Calculus 2
Lia Vas
Arc Length. Surface Area.
Arc Length. Suppose that y = f (x) is a continuous function with a continuous derivative on
[a, b]. The arc length L of f (x) for a ≤ x ≤ b can be obtained by integrating the length element
ds from a to b. The length element ds on a sufficiently small interval can be approximated by the
hypotenuse of a triangle with sides dx and dy.
Thus ds2 = dx2 + dy 2 ⇒ ds =
L=
Z b
ds =
Note that
dx2
=
dy
dx
2
dx2 + dy 2 and so
dx2 + dy 2 =
a
a
dy 2
Z bq
√
Z b
s
Z b
dy 2
2
(1 + 2 )dx =
dx
a
a
s
(1 +
dy 2
)dx.
dx2
= (y 0 )2 . So the formula for the arc length becomes
L=
Z bq
1 + (y 0 )2 dx.
a
Area of a surface of revolution. Suppose that y = f (x) is a continuous function with a
continuous derivative on [a, b]. To compute the surface area Sx of the surface obtained by rotating
f (x) about x-axis on [a, b], we can integrate the surface area element dS which can be approximated
as the product of the circumference 2πy
q of the circle with radius y and the height that is given by
the arc length element ds. Since ds is 1 + (y 0 )2 dx, the formula that computes the surface area is
Sx =
Z b
q
2πy 1 + (y 0 )2 dx.
a
If y = f (x) is rotated about y-axis on [a, b], then dS is the product of the circumference 2πx of
the circle with radius x and the height that is given by the arc length element ds. Thus, the formula
that computes the surface area is
Sy =
Z b
q
2πx 1 + (y 0 )2 dx.
a
Practice Problems.
1
1. Find the length of the curve y = x3/2 , 1 ≤ x ≤ 4.
√
2. Find the length of the curve y = 1 − x2 , −1 ≤ x ≤ 1.
3. Use the Left-Right sum calculator program to approximate the length of the curve y = x3 , for
0 ≤ x ≤ 1 to two digits.
4. Use the Left-Right sum calculator program to approximate the length of the curve y = sin x,
for 0 ≤ x ≤ π to five digits.
5. Use the Left-Right sum calculator program with n = 100 subintervals to approximate the
length of the curve y = ex , for 0 ≤ x ≤ 1.
6. Find the area of the surface obtained by rotating y = x3 , for 0 ≤ x ≤ 2 about the x-axis.
√
7. Find the area of the surface obtained by rotating y = x, for 4 ≤ x ≤ 9 about the x-axis.
8. Find the area of the surface obtained by rotating y = x2 , for 1 ≤ x ≤ 2 about the y-axis.
9. Prove the formula 4r2 π computes the surface area of a sphere with radius r.
10. Use the Left-Right sum calculator program to approximate the surface area obtained by rotating
the curve y = sin x, for 0 ≤ x ≤ π about x-axis to four digits.
11. Use the Left-Right sum calculator program with 100 subintervals to find the Left sum which
2
approximates the surface area of the surface obtained by rotating y = ex +1 0 ≤ x ≤ 1, about
x-axis.
12. Use the Left-Right sum calculator program with 100 subintervals to find the Right sum which
approximates the surface area of the surface obtained by rotating y = ln(x3 + 1) 0 ≤ x ≤ 1,
about y-axis.
13. A solid with infinite surface area that encloses a finite volume. The surface of revolution obtained by revolving y = x1 for 1 ≤ x ≤ ∞ is known as the Gabriel’s Horn or
Torricelli’s trumpet. Using the inequality 1 + x14 > 1, demonstrate that this surface has
infinite surface area. Then find the volume enclosed by this surface and show it is finite.
In Calculus 3, we will encounter another example of a similar phenomenon: a fractal object
called Koch snowflake with infinite perimeter that encloses a finite area.
Solutions.
2
1. y 0 = 23 x1/2 so (y 0 )2 =
u = 1 + 49 x and obtain
q
R
9
x L = 14 1 +
4
42
(1 + 94 x)3/2 |41 =
93
9
xdx. Evaluate this
4
8
(103/2 − ( 13
)3/2 ) =
27
4
integral using the substitution
7.6337.
q
2. The key step in this problem is to simplify the formula 1 + (y 0 )2 . The derivative is y 0 =
−x
x2
1
1−x2 +x2
1
0 2
(1 − x2 )−1/2 (−2x) = √1−x
= 1 + 1−x
= 1−x
2 =
2 . Thus, the length is
2 so 1 + (y )
2
1−x2
L=
R1 q 1
−1
1−x2
dx =
R1
−1
√ 1
dx
1−x2
= sin−1 x|1−1 = sin−1 (1) − sin−1 (−1) =
π
2
+
π
2
= π.
3. Careful: first write down the integral that you need to evaluate using the formula for the arc
length, then use the calculator. Do not enter x3 in Y1 because in that case the program would
give you the area under the curve, not the length.
R1√
4
y = x3 ⇒ y 0 = 3x2 . So the integral
√ L = 0 1 + 9x dx computes the arc length. To evaluate
this integral, enter the function 1 + 9x4 as Y1 in your calculator and use the program for left
and right sums. With n = 300, you obtain that the length is approximately 1.5.
4. The problems is asking for the arc length not the area under the curve so, as in the previous
problem, you need to use the formula for theR arc
function in
√ length first, before entering any √
π
0
2
the calculator. y = sin x ⇒ y = cos x. L = 0 1 + cos xdx. Enter the function 1 + cos2 x
as Y1 in your calculator and use the program for left and right sums. With n = 100, you obtain
that the length is approximately 3.8202.
√
R √
R q
5. y = ex ⇒ y 0 = ex . L = 01 1 + (ex )2 dx = 01 1 + e2x dx. Enter 1 + e2x as y1 and use the
Left-Right Sums program with a = 0, b = 1 and n = 100. Obtain the length of approximately
2.00.
q
√
R
R
6. y = x3 ⇒ y 0 = 3x2 . Sx = 02 2πy 1 + (y 0 )2 dx = 2π 02 x3 1 + 9x4 dx. Evaluate this integral
1 2
π
using the substitution u = 1 + 9x4 . Obtain 2π 36
(1 + 9x4 )3/2 |20 = 27
(1453/2 − 1) = 203.04.
3
7. y =
√
R √
R
1
1
√
. S = 49 2πy 1 + (y 0 )2 dx = 2π 49 x 1 + 4x
dx. Simplify
2 xq x
√
√
√
R
R
R9√
9
9
√ dx = π
dx = 2π 4 x 24x+1
4x + 1dx. Evaluate
Obtain 2π 4 x 4x+1
4
4x
x
12
π
3/2 9
3/2
3/2
u = 1 + 4x. Obtain π 4 3 (1 + 4x) |4 = 6 (37 − 17 ) = 81.14.
q
x = x1/2 ⇒ y 0 = 12 x−1/2 =
the function first.
this integral using
q
q
√
R
R
8. y = x2 ⇒ y 0 = 2x. Sy = 12 2πx 1 + (y 0 )2 dx = 2π 12 x 1 + 4x2 dx. Evaluate this integral using
the substitution u = 1 + 4x2 . Obtain 2π 18 23 (1 + 4x2 )3/2 |21 = π6 (173/2 − 53/2 ) = 30.85.
9. You can represent the sphere as√
the surface of revolution of the upper part of the circle x2 +y 2 =
r2 around x-axis. So, y = ± r2 − x2 . The upper half is given by the positive root. The
bounds for x are −r and r. The derivative is y 0 = 21 (r2 − x2 )−1/2 (−2x) = √r−x
2 −x2 . Similarly
to problem 2. in part a), the key step in this problem is to simplify the formula
x2
r2 −x2 +x2
r2
1 + (y 0 )2 = 1 + r2 −x2 = r2 −x2 = r2 −x2 . Thus, the surface area is Sx =
√
Rr
Rr
2
2√ r
dx = −r
2πrdx = 2πrx|r−r = 2πr(r + r) = 4r2 π.
−r 2π r − x
r2 −x2
Rr
−r
2πy
q
q
1 + (y 0 )2 .
r2
dx
r2 −x2
=
10. Careful: first write down the integral that you need to evaluate using the formula for the surface
area, then use the calculator. Do not enter sin x in Y1 because the program would give you the
area under the curve in that case, not the surface area of the surface of revolution.
3
√
√
R
y = sin x ⇒ y 0 = cos x. Sx = 0π 2π sin x 1 + cos2 xdx. Enter the function 2π sin x 1 + cos2 x
as Y1 in your calculator and use the program for left and right sums. With n = 100, obtain
that the surface area is approximately 14.42.
q
11. The problems is asking for the surface area Sx = ab 2πy 1 + (y 0 )2 dx. Find the derivative
2
x2 +1
of the function
and plug it in the formula first.
⇒ y 0 = ex +1 2x ⇒ Sx =
q y = e
q
R1
2
x2 +1
1 + (2xex2 +1 )2 dx. Then enter 2πex +1 1 + (2xex2 +1 )2 as y1 (careful with the paren0 2πe
thesis) and use the program with a = 0, b = 1 and n = 100. Obtain that the surface are is
approximately 152.9.
R
q
12. The problems is asking for the surface area Sy = ab 2πx 1 + (y 0 )2 dx. Find the derivative
2
of the function and plug it in the formula first. y = ln(x3 + 1) ⇒ y 0 = x3x
3 +1 ⇒ Sx =
R
R1
q
2
q
q
2
4
3x
3x
9x
2
2
0 2πx 1 + ( x3 +1 ) dx. Then enter 2πx 1 + ( x3 +1 ) or its simplified form 2πx 1 + (x3 +1)2 as
y1 (careful with the parenthesis) and use the program with a = 0, b = 1 and n = 100. Obtain
that the surface are is approximately 4.54.
q
. The surface area is Sx = 1∞ 2π x1 1 + x14 dx. Using the given
13. y = x1 ⇒ y 0 = −x−2 = −1
x2
√
R
R
inequality, this integral is larger than 1∞ 2π x1 1dx = 2π 1∞ x1 dx = 2π ln x|∞
1 = ∞. So, the
surface area is larger than the value of this divergent integral. So, Sx is infinite as well.
R
Volume, on the other hand, is computed as Vx =
π( −1
− (−1)) = π.
∞
4
R∞
1
π
2
1
x
dx = π
R∞ 1
1
x2
dx = π −1
|∞ =
x 1