where a_n are all positive.

For verification, Dx(1r+1(g(x))r+1)=1r+1Dx[(g(x))r+1]=1r+1(r+1)(g(x))rg′(x)=(g(x))rg′(x) by the Power Chain Rule.

In this case, we had to insert a factor of 2 in the integrand in order to use Quick Formula I.

Law 9. Substitution Method

where u is replaced by g(x) after the right-hand side is evaluated. The "substitution" is carried out on the left-hand side by letting u = g(x) and du = g′(x) dx. (For justification, see Problem 21.)

Observe that Quick Formula I is just a special case of the Substitution Method, with u = g(x). The advantage of Quick Formula I is that we save the bother of carrying out the substitution.

The known formulas for derivatives of trigonometric and inverse trigonometric functions yield the following formulas for antiderivatives:

Figure 29-1  

Now consider a region like that shown in Fig. 29-2, bounded on the left by the y-axis, on the right by a curve x = g(y), and lying between y = c and y = d. Then by an argument similar to that for the case shown in Fig. 29-1, the area of the region is the definite integral ∫cdg(y) dy.

Figure 29-2  

EXAMPLE 29.1:

Consider the region bounded on the right by the parabola x = 4 − y², on the left by the y-axis, and above and below by y = 2 and y = −1. See Fig. 29-3. Then the area of this region is ∫−12(4−y2) dy. By the Fundamental Theorem of Calculus, this is

(4y−13y3)]2−1=(8−83)−(−4−(−13))=12−93=12−3=9

Figure 29-3  

• 1. Find the area S of the surface of revolution generated by revolving about the x-axis the arc of the parabola y² = 12x from x = 0 to x = 3.

  By implicit differentiation,

  dydx=6y and 1+(dydx)2=y2+36y2

  By (36.1),

  S=2π∫03yy2+36y dx=2π∫0312x+36 dx=2π(8(12x+36)3/2]03=24(22−1)π

• 2. Find the area S of the surface of revolution generated by revolving about the y-axis the arc of x = y³ from y = 0 to y = 1.

  dxdy=3y2 and 1+(dxdy)2=1+9y4. So, by (36.2),S=2π∫01x1+9y4dy=2π∫01y31+9y4dy=π18∫01(1+9y4)12 (36y3) dy=π1823(1+9y4)3/2]01=π27(1010−1)

• 3. Find the area of the surface of revolution generated by revolving about the y-axis the arc of y² + 4x = 2 ln y from y = 1 to y = 3.

  S=2π∫edy1+(dxdy)2dy=2π∫13y1+y22y dy=π∫13(1+y2) dy=323π

• 4. Find the area of the surface of revolution generated by revolving the top half of the loop of the curve 8a²y² = a²x² − x⁴ about the x-axis. (See Fig. 36-1.)

  Figure 36-1  

  

  Here

  dydx=a2x−2x38a2y and 1+  (dydx)2=1+(a2−2x2)28a2(a2−x2)=(3a2−2x2)28a2(a2−x2)

  Hence

  S=2π∫0ay1+(dydx)2dx=2π∫0axa2−x22a23a2−2x22a2a2−x2dx=π4a2∫0a(3a2−2x2)x dx=14πa2

• 5. Find the area of the surface of revolution generated by revolving about the x-axis the bottom half of the ellipse x216+y24=1.

  S=2π∫−44y16y2+x24ydx=π2∫−4464−3x2dx=π23(x3264−3x2+32sin−1(x38))]−44=8π(1+439π)

• 6. Find the area of the surface of revolution generated by revolving about the x-axis the hypocycloid x = a cos³ θ, y = a sin³ θ.

  The required surface is generated by revolving the arc from θ = 0 to θ = π. We have dxdθ=−3acos2θsinθ,  dydθ=3asin2θcosθ, and (dxdθ)2+(dydθ)2=9a2cos2θsin2θ. Then

  S=2(2π)∫0π/2y(dxdθ)2+(dydθ)2dθ=2(2π)∫0π/2(asin3θ)3acos θsinθ dθ=12a2π5  (square units)

• 7. Find the area of the surface of revolution generated by revolving about the x-axis the cardioid x = 2 cos θ − cos 2θ, y = 2 sin θ − sin 2θ.

  The required surface is generated by revolving the arc from θ = 0 to θ = π. (See Fig. 36-2.) We have

  dxdθ=−2sinθ+2sin2θ, dydθ=2cosθ−2cos2θ,
  Figure 36-2  

  

  and

  (dxdθ)2+(dydθ)2=8(1−sinθsin2θ−cosθcos2θ)=8(1−cosθ)

  Then

  S=2π∫0π(2sinθ−sin2θ)(221−cosθ) dθ=82π∫0πsin θ(1−cos θ)32dθ=(1625π(1−cosθ)52))]0π=128π5  (square units)

• 8. Show that the surface area of a cylinder of radius r and height h is 2πrh.

  The surface is generated by revolving about the x-axis the curve y = r from x = 0 to x = h. Since dydx=0, 1+(dydx)2=1. Then by (36.1),

  S=2π∫0hrdx=2π(rx)]0h=2πrh

• 9. Show that the surface area of a sphere of radius r is 4πr².

  The surface area is generated by revolving about the x-axis the semicircle y=r2−x2 from x = − r to x = r. By symmetry, this is double the surface area from x = 0 to x = r. Since y² = r² − x²,

  2ydydx=−2x and, therefore, dydx=−xy and 1+(dydx)2=1+x2y2=x2+y2y2=r2y2

  Hence, by (36.1),

  S=2⋅2π∫0ryr2y2 dx =  4πr∫0r1  dx=4πr​ x]0r=4πr2

• 10. 

  1. Show that the surface area of a cone with base of radius r and with slant height s (see Fig. 36-3) is πrs.

     Figure 36-3  

     

  2. Show that the surface area of a frustum of a cone having bases of radius r₁ and r₂ and slant height u (see Fig. 36-4) is π(r₁ + r₂)u. (Note that the frustum is obtained by revolving the right-hand segment of the slant height around the base of the triangle.)

     Figure 36-4  

     

  1. Cut open the cone along a slant height and open it up as part of a circle of radius s (as shown in Fig. 36-5). Note that the portion of the circumference cut off by this region is 2πr (the circumference of the base of the cone). Now the desired area S is the difference between πs² (the area of the circle in Fig. 36-5) and the area A₁ of the circular sector with central angle θ. This area A₁ is θ2π(πs2)=12θs2. Since the arc cut off by θ is 2πs − 2πr, we get θ=2πs−2πrs. Thus, A₁ = π(s − r)s. Hence, S = πs² − π(s − r)s = πrs square units.

     Figure 36-5  

     

  2. From the similar triangles in Fig. 36-4, we get u1r1=u1+ur2. Then r₂ u₁ = r₁ u_l + r₁u. So, u1=r1ur2−r1. Now by part (a), the surface area of the frustum is πr₂(u₁ + u) − πr_l u_l = π(r₂ − r_l)u_l + πr₂ u = πr₁ u + πr₂ u = π(r₁ + r₂)u square units.

• 11.  Sketch a derivation of formula (36.1).

  Assume that [a, b] is divided into n equal subintervals, [x_k−1, x_k], each of length Δx=b−an. The total surface area S is the sum of the surface areas S_k generated by the arcs between the points (x_k−1,f(x_k−1)) and (x_k, f(x_k)), each of which is approximated by the surface area generated by the line segment between (x_k−1, f(x_k−1)) and (x_k, f(x_k)). The latter is the area of a frustum of a cone. In the notation of Fig. 36-6, this is, by virtue of Problem 10(b):

  π(f(xk−1)+f(xk))(Δx)2+(Δy)2=2π(f(xk−1)+f(xk)2)(Δx)2+(Δy)2
  Figure 36-6  

  

  Now f(xk−1)+f(xk)2, being the average of f(x_k−1) and f(x_k), is between those two values and, by the Intermediate Value Theorem, is equal to f(xk*) for some xk* in (x_k-1, x_k). Also, (Δx)2+(Δy)2=1+(ΔyΔx)2Δx. By the Mean Value Theorem ΔyΔ x=f′(xk#) for some xk# in (x_k−1, x_k). Thus, S is approximated by the sum

  ∑k=1n2πf(xk*)1+(f′(xk#))2Δx

  and it can be shown that this sum can be made arbitrarily close to 2π∫abf(x)1+(f′(x))2dx.^[1] Hence, the latter is equal to S.

Figure 30-2  

See Problem 9 for a sketch of the proof of this formula.

Similarly, when the axis of rotation is the y-axis and the region that is revolved lies between the y-axis and a curve x = g(y) and between y = c and y = d (see Fig. 30-3), then the volume V of the resulting solid of revolution is given by the formula

Figure 30-3  

EXAMPLE 30.1:

Consider the solid of revolution obtained by revolving about the x-axis the region in the first quadrant bounded by the parabola y² = 8x and the line x = 2. (See Fig. 30-4.) By the disk formula, the volume is

V=π∫02y2dx= π∫028x dx=π(4x2)]02=π(16−0)=16π

Figure 30-4  

EXAMPLE 30.2:

Consider the solid of revolution obtained by revolving about the y-axis the region bounded by the parabola y = 4x² and the lines x = 0 and y = 16. (See Fig. 30-5.) To find its volume, we use the version of the disk formula in which we integrate along the y-axis. Thus,

V=π∫016x2dy= π∫016y4dy=π8y2]016=π8(256−0)=32π

Figure 30-5  

In polar coordinates

where θ = α, θ = β, ρ = ρ₁(θ), and ρ = ρ₂(θ) are chosen as boundaries of the region R.

Thus, P lies on the circle if and only if

(4.1)

Equation (4.1) is called the standard equation of the circle with center at (a, b) and radius r.

Figure 4-1  

1. f(x₀) is defined;

2. limx→x0f(x) exists;

3. limx→x0f(x)=f(x0).

For example, f(x) = x² + 1 is continuous at 2, since limx→2f(x)=5=f(2). Condition (i) implies that a function can be continuous only at points of its domain. Thus, f(x)=4−x2 is not continuous at 3 because f(3) is not defined.

Let f be a function that is defined on an interval (a, x₀) to the left of x₀ and/or on an interval (x₀, b) to the right of x₀. We say that f is discontinuous at x₀ if f is not continuous at x₀, that is, if one or more of the conditions (i) to (iii) fails.

The discontinuity at 2 in Example 8.1(b) is said to be removable because if we extended the function f by defining its value at x = 2 to be 4, then the extended function g would be continuous at 2. Note that g(x) = x + 2 for all x. The graphs of f(x)=x2−4x−2 and g(x) = x + 2 are identical except at x = 2, where the former has a "hole." (See Fig. 8-2.) Removing the discontinuity consists simply of filling the "hole."

Figure 8-2  

The discontinuity at 2 in Example 8.1(a) is not removable. Redefining the value of f at 2 cannot change the fact that limx→21x−2 does not exist.

We also call a discontinuity of a function f at x₀ removable when f(x₀) is defined and changing the value of the function at x₀ produces a function that is continuous at x₀.

EXAMPLE 8.3:

Let f be the function such that f(x)=|x|x for all x ≠ 0. The graph of f is shown in Fig. 8-3. f is discontinuous at 0 because f(0) is not defined. Moreover,

limx→0+f(x)=limx→0+xx=1    and    limx→0−f(x)=limx→0−−xx=−1

Thus, limx→0−f​(x)≠limx→0+f​(x). Hence, the discontinuity of f at 0 is not removable.

Figure 8-3  

The kind of discontinuity shown in Example 8.3 is called a jump discontinuity. In general, a function f has a jump discontinuity at x₀ if limx→x0− f(x) and limx→x0+f(x) both exist and limx→x0−f(x)≠limx→x0+f(x). Such a discontinuity is not removable.

We say that a function f is continuous on a set A if f is continuous at every point of A. Moreover, if we just say that f is continuous, we mean that f is continuous at every real number.

The original intuitive idea behind the notion of continuity was that the graph of a continuous function was supposed to be "continuous" in the intuitive sense that one could draw the graph without taking the pencil off the paper. Thus, the graph would not contain any "holes" or "jumps." However, it turns out that our precise definition of continuity goes well beyond that original intuitive notion; there are very complicated continuous functions that could certainly not be drawn on a piece of paper.

This is a consequence of Theorems 8.1(a–e) and 8.2.

This follows from Theorems 8.1(f) and 8.3. As examples, the function H(x)=xx2−1 is continuous at all points except 1 and −1, and the function G(x)=x−7x2+1 is continuous at all points (since x² + 1 is never 0).

We shall use a special notion of continuity with respect to a closed interval [a, b]. First of all, we say that a function f is continuous on the right at a if f(a) is defined and limx→a+f(x) exists, and limx→a+f(x)=f(a). We say that f is continuous on the left at b if f(b) is defined and limx→b−f(x) exists, and limx→b−f(x)=f(b).

Definition: f is continuous on [a, b] if f is continuous at each point on the open interval (a, b), f is continuous on the right at a, and f is continuous on the left at b.

Note that whether f is continuous on [a, b] does not depend on the values of f, if any, outside of [a, b]. Note also that every continuous function (that is, a function continuous at all real numbers) must be continuous on any closed interval. In particular, every polynomial function is continuous on any closed interval.

We want to discuss certain deep properties of continuous functions that we shall use but whose proofs are beyond the scope of this book.

Figure 8-4(a) is an illustration of Theorem 8.5. Fig. 8-5 shows that continuity throughout the interval is essential for the validity of the theorem. The following result is a special case of the Intermediate Value Theorem.

Figure 8-4  

Figure 8-5  

Corollary 8.6: If f is continuous on [a, b] and f(a) and f(b) have opposite signs, then the equation f(x) = 0 has at least one root in the open interval (a, b), and therefore, the graph of f crosses the x-axis at least once between a and b. [See Fig. 8-4(b).]

As an illustration of the Extreme Value Theorem, look at Fig. 8-6(a), where the minimum value m occurs at x = c and the maximum value M occurs at x = d. In this case, both c and d lie inside the interval. On the other hand, in Fig. 8-6(b), the minimum value m occurs at the endpoint x = a and the maximum value M occurs inside the interval. To see that continuity is necessary for the Extreme Value Theorem to be true, consider the function whose graph is indicated in Fig. 8-6(c). There is a discontinuity at c inside the interval; the function has a minimum value at the left endpoint x = a, but the function has no maximum value.

Figure 8-6  

Another useful property of continuous functions is given by the following result.

This theorem is illustrated in Fig. 8-7. For a proof, see Problem 3.

Figure 8-7