Arc Length. Surface Area

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 + dy2 ⇒ ds = dx2 + dy2 and so s s Z b Z b q Z b dy2 Z b dy2 L = ds = dx2 + dy2 = (1 + )dx2 = (1 + )dx. a a a dx2 a dx2 2 dy2 dy 0 2 Note that dx2 = dx = (y ) . So the formula for the arc length becomes Z b q L = 1 + (y0)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 of the circle with radius y and the height that is given by q the arc length element ds. Since ds is 1 + (y0)2dx, the formula that computes the surface area is Z b q 0 2 Sx = 2πy 1 + (y ) 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 Z b q 0 2 Sy = 2πx 1 + (y ) 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 4r2π 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 approximates the surface area of the surface obtained by rotating y = ex2+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 rev- 1 olution obtained by revolving y = x for 1 ≤ x ≤ ∞ is known as the Gabriel’s Horn or 1 Torricelli’s trumpet. Using the inequality 1 + x4 > 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 0 3 1/2 0 2 9 R 4 q 9 1. y = 2 x so (y ) = 4 x L = 1 1 + 4 xdx. Evaluate this integral using the substitution 9 4 2 9 3/2 4 8 3/2 13 3/2 u = 1 + 4 x and obtain 9 3 (1 + 4 x) |1 = 27 (10 − ( 4 ) ) = 7.6337. q 2. The key step in this problem is to simplify the formula 1 + (y0)2. The derivative is y0 = 2 2 2 1 (1 − x2)−1/2(−2x) = √−x so 1 + (y0)2 = 1 + x = 1−x +x = 1 . Thus, the length is 2 1−x2 1−x2 1−x2 1−x2 q L = R 1 1 dx = R 1 √ 1 dx = sin−1 x|1 = sin−1(1) − sin−1(−1) = π + π = π. −1 1−x2 −1 1−x2 −1 2 2 3. Careful: ﬁrst write down the integral that you need to evaluate using the formula for the arc 3 length, then use the calculator. Do not enter x in Y1 because in that case the program would give you the area under the curve, not the length. √ 3 0 2 R 1 4 y = x ⇒ y = 3x . So the integral√ L = 0 1 + 9x dx computes the arc length. To evaluate 4 this integral, enter the function 1 + 9x 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 the arc√ length ﬁrst, before entering any√ function in 0 R π 2 2 the calculator. y = sin x ⇒ y = cos x. L = 0 1 + cos xdx. Enter the function 1 + cos 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. q √ √ x 0 x R 1 x 2 R 1 2x 2x 5. y = e ⇒ y = e .L = 0 1 + (e ) dx = 0 1 + e dx. Enter 1 + e 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 √ 3 0 2 R 2 0 2 R 2 3 4 6. y = x ⇒ y = 3x .Sx = 0 2πy 1 + (y ) dx = 2π 0 x 1 + 9x dx. Evaluate this integral 4 1 2 4 3/2 2 π 3/2 using the substitution u = 1 + 9x . Obtain 2π 36 3 (1 + 9x ) |0 = 27 (145 − 1) = 203.04. √ q √ q 1/2 0 1 −1/2 √1 R 9 0 2 R 9 1 7. y = x = x ⇒ y = 2 x = 2 x .Sx = 4 2πy 1 + (y ) dx = 2π 4 x 1 + 4x dx. Simplify √ q √ √ √ R 9 4x+1 R 9 4√x+1 R 9 the function ﬁrst. Obtain 2π 4 x 4x dx = 2π 4 x 2 x dx = π 4 4x + 1dx. Evaluate 1 2 3/2 9 π 3/2 3/2 this integral using u = 1 + 4x. Obtain π 4 3 (1 + 4x) |4 = 6 (37 − 17 ) = 81.14. q √ 2 0 R 2 0 2 R 2 2 8. y = x ⇒ y = 2x. Sy = 1 2πx 1 + (y ) dx = 2π 1 x 1 + 4x dx. Evaluate this integral using 2 1 2 2 3/2 2 π 3/2 3/2 the substitution u = 1 + 4x . Obtain 2π 8 3 (1 + 4x ) |1 = 6 (17 − 5 ) = 30.85.

2 2 9. You can represent the sphere as√ the surface of revolution of the upper part of the circle x +y = 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 y0 = 1 (r2 − x2)−1/2(−2x) = √ −x . Similarly 2 r2−x2 q to problem 2. in part a), the key step in this problem is to simplify the formula 1 + (y0)2. 2 2 2 2 2 q 1 + (y0)2 = 1 + x = r −x +x = r . Thus, the surface area is S = R r 2πy r2 dx = √ r2−x2 r2−x2 r2−x2 x −r r2−x2 R r 2π r2 − x2 √ r dx = R r 2πrdx = 2πrx|r = 2πr(r + r) = 4r2π. −r r2−x2 −r −r 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 √ √ 0 R π 2 2 y = sin x ⇒ y = cos x. Sx = 0 2π sin x 1 + cos xdx. Enter the function 2π sin x 1 + cos 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 R b 0 2 11. The problems is asking for the surface area Sx = a 2πy 1 + (y ) dx. Find the derivative x2+1 0 x2+1 of the function and plug it in the formula first. y = e ⇒ y = e 2x ⇒ Sx = q q R 1 x2+1 x2+1 2 x2+1 x2+1 2 0 2πe 1 + (2xe ) dx. Then enter 2πe 1 + (2xe ) 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 152.9. q R b 0 2 12. The problems is asking for the surface area Sy = a 2πx 1 + (y ) dx. Find the derivative 3 0 3x2 of the function and plug it in the formula first. y = ln(x + 1) ⇒ y = x3+1 ⇒ Sx = q q R 1 3x2 2 3x2 2 q 9x4 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 13. y = 1 ⇒ y0 = −x−2 = −1 . The surface area is S = R ∞ 2π 1 1 + 1 dx. Using the given x x2 √ x 1 x x4 R ∞ 1 R ∞ 1 ∞ inequality, this integral is larger than 1 2π x 1dx = 2π 1 x dx = 2π ln x|1 = ∞. So, the surface area is larger than the value of this divergent integral. So, Sx is infinite as well. 2 R ∞ 1 R ∞ 1 −1 ∞ Volume, on the other hand, is computed as Vx = 1 π x dx = π 1 x2 dx = π x |1 = −1 π( ∞ − (−1)) = π.