FURTHER APPLICATIONS OF INTEGRATION

1. 8 FURTHER APPLICATIONS OF INTEGRATION

2. FURTHER APPLICATIONS OF INTEGRATION 8.2Area of a Surface of Revolution • In this section, we will learn about: • The area of a surface curved out by a revolving arc.

3. SURFACE OF REVOLUTION • A surface of revolution is formed when a curve is rotated about a line. • Such a surface is the lateral boundary of a solid of revolution of the type discussed in Sections 6.2 and 6.3

4. AREA OF A SURFACE OF REVOLUTION • We want to define the area of a surface of revolution in such a way that it corresponds to our intuition. • If the surface area is A, we can imagine that painting the surface would require the same amount of paint as does a flat region with area A .

5. AREA OF A SURFACE OF REVOLUTION • Let’s start with some simple surfaces.

6. CIRCULAR CYLINDERS • The lateral surface area of a circular cylinder with radius r and height h is taken to be: A = 2πrh • We can imagine cutting the cylinder and unrolling it to obtain a rectangle with dimensions of 2πrh and h.

7. CIRCULAR CONES • We can take a circular cone with base radius r and slant height l, cut it along the dashed line as shown, and flatten it to form a sector of a circle with radius and central angle θ= 2πr/l.

8. CIRCULAR CONES • We know that, in general, the area of a sector of a circle with radius l and angle θ is ½l2 θ.

9. CIRCULAR CONES • So, the area is: • Thus, we define the lateral surface area of a cone to be A =πrl.

10. AREA OF A SURFACE OF REVOLUTION • What about more complicated surfaces of revolution?

11. AREA OF A SURFACE OF REVOLUTION • If we follow the strategy we used with arc length, we can approximate the original curve by a polygon. • When this is rotated about an axis, it creates a simpler surface whose surface area approximates the actual surface area. • By taking a limit, we can determine the exact surface area.

12. BANDS • Then, the approximating surface consists of a number of bands—each formed by rotating a line segment about an axis.

13. BANDS • To find the surface area, each of these bands can be considered a portion of a circular cone.

14. BANDS Equation 1 • The area of the band (or frustum of a cone) with slant height land upper and lower radii r1 and r2 is found by subtracting the areas of two cones:

15. BANDS • From similar triangles, we have: • This gives:

16. BANDS Formula 2 • Putting this in Equation 1, we get • or • where r =½(r1 + r2) is the average radius of the band.

17. AREA OF A SURFACE OF REVOLUTION • Now, we apply this formula to our strategy.

18. SURFACE AREA • Consider the surface shown here. • It is obtained by rotating the curve y =f(x), a ≤ x ≤ b, about the x-axis, where f is positive and has a continuous derivative.

19. SURFACE AREA • To define its surface area, we divide the interval [a, b] into n subintervals with endpoints x0, x1, . . . , xn and equal width Δx, as we did in determining arc length.

20. SURFACE AREA • If yi= f(xi), then the point Pi(xi, yi) lies on the curve. • The part of the surface between xi–1 and xi is approximated by taking the line segment Pi–1 Piand rotating it about the x-axis.

21. SURFACE AREA • The result is a band with slant height l = | Pi–1Pi | andaverage radius r = ½(yi–1 + yi). • So, by Formula 2, its surface area is:

22. SURFACE AREA • As in the proof of Theorem 2 in Section 8.1, we have • where xi* is some number in [xi–1, xi].

23. SURFACE AREA • When Δxis small, we have yi = f(xi) ≈ f(xi*) and yi–1 = f(xi–1) ≈ f(xi*), since f is continuous. • Therefore,

24. SURFACE AREA Formula 3 • Thus, an approximation to what we think of as the area of the complete surface of revolution is:

25. SURFACE AREA • The approximation appears to become better as n → ∞.

26. SURFACE AREA • Then, recognizing Formula 3 as a Riemann sum for the function we have:

27. SURFACE AREA—DEFINITION Formula 4 • Thus, in the case where f is positive and has a continuous derivative, we define the surface areaof the surface obtained by rotating the curve y = f(x), a ≤x≤b, about the x-axis as:

28. SURFACE AREA Formula 5 • With the Leibniz notation for derivatives, this formula becomes:

29. SURFACE AREA Formula 6 • If the curve is described as x = g(y), c ≤y ≤d, then the formula for surface area becomes:

30. SURFACE AREA Formula 7 • Then, both Formulas 5 and 6 can be summarized symbolically—using the notation for arc length given in Section 8.1—as:

31. SURFACE AREA Formula 8 • For rotation about the y-axis, the formula becomes: • Here, as before, we can use either • or

32. SURFACE AREA—FORMULAS • You can remember these formulas in the following ways.

33. SURFACE AREA—FORMULAS • Think of 2πy as the circumference of a circle traced out by the point (x, y) on the curve as it is rotated about the x-axis.

34. SURFACE AREA—FORMULAS • Think of 2πx s the circumference of a circle traced out by the point (x, y) on the curve as it is rotated about the y-axis.

35. SURFACE AREA Example 1 • The curve , –1 ≤ x≤ 1, is an arc of the circle x2 + y2 = 4 . • Find the area of the surface obtained by rotating this arc about the x-axis. • The surface is a portion of a sphere of radius 2.

36. SURFACE AREA Example 1 • We have:

37. SURFACE AREA Example 1 • So, by Formula 5, the surface area is:

38. SURFACE AREA Example 2 • The arc of the parabola y = x2 from (1, 1) to (2, 4) is rotated about the y-axis. • Find the area of the resulting surface.

39. SURFACE AREA E. g. 2—Solution 1 • Using y = x2 and dy/dx = 2x, from Formula 8,we have:

40. SURFACE AREA E. g. 2—Solution 1 • Substituting u = 1 + 4x2, we have du = 8x dx. • Remembering to change the limits of integration, we have:

41. SURFACE AREA E. g. 2—Solution 2 • Using x = and dx/dy = , • we have the following solution.

42. SURFACE AREA E. g. 2—Solution 2

43. SURFACE AREA Example 3 • Find the area of the surface generated by rotating the curve y = ex, 0 ≤x≤ 1, about the x-axis.

44. SURFACE AREA Example 3 • Using Formula 5 with y = exand dy/dx = ex,we have:

45. SURFACE AREA Example 3

46. SURFACE AREA Example 3 • Since tanα = e , we have: • sec2α = 1 + tanα = 1 + e2 • Thus,