7.4 Arc Length and Area of a Surface of Revolution

Home , Solid of revolution, Surface of revolution

Arkansas Tech University MATH 2924: Calculus II Dr. Marcel B. Finan

7.4 Arc Length and Surfaces of Revolution The deﬁnite integral can also be used to compute the length of a smooth curve (i.e. a curve with no corner points) and the area of a surface of revolution.

Arc Length Recall that when using the integral to find the area of a region one approximates the region by rectangles the sum of whose areas approximates the area of the region. In finding the length of an arc one approximates the arc by a finite set of straight line segments. An approximation of the length of the arc is made by using the well known formula for the length of a line segment and taking a sum . A limiting process then yields the definite integral which is equal to the length of the arc. To elaborate the above statement, if an arc is just a line segment with end- points (x1, y1) and (x2, y2) then its length can be found by the Pythagorean theorem:

p 2 2 p 2 2 s = (x2 − x1) + (y2 − y1) = (∆x) + (∆y) .

Now, if the arc is the graph of a function f(x) deﬁned on an interval [a, b], then we divide the interval into n equal subintervals. See Figure 7.4.1. The corresponding points on the arc have coordinates (xi, f(xi)), so two consecutive points are seperated by a distance equal to

p 2 2 si = (xi − xi−1) + (f(xi) − f(xi−1)) .

1 ∗ But by the Mean Value Theorem there is a point xi in the interval [xi−1, xi] such that

0 ∗ 0 ∗ f(xi) − f(xi−1) = f (xi )(xi − xi−1) = f (xi )∆x.

Hence, q q 2 0 ∗ 2 0 ∗ 2 si = (∆x) + [f (xi )∆x] = 1 + [f (xi )] ∆x.

1If f is continuous in [a, b] and diﬀerentiable in (a, b) then there is a number c in the interval (a, b) such that f(b) − f(a) = f 0(c)(b − a).

1 The total length of the arc is

n n q X X 0 ∗ 2 s ≈ si = 1 + [f (xi )] ∆x. i=1 i=1 Again, we recognize the sum on the right-hand side as a Riemann sum which converges to the following integral s n b b 2 X q Z p Z dy s = lim 1 + [f 0(x∗)]2∆x = 1 + [f 0(x)]2dx = 1 + dx. n→∞ i dx i=1 a a (1)

Figure 7.4.1

Example 7.4.1 3 Find the length of the arc deﬁned by the curve y = x 2 between the points (0, 0) and (1, 1).

Solution. Using the arc length formula we have

1 r 1 Z dy Z q 3 s = 1 + ( )2dx = 1 + [(x 2 )0]2dx 0 dx 0 1 r 1 r Z 3 1 Z 9x = 1 + ( x 2 )2dx = 1 + dx 0 2 0 4 1 1 3 1 3 = (4 + 9x) 2 = (13 2 − 8) unit length 27 0 27 Example 7.4.2 π Find the arc length of the graph of y = ln (cos x) from x = 0 to x = 4 .

2 Solution. We have

π π Z 4 p Z 4 p s = 1 + [− tan x]2dx = 1 + tan2 xdx 0 0 π π Z 4 √ Z 4 = sec2 xdx = sec xdx 0 0 π √ = [ln | sec x + tan x|] 4 = ln ( 2 + 1) − ln 1 √ 0 = ln ( 2 + 1)

If a curve has the equation x = g(y), c ≤ y ≤ d then by interchanging the role of x and y above we ﬁnd s d d 2 Z p Z dx s = 1 + [g0(y)]2dy = 1 + dy. c c dy

Example 7.4.3 3 2 2 Find the length of the arc of the curve x = 3 (y − 1) for 1 ≤ y ≤ 4.

Solution. We have s 4 2 Z 2 3 1 s = 1 + (y − 1) 2 dy 1 3 2 Z 4 Z 4 √ = p1 + y − 1dy = ydy 1 1 4 2 3 14 = y 2 = 3 1 3 Example 7.4.4 Find the length of the arc of the curve x2 = (y − 1)3 for 0 ≤ x ≤ 8.

Solution. 2 If we solve y in terms of x, we ﬁnd y = x 3 +1. The derivative of this function dy 2 dy is dx = 1 . Since dx is undeﬁned at x = 0, the arc length formula (1) can 3x 3 3 not be used. Hence, we express x in terms of y obtaining x = ±(y − 1) 2 .

3 3 But 0 ≤ x ≤ 8 implies x = (y − 1) 2 . Hence,

Z 5 r 9 Z 5 r9 5 s = 1 + (y − 1)dy = y − dy 1 4 1 4 4 5 Z 5 " 3 # 1 p 1 (9y − 5) 2 = 9y − 5dy = 3 2 1 18 2 1 1 3 3 1 3 = (40 2 − 4 2 ) = (40 2 − 8) 27 27 Example 7.4.5 An electric cable is hung between two towers that are 200 ft apart. The cable takes the shape of a catenary whose equation is

x − x y = 75(e 150 + e 150 ).

Find the arc length of the cable between the two towers.

Solution. Taking the derivative of y, we ﬁnd 0 1 x 1 − x 1 x − x y = 75 e 150 − e 150 = [e 150 − e 150 ]. 150 150 2

Thus, 0 2 1 h x − x i (y ) = e 75 − 2 + e 75 4 and 0 2 1 x 1 1 − x 1 x − x 2 1 + (y ) = 1 + e 75 − + e 75 = (e 150 + e 150 ) . 4 2 4 4 The arc length of the cable is

Z 100 100 1 x − x h x − x i s = (e 150 + e 150 )dx = 75 e 150 − e 150 −100 2 −100 2 − 2 − 2 2 =75(e 3 − e 3 ) − 75(e 3 − e 3 ) 2 − 2 =150(e 3 − e 3 )

4 Area of a Surface of Revolution The surface of a solid of revolution is the lateral boundary of the solid. It is formed when a curve is rotated about a line. In this section, we want to ﬁnd the area of such a surface.

Example 7.4.6 Find the surface area of a circular cone with base radius r and slant height l.

Solution. Recall from trigonometry that the area of a circular sector of radius l and 1 2 central angle θ is given by A = 2 l θ. Cutting the cone along the dashed line and ﬂattening out we obtain a circular sector with radius l and central angle θ as shown in Figure 7.4.2.

Figure 7.4.2

The area of this circular sector which is the area of the cone is 1 1 2πr l2θ = l2 = πrl 2 2 l

Example 7.4.7 Find the surface area of a portion of a cone as shown in Figure 7.4.3. Such a solid is called a frustum.

5 Figure 7.4.3

Solution. Let S be the surface area of the portion of the cone. Then

S = πr2l1 + l) − πr1l1 = πr2l1 + πr2l − πr1l1.

From similar triangles we have

l1 l1 + l = =⇒ r2l1 = r1l1 + r1l. r1 r2 Thus, S = π(r1 + r2)l

Now, consider the surface shown in Figure 7.4.4 which is obtained by rotating the function f(x), a ≤ x ≤ b about the x−axis.

Figure 7.4.4

6 b−a Divide the interval [a, b] into n subintervals each of length ∆x = n using the partition points xi = a + i∆x, i = 0, 1, ··· , n. We approximate the arc from the point Pi−1(xi−1, f(xi−1)) to Pi(xi, f(xi)) with the line segment from these two points. When we rotate this line segment about the x−axis we obtain a portion of a cone as in Example 7.4.7. Thus, the surface area from x = xi−1 to x = xi is estimated by q ∗ 2 Si ≈ π(f(xi) + f(xi−1))|Pi−1Pi| = π(f(xi) + f(xi−1)) 1 + [f(xi )] ∆x.

∗ ∗ For small ∆x, we have f(xi−1) ≈ f(xi ) and f(xi) ≈ f(xi ). Hence, q ∗ ∗ 2 Si ≈ 2πf(xi ) 1 + [f(xi )] ∆x.

Hence, the surface area of the solid of revolution is approximated by

n q X ∗ ∗ 2 S ≈ 2πf(xi ) 1 + [f(xi )] ∆x. i=1 This approximation appears to become better when n → ∞ giving

Z b p S = 2πf(x) 1 + [f 0(x)]2dx. a Note that in the above formula, 2πf(x) is the circumference of the circle with radius f(x) and p1 + [f 0(x)]2 is an element of arc length. In general, if a region is rotated about the x−axis its surface area is

Z upper limit S = 2πyds lower limit where s dy 2 ds = 1 + dx dx if y = f(x) and s dx2 ds = 1 + dy dy if x = g(y). Likewise, if a region is rotated about the y−axis its surface area is Z upper limit S = 2πxds lower limit

7 where s dy 2 ds = 1 + dx dx if y = f(x) and s dx2 ds = 1 + dy dy if x = g(y).

Example 7.4.8 √Find the surface area of the solid of revolution generated by rotating y = 4x + 1, 1 ≤ x ≤ 5 about the x−axis.

Solution. We have s Z 5 √ 2 2 S = 2π 4x + 1 1 + √ dx 1 4x + 1 Z 5 √ =2π 5 + 4xdx 1 25 25 π Z √ π 2 3 = udu = u 2 2 9 2 3 9 π 3 3 98 = [25 2 − 9 2 ] = π 3 3 Example 7.4.9 Find the surface area of the solid of revolution generated by rotating x = 3 1 2 2 3 (y + 2) , 1 ≤ y ≤ 2 about the x−axis.

Solution. We have

Z 2 p Z 2 p S = 2πy 1 + y2(y2 + 2)dy = 2π y y4 + 2y2 + 1dy 1 1 Z 2 p Z 2 =2π y (y2 + 1)2dy = 2π y(y2 + 1)dy 1 1 Z 2 y4 y2 2 21 =2π (y3 + y)dy = 2π + = π 1 4 2 1 2

8 Example 7.4.10 Find the surface area of the solid of revolution generated by rotating y = x2, 1 ≤ x ≤ 2 about the y−axis.

Solution. We have

Z 2 p π Z 17 √ S = 2πx 1 + (2x)2dx = udu 1 4 5 π h 3 i17 π √ √ = u 2 = (17 17 − 5 5) 6 5 6 where u = 1 + 4x2

Example 7.4.11 Find the exact surface area of the region obtained by rotating x = p2y − y2, 0 ≤ y ≤ 1 about the y−axis.

Solution. We have

v 2 Z 1 u ! p 2u 1 − y S = 2π 2y − y t1 + p dy 0 2y − y2 s Z 1 2 p 2 1 − 2y + y = 2π 2y − y 1 + 2 dy 0 2y − y Z 1 r p 2 1 = 2π 2y − y 2 dy 0 2y − y Z 1 = 2πdy = 2π 0