Length of Curves

1. Raymund T. de la Cruz
MAEd – Mathematics
Math Analysis II
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APPLICATIONS OF INTEGRATION
LENGTH OF CURVE
Let 𝑓 be differentiableon [𝑎, 𝑏]. Considerthe partofthe graphof 𝑓 from(𝑎, 𝑓( 𝑎))to (𝑏, 𝑓( 𝑏)).
Let us find a formula for the length 𝐿 of this curve. Divide [𝑎, 𝑏] into 𝑛 equal subintervals, each of
length ∆𝑥. To each point 𝑥 𝑘 in this subdivision there corresponds a point 𝑃 𝑘(𝑥 𝑘, 𝑓( 𝑥 𝑘))on the curve.
For large 𝑛, the sum 𝑃0 𝑃1 + 𝑃1 𝑃2 + ⋯+ 𝑃 𝑛−1 𝑃𝑛 = ∑ 𝑃 𝑘−1 𝑃 𝑘
𝑛
𝑘=1 of the lengths of the line segments
𝑃 𝑘−1 𝑃 𝑘 is an approximation to the length of the curve.
By the distance formula,
𝑃 𝑘−1 𝑃 𝑘 = √(𝑥 𝑘 − 𝑥 𝑘−1)2 + (𝑓( 𝑥 𝑘)− 𝑓(𝑥 𝑘−1))2
Now, 𝑥 𝑘 − 𝑥 𝑘−1 = ∆𝑥 and, by the law of mean,
𝑓( 𝑥 𝑘)− 𝑓( 𝑥 𝑘−1) = ( 𝑥 𝑘 − 𝑥 𝑘−1) 𝑓′( 𝑥 𝑘
∗ ) = (∆𝑥)𝑓′( 𝑥 𝑘
∗ )
for some 𝑥 𝑘
∗
in ( 𝑥 𝑘 − 𝑥 𝑘−1). Thus,
𝑃 𝑘−1 𝑃 𝑘 = √(∆𝑥)2 + (∆𝑥)2 𝑓′(𝑥 𝑘
∗
)2
= √1 + (𝑓′(𝑥 𝑘
∗
))2
√(∆𝑥)2
= √1 + (𝑓′(𝑥 𝑘
∗
))2∆𝑥
So, ∑ 𝑃 𝑘−1 𝑃 𝑘
𝑛
𝑘=1 = ∑ √1 + (𝑓′(𝑥 𝑘
∗
))2∆𝑥𝑛
𝑘=1
The right-hand sum is an approximating sum for the definite integral ∫ √1 + (𝑓′(𝑥 𝑘
∗
))2 𝑑𝑥.
𝑏
𝑎
Therefore, letting 𝑛 → +∞, we get the 𝑎𝑟𝑐 𝑙𝑒𝑛𝑔𝑡ℎ 𝑓𝑜𝑟𝑚𝑢𝑙𝑎:
𝐿 = ∫ √1 + (𝑓′(𝑥 𝑘
∗
))2 𝑑𝑥 = ∫ √1 + (𝑦′)2 𝑑𝑥.
𝑏
𝑎
𝑏
𝑎

2. EXAMPLE 1: Find the arc length 𝐿 of the curve 𝑦 = 𝑥3 2⁄
from 𝑥 = 0 to 𝑥 = 5.
Since 𝑦′
=
3
2
𝑥1 2⁄
,
𝐿 = ∫ √1 + (𝑦′)2 𝑑𝑥
5
0
= ∫ √1 +
9
4
𝑥𝑑𝑥
5
0
=
4
9
∫ (1 +
9
4
𝑥)1 2⁄
(
9
4
) 𝑑𝑥 =
4
9
2
3
(1 +
9
4
𝑥)3 2⁄5
0
]0
5
=
8
27
((
49
4
)3 2⁄
− 13 2⁄
) =
8
27
(
343
8
− 1) =
335
27
EXAMPLE 2: Find the arc length of the curve 𝑥 = 3𝑦3 2⁄
− 1 from 𝑦 = 0 to 𝑦 = 4.
We can reverse the role of 𝑥 and 𝑦 in the arc length formula.
Since 𝑥′
=
9
2
𝑦1 2⁄
𝐿 = ∫ √1 + (𝑥′)2 𝑑𝑦
4
0
= ∫ √1 +
81
4
𝑦𝑑𝑦
4
0
=
4
81
∫ (1 +
81
4
𝑥)1 2⁄
(
81
4
) 𝑑𝑥 =
4
81
2
3
(1 +
81
4
𝑥)3 2⁄5
0
]0
4
=
8
243
((82)3 2⁄
− 13 2⁄
) =
8
243
(82√82 − 1)
EXERCISES:
1. Find the arc length of the curve 24𝑥𝑦 = 𝑥4
+ 48 from 𝑥 = 2 to 𝑥 = 4. Ans.
17
6
2. Find the arc length of the curve 6𝑥𝑦 = 𝑥4
+ 3 from 𝑥 = 1 to 𝑥 = 2. Ans.
17
12
3. Find the arc length of the curve 27𝑦2
= 4(𝑥 − 2)3
from (2,0) to (11,6√3). Ans. 14
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SURFACE AREA
Surface Area Formulas
𝑆 = ∫ 2𝜋𝑦𝑑𝑠 rotation about the 𝑥 − axis
𝑆 = ∫ 2𝜋𝑥𝑑𝑠 rotation about the 𝑦 − axis
where,
𝑑𝑠 = √1 + (𝑦′)2 𝑑𝑥 if 𝑦 = 𝑓( 𝑥), 𝑎 ≤ 𝑥 ≤ 𝑏
𝑑𝑠 = √1 + (𝑥′)2 𝑑𝑦 if 𝑥 = ℎ( 𝑦), 𝑐 ≤ 𝑦 ≤ 𝑑
There are couple of things to note about these formulas. First, notice that the variable in the
integral itself is always the opposite variable from the one we’re rotating about. Second, we are
allowed to use either 𝑑𝑠 in either formula. This means that there are, in some way, four formulas
here. We will choose the 𝑑𝑠 based upon which is the most convenient for a given function and
problem.

3. EXAMPLE 1: Determine the surface area of the solid obtained by rotating 𝑦 = √9 − 𝑥2,−2 ≤ 𝑥 ≤ 2
about the 𝑥 − axis.
The formula that we’ll be using here is 𝑆 = ∫ 2𝜋𝑦𝑑𝑠.
𝑦′
=
1
2
(9 − 𝑥2
)
−
1
2 (−2𝑥) = −
𝑥
(9−𝑥2)
1
2
𝑑𝑠 = √1 +
𝑥2
9−𝑥2 = √
9
9−𝑥2 =
3
√9−𝑥2
𝑆 = ∫ 2𝜋√9 − 𝑥22
−2
3
√9−𝑥2 𝑑𝑥
= ∫ 6𝜋𝑑𝑥
2
−2
= 24𝜋
EXAMPLE 2: Determine the surface area of the solid obtained by rotating 𝑦 = √ 𝑥3
,1 ≤ 𝑦 ≤ 2 about
the 𝑦 − axis.
The formula that we’ll be using here is 𝑆 = ∫ 2𝜋𝑥𝑑𝑠.
𝑥 = 𝑦3
𝑥′
= 3𝑦2
𝑑𝑠 = √1 + 9𝑦4
𝑆 = ∫ 2𝜋𝑦3
√1 + 9𝑦4 𝑑𝑦
2
1
𝑢 = 1 + 9𝑦4
=
𝜋
18
∫ √ 𝑢𝑑𝑢
145
10
=
𝜋
27
(145
3
2 − 10
3
2 ) = 199.48
EXERCISES:
1. Determine the surface area of the solid obtained by rotating 𝑦 = 2𝑥, 0 ≤ 𝑥 ≤ 1 about the
𝑥 − axis. Ans. 2𝜋√5
2. Determine the surface area of the solid obtained by rotating 𝑦 = 2𝑥, 0 ≤ 𝑥 ≤ 1 about the
𝑦 − axis. Ans. 𝜋√5
3. Find the surface area of the solid obtained by rotating 𝑦 = √ 𝑥,2 ≤ 𝑥 ≤ 6 about the
𝑥 −axis. Ans.
49𝜋
3
4. Find the surface area of the solid obtained by rotating 𝑦 = 𝑥 + 1,0 ≤ 𝑥 ≤ 3 about the
𝑦 − axis. Ans. 9𝜋√2

4. CENTROID OF A VOLUME
The Triple Integral
Let 𝑓(𝑥, 𝑦, 𝑧) be a continuous function on a three-dimensional region 𝑅. The definition of the double
integral can be extended in an obvious way to obtain the definition of the integral ∭ 𝑓( 𝑥, 𝑦, 𝑧) 𝑑𝑉.
If 𝑓( 𝑥, 𝑦, 𝑧) = 1, then ∭ 𝑓( 𝑥, 𝑦, 𝑧) 𝑑𝑉 may be interpreted as measuring the volume of the
region 𝑅.
Evaluation of Triple Integral
Example 1: Evaluate ∫ ∫ ∫ 𝑥𝑦𝑧 𝑑𝑧 𝑑𝑦 𝑑𝑥
2−𝑥
0
1−𝑥
0
1
0
= ∫ [∫ (∫ 𝑥𝑦𝑧 𝑑𝑧
2−𝑥
0
) 𝑑𝑦
1−𝑥
0
] 𝑑𝑥
1
0
= ∫ [∫
𝑥𝑦(2−𝑥)2
2
𝑑𝑦
1−𝑥
0
] 𝑑𝑥
1
0
= ∫ [
𝑥(1−𝑥)2
(2−𝑥)2
4
] 𝑑𝑥 =
1
0
1
4
∫ (4𝑥 − 12𝑥2
+ 13𝑥3
− 6𝑥4
+ 𝑥5) 𝑑𝑥 =
13
240
1
0
Centroid of Volume Problem
Example: Compute the triple integral of 𝐹( 𝑥, 𝑦, 𝑧) = 𝑧 over the region 𝑅 in the first octant bounded
by the planes 𝑦 = 0, 𝑧 = 0, 𝑥 + 𝑦 = 2, 2𝑦 + 𝑥 = 6,and the cylinder 𝑦2
+ 𝑧2
= 4.
Integrate first with respect to 𝑧 from 𝑧 = 0 (the 𝑥𝑦 plane)to 𝑧 = √4 − 𝑦2 (the cylinder), then
with respect to 𝑥 from 𝑥 = 2 − 𝑦 to 𝑥 = 6 − 2𝑦, and finally with respect to 𝑦 from 𝑦 = 0 to
𝑦 = 2. This yields
∫ ∫ ∫ 𝑧 𝑑𝑉 = ∫ ∫ ∫ 𝑧 𝑑𝑧 𝑑𝑥 𝑑𝑦 =
√4−𝑦2
0
6−2𝑦
2−𝑦
2
0
∫ ∫ [
𝑧2
2
]0
√4−𝑦2
𝑑𝑥 𝑑𝑦
6−2𝑦
2−𝑦
2
0
=
1
2
∫ ∫ (4 − 𝑦2) 𝑑𝑥 𝑑𝑦 =
6−2𝑦
2−𝑦
2
0
1
2
∫ [(4− 𝑦2) 𝑥
2
0
]2−𝑦
6−2𝑦
𝑑𝑦 =
26
3

5. EXAMPLE 2: Compute the triple integral of 𝑓( 𝑟, 𝜃, 𝑧) = 𝑟2
over the region 𝑅 bounded by the
paraboloid 𝑟2
= 9 − 𝑧 and the plane 𝑧 = 0.
Integrate first with respect to 𝑧 from 𝑧 = 0 to 𝑧 = 9 − 𝑟2
, then with respect to 𝑟 from 𝑟 = 0 to
𝑟 = 3, and finally with respect to 𝜃 from 𝜃 = 0 to 𝜃 = 2𝜋. This yields
∫ ∫ ∫ 𝑟2
𝑑𝑉 =∫ ∫ ∫ 𝑟2( 𝑟 𝑑𝑧 𝑑𝑟 𝑑𝜃)
9−𝑟2
0
3
0
2𝜋
0
= ∫ ∫ 𝑟3(9 − 𝑟2)
3
0
2𝜋
0
𝑑𝑟 𝑑𝜃
= ∫ [
9
4
𝑟4
−
1
6
𝑟6
]
0
3
𝑑𝜃 = ∫
243
4
𝑑𝜃 =
243
2
𝜋
2𝜋
0
2𝜋
0