Problem 2 (5 pts). Find the volume of the solid formed by rotating the region bounded
by x = 0, y = ln(x), y = 0, and y = 2 about the y-axis.
(a)
π 4
e
2
(b)
π 4
e
2
−1
(c) πe4 − 1
(d)
π
2
(e)
π 4
(e
2
− 1)
Solution. The correct answer choice is (e). The region to be rotated is shown below.
Cross sections of the resulting solid perpendicular to the y-axis are discs with radius x.
Hence the area of each disc is πx2 . Since the cross sections are perpendicular to the
y-axis, we need to find the area in terms of y. To do so, note that if y = ln(x), then
x = ey . Hence the are of each disc, in terms of y, is πe2y . Thus the volume of the solid
is given by
Z 2
π 2y 2 π 4
2y
πe dy = e = (e − 1).
2
2
0
0
MATH 152:561-563 – Spring 2015
Quiz #2
3
Problem 3 (5 pts). Which of the following represents the area between the curves y =
and y = 6 on the interval from x = 1 to x = 4?
R 4 12
(a) 1 x − 6 dx
(b)
R2
(c)
R4
(d)
R 12 12
(e)
R6
1
1
3
3
6−
12
x
6−
12
x
y
dx +
R 4 12
2
x
12
x
− 6 dx
dx
− 1 dy
4−
12
y
dy +
R 12 12
6
y
− 1 dy
Solution. The correct answer choice is (e). This can be deduced by a process of eliminationg. The graph of the two curves is given below.
Clearly (a) cannot be the correct answer, since the function y = 12
is not always the “top”
x
function. For similar reasons, (c) cannot be the correct answer. Choice (b) looks like it
might be correct, but on the interval from x = 1 to x = 2, y = 12
is the top function,
x
not 6, so this cannot be the right choice. This leaves choices (d) and (e), both of which
are integration over the y-axis. However, (d) cannot be the correct choice, because even
is not always the “top” function.
when integrating over the y-axis, the function x = 12
y
This leaves only choice (e).
MATH 152:561-563 – Spring 2015
Quiz #2
4
Problem 4 (5 pts). Find the area of the region bounded by y = cos(x), y = −1, x = 0
and x = π2 .
(a) π
π
2
(b)
(c) 1 +
π
2
(d) −1 +
π
2
(e)
π
2
+2
Solution. The correct choice is (c). We can split the computation into two parts. First
we find the area under y = cos(x) from x = 0 to x = π2 . Then we find the area “under”
the curve y = −1. The total area will be the sum of these two. That is, the are of the
region is given by
Z
π/2
Z
cos(x)dx +
0
0
π/2
π/2
π
π
π
dy = (sin(x) + x) = sin( ) + − (0 + 0) = 1 + .
2
2
2
0