Math 166
Quiz 2
Name:
Directions: This quiz is worth a total of 10 points. To receive full credit, all work must be shown.
1. Recall that the length L of a curve given by x = g(y) for a ≤ y ≤ b is
Z
b
L=
p
1 + g 0 (y)2 dy.
a
Z
2
r
1+
Use this to find a curve through the point (0, 1) whose length is given by the integral
1
1
dy.
y4
Solution. Note that the curve x = g(y) whose length is the integral given above must satisfy the
equation
1
g 0 (y)2 = 4 .
y
Taking the square root of both sides gives
g 0 (y) = ±
1
.
y2
Therefore,
1
g(y) = ∓ + C
y
for some constant C. Note that since the curve must pass through (0, 1), it must be that g(1) = 0.
That is,
0 = g(1) = ∓1 + C
which implies C = ±1. Therefore, there are two possible curves,
g(y) =
1
−1
y
1
or g(y) = − + 1,
y
both with y values in the range 1 ≤ y ≤ 2 which work.
√
2. Find the area of the surface generated by revolving the curve y =
x + 1, 1 ≤ x ≤ 5, about the x-axis.
Solution. We use here the formula for surface area
Z 5
p
A=
2πf (x) 1 + f 0 (x)2 dx,
1
where f (x) =
√
x + 1. To apply the formula, we first note that
f 0 (x) =
1
(x + 1)−1/2 ,
2
which implies
1
1 + f 0 (x)2 = 1 + (x + 1)−1 .
4
Therefore
p
√
f (x) 1 + f 0 (x)2 = x + 1
r
1
1 + (x + 1)−1 =
4
r
1
x+1+ =
4
Thus
Z
=
=
=
=
5
r
5
dx
4
1
3/2 4π
5
5
x+
3
4
1
" 3/2 #
3/2
25
9
4π
−
3
4
4
4π 125 27
−
3
8
8
49π
.
3
A = 2π
x+
r
5
x+ .
4