Area surfaces of revolution

AREA OF SURFACES OF REVOLUTION

When you jump rope, the rope sweeps out a surface in the space around you similar to what is called a surface of revolution. The surface surrounds a volume of revolution, and many applications require that we know the area of the surface rather than the volume it encloses. In this section we define areas of surfaces of revolution.

DEFINING SURFACE AREA

If you revolve a region in the plane that is bounded by the graph of a function over an interval, it sweeps out a solid of revolution, as we saw earlier in the chapter. However, if you revolve only the bounding curve itself, it does not sweep out any interior volume but rather a surface that surrounds the solid and forms part of its boundary. Just as we were interested in defining and finding the length of a curve in the last section, we are now interested in defining and finding the area of a surface generated by revolving a curve about an axis.

Before considering general curves, we begin by rotating horizontal and slanted line segments about the x‒axis. If we rotate the horizontal line segment AB having length about the x‒axis, we generate a cylinder with surface area 2πу∆x. This area is the same as that of a rectangle with side lengths ∆x and 2πу. The length 2πy is the circumference of the circle of radius y generated by rotating the point (x, y) on the line AB about the x‒axis.

(a) A cylindrical surface generated by rotating the horizontal line segment AB of length ∆x about the x axis has area 2πу∆x.

(b) The cut and rolled out cylindrical surface as a rectangle.

Suppose the line segment AB has length L and is slanted rather than horizontal. Now when AB is rotated about the x‒axis, it generates a frustum of a cone. From classical geometry, the surface area of this frustum is 2πу*L, where y* = (y₁+y₂)/2 the average height of the slanted segment AB above the x‒axis. This surface area is the same as that of a rectangle with side lengths L and 2πу*.

Let's build on these geometric principles to define the area of a surface swept out by revolving more general curves about the x‒axis. Suppose we want to find the area of the surface swept out by revolving the graph of a nonnegative continuous function y = f(x), a≤x≤ b about the x‒axis. We partition the closed interval [a, b] in the usual way and use the points in the partition to subdivide the graph into short arcs. Figure shows a typical arc PQ and the band it sweeps out as part of the graph of f.

(a) The frustum of a cone generated by rotating the slanted line segment AB of length L about x‒area has area 2лу*L. (b) The area of the rectangle for y* = [ y₁+y₂ ] / 2, the average height of AB above the x‒axis.

The surface generated by revolving the graph of a nonnegative function y = f(x), a≤ x ≤ b about the axis. The surface is a union of bands like the one swept out by the arc PQ.

As the arc PQ revolves about the x‒axis, the line segment joining P and Q sweeps out a frustum of a cone of a cone whose axis along the x‒axis. The surface area of the frustum approximates the surface area of the band swept the are PQ. The surface area of the frustum of the cone shown in below Figure is 2πу*L. where y* is the average height of the line segment joining P and Q, and L is its length (just as before).

The line segment joining P and Q sweeps out a frustum of a cone

Since f ≥ 0 from below Figure we see that the average height of line segment is y*= [f(x_k‒1)+f(x_k) ] / 2, and the slant length is L = √[(∆x_k)² + (∆у_k)²].

Dimensions associated with the arc and line segment PQ

Therefore,

Frustum surface area = 2π . [f(x_k‒1) + f(x_k)]/2 . √[(∆x_k)² + (∆y_k)²]

= π(f(x_k‒1) + f(x_k)) . √[(∆x_k)² + (∆y_k)²]

 The area of the original surface, being the sum of the areas of the bands swept out by arcs like are PQ, is approximated by the frustum area sum

We expect the approximation to improve as the partition of [a, b] becomes finer. Moreover, if the function is differentiable, then by the Mean Value Theorem, there is a point (C_k, f(c_k)) on the curve between P and Q where the tangent is parallel to the segment PQ.

If f is smooth, the Mean Value Theorem guarantees the existence of a point c_k where the tangent is parallel to segment PQ.

At this point,

 f '(c_k) = ∆y_k / ∆x_k

  ∆y_k = f '(c_k) ∆x_k

With this substitution for ∆y_k the sums in Equation (1) take the for

These sums are not the Riemann sums of any function because the points x_k‒1,x_k, and c_k are not the same. However, it can be proved that as the norm of the partition [a, b] goes to zero, the sums in Equation (2) converge to the integral

_a∫^b 2π f(x)√[1 + (f '(x))²] dx

We therefore define this integral to be the area of the surface swept out by the graph of f from a to b.

Definition:

If the function f(x) ≥ 0 is continuously differentiable on [a, b] the area of the surface revolving the graph of y = f(x) about the x‒axis is

Parametric Form:

Suppose that the curve is given in the parametric form x = f(t), y = ϕ(t)

Polar Form:

Suppose that the curve is given in the polar form r = f(θ)

Example 122. Find the area of the surface generated by revolving the curve y = 2√x, 1 ≤ x ≤ 2, about the x‒axis.

Example 123. The curve y = √[4‒x²],‒1 ≤ x ≤ 1, is an arc of the circle x² + y² = 4. Find the area of the surface obtained by rotating this arc about x‒ axis. (The surface is a portion of sphere of radius 2.)

Solution:

Given y = √[4 − x²],

S=8π

Example 124. Find the area of the surface that is generating by revolving the portion of the curve y = x³ between x = 0 and x = 1 about the x‒axis.

Solution:

Given y = x³

dy/dx = 3x²

The surface area is given by

Surface Area for Revolution About the y‒Axis

If x = g(y) ≥ 0 is continuously differentiable on [c, d], the area of the surface generated by revolving the graph of x = g(y) about the y‒axis is

Example 125. Find the area of the surface that is generating by revolving the portion of the curve y = x² between x = 1 and x = 4 about the y‒axis.

Solution:

Given y = x²

Since we are revolving portion of the parabola about y‒axis, we find x in terms of y.

 x = √y

 dx/dy = 1 / 2√y

The surface area is given by

S = 30.83

Example 126. The line segment x=1‒y, 0 ≤ y ≤ 1 is revolved about the y‒axis to generate the cone. Find its lateral surface area (which excludes the base area).

Solution:

Revolving line segment AB about the y axis generates a cone whose lateral surface area we can now calculate in two different ways.

First way:

Here we have a calculation we can check with a formula from geometry:

Lateral surface area = [base circumference/2] × slant height = π√2.

 x = 1−y, dx/dy = ‒1.

The surface area is given by

S = π√2

Resolving the curve about initial line when the curve is given in

Suppose that the curve is given in the polar form r = f(θ)

Example 127. Find the surface of the solid formed by revolving the cardioid r = a (1+ cosθ) about the initial line.

Solution:

The cardioid is symmetrical about the initial line and for its upper half, θ varies from 0 to π.

 S = 32πα² / 5

Resolving a curve about y‒axis, when it is given in polar form

Suppose that the curve is given in the parametric form x = f(t), y = ϕ(t)

Example 128. Find the surface area of the solid generated by the revolution of the asteroid x = a cos³t, y = a sin³t about the y‒axis.

Solution:

The asteroid is symmetrical about the x‒axis, and for its portion in the first quadrant t varies from 0 to π /2.

x = a cos³t,

y = a sin³t

 S = 12πα² / 5

EXERCISE

16. Find the surface of the reel formed by the revolution round the tangent at the vertex of an arch of the cycloid

 x = a(θ + sinθ), y = a(1 ‒ cosθ)

17. Find the surface area of the solid generated by the revolving the cycloid  x = a(t‒sint), y = a(1 ‒ cost) about the base.

18. Find the exact area of the surface obtained the curve about the x-axis

(1) y = sinπx, 0 ≤ x ≤1

(2) y = √[1+e^x], 0 ≤x≤1

(3) y = √[1+ 4x], 1≤x≤5

(4) y=x³, 0≤x≤2

(5) 9x y²+18, 2≤x≤6

19. The given curve is rotated about the y‒axis. Find the area of the resulting surface.

(1) y = ³√x, 1≤y≤2

(2) x = √[a²‒y²], 0≤y≤a/2

(3) y = √[1‒x²], 0≤x≤1