Quiz 4. Time: 15 minutes
KEY
Name:
1
Math 142, Calculus II, Fall 2016
Section:
ID#:
Note: Write all needed details in an organized manner to receive full credit.
Problem 1.(5 points) Consider the curves x = 2y and x + 1 = (y − 1)2 . (a) (2 pts) Find the intersection points,
(b) (1 pt) sketch the region enclosed by the curves (Hint: x = 2y is easy. To graph the second curve you need to
appropriately transform the graph x = y 2 ), and (c) (2 pts) compute the region area.
Solution:
(a) Note that the second curve can be written as x = (y − 1)2 − 1. Set x = 2y and x = (y − 1)2 − 1 equal to each
other and solve for y. We have (y − 1)2 − 1 = 2y or y 2 − 2y + 1 − 1 = 2y or y 2 − 4y = 0 or y(y − 4) = 0. Thus,
the curves intersect when y = 0 and y = 4.
4
(b)
x = 2y
2
−2
2
−2
4
6
8
x = (y − 1)2 − 1
x = y2
−4
(c) The area of the shaded region is
Z 4
Z 4
2
A=
[(2y) − ((y − 1) − 1)] dy =
[2y − (y 2 − 2y)] dy
0
0
Z
=
4
(4y − y 2 ) dy
0
32
64
1 i4
=
.
= 2y 2 − y 3 = 32 −
3
3
3
0
J
Problem 2.(5 points) Consider the solid whose base is the unit circle x2 + y 2 = 1 and its cross sections perpendicular to the x-axis are triangles whose height and base are equal. (a) (3 pts) Find the area of an arbitrary cross
section at x (It should be in terms of x) and (b) (2 pts) Find the volume of the solid by integrating the area in
part (a) (You need the integral boundaries).
Solution:
√
(a) The circle is two semicircles described by y = ± 1 − x2 , where
√
√
y = + 1 − x2 is the semicircle above the x−axis and y = − 1 − x2
is the semicircle below the x−axis. Let A(x) be the area of the
triangular cross section (See the figure to imagine the triangles).
Then the volume of the solid will be
Z 1
V =
A(x) dx
−1
Let’s find this area. The base of the triangle at an arbitrary x,
between −1 and 1, is a vertical line of length 2y (See the second
figure: Red segment is the base of an arbitrary triangle at x), where
√
y = 1 − x2 . Since the base of each triangle is equal to its height,
A(x) =
1
(2y) (2y) = 2y 2 = 2(1 − x2 ).
2
sec
TEC
Math 142, Calculus II,
FallVisual
2016 6.2C shows how the solid 2
Quiz 4. Time: 15 minutes
in Figure 12 is generated.
SOL
sec
2 y
y=
√
1 − x2
1
y
x
−2
x
−1
√
y =− 1−x
1
2
−y
−1
2
y
_1
−2
x
(b) Therefore,
Z
1
V =
h FIGURE
1 i1 128
(1 − x2 ) dx = x − x3
= .
Computer-generated
3
3
−1
−1
Z
A(x) dx = 2
−1
FIG
1
picture
of the solid in Example 7
J
Problem 3.(BONUS
3 points) Show that a pyramid of height h whose base is an equilateral triangle of side s
√
has volume 123 hs2 .
Solution:
Taking a horizontal slice at an arbitrary height y, then this slice is an equilateral triangle. Let A(y) be the area
of this horizontal triangle (The green triangle in the figure).
Let the side of our arbitrary triangle be s0 , then its height is found
by using Pythagorean theorem as
0 2
s
= s02
2
q
q
√
02
02
Solve for the height, then height = s02 − s4 = 3s4 = 23 s0 . The
base of our arbitrary triangle is just s0 . Thus its area is
√ !
√
1
3 0 0
3 02
1
s s =
s .
A(y) = (height) (base) =
2
2
2
4
(height)2 +
Therefore, we need to find s0 in terms of y. We use the simple trick of similar triangles. Lets consider√ the
similar triangles in black in the figure, then base of the small one is just (height of our√arbitrary triangle)/2 = 43 s0 .
The height of this small triangle is h − y. Similarly, the big triangle has base = 43 s and height h. Now since
these two triangles are similar,
h
h−y
s
√
= √
=⇒ s0 = (h − y).
3
3 0
h
s
s
4
Thus,
4
√
A(y) =
√ h
√ 2
i2
3 02
3 s
3s
s =
(h − y) =
(h2 − 2hy + y 2 ).
4
4 h
4 h2
Now the volume is simply
√ 2Z h
√ 2
√
Z h
3s
3s
1 3 ih
3 2
2
2
2
2
V =
A(y) dy =
(h
−
2hy
+
y
)
dy
=
(h
y
−
hy
+
y
)
=
hs .
2
2
4
h
4
h
3
12
0
0
0
J
is
ⱍ