AREAS OF PLANE CURVES - Integration as a Process of Summation - The Calculus Primer

The Calculus Primer (2011)

Part XV. Integration as a Process of Summation

Chapter 57. AREAS OF PLANE CURVES

15—5. Curves in Rectangular Coordinates. We have already learned how an area may be determined by using a definite integral. We shall illustrate the use of the summation idea of integration to find areas, first of curves in rectangular coordinates, and then of curves in parametric equation form and in polar coordinate form.

As we have seen, the areas (I) and (II) below are given by the corresponding definite integrals, where the cross-hatched rectangular strip may be regarded as the type of similar strips which are summed up; the area of the strip shown being equal to y·dx, or x·dy, respectively.

EXAMPLE 1. Find the area between the curve x2 = y3, the line y = 4, and the Y-axis.

Solution.

EXAMPLE 2. Find the area included between the curves x2 = 4y and x2 = 5 − y.

Solution. The required area ABCOA = area DECBAD − area DOECOAD. Solving the given equations simultaneously, the points of

intersection A and C are given by (2,1) and (−2,1). The area OBCE under the curve ABC is equal to:

Area OEC under curve AOC equals:

Hence, area OCBO equals image

and area ABCOA equals image

An alternative analysis would give:

EXERCISE 15—1

1. Find the area bounded by the curve x½ + y½ = a½ and the coordinate axes.

2. Find the area bounded by the curve y = log x, the line x = 5, and the X-axis. Hint: Remember that log 1 = 0; why does the lower limit of integration equal 1?

3. Find the area under one arch of the sine curve y = sin x, i.e., from x = 0 to x = π. Find also the area under this curve from x = π to x = 2π, What is the meaning of the minus sign? What is the total area under the curve from x = 0 to x = 2π?

4. Find the area bounded by y = x3, the X-axis, and the lines x = a and x = b.

5. Find the area between the curve y = x3 — ax2 and the X-axis.

6. Find the area between the parabola y2 = ax and the circle y2 = 2ax — x2.

15—6. Area under a Curve Given by Parametric Equations. If the equations of a curve are given in parametric form, say x — f(t) and y = ø(t), it can be proved that, since dx = f′ (t) dt, the area under the curve from x = a to x = b equals

where t = t1 when x = a and t = t2 when x = b.

EXAMPLE 1. Find the area of the circle

Solution. We find the area of the first quadrant from x = 0 to x = r. Thus, when x = 0, cos t = 0, and t1 = image; when x = r, cos t = 1, and t2 = 0. Here ø(t) = r sin t; f(t) = r cos t, and f′ (t) = −r sin t. Then, from equation [1]:

Therefore the area of entire circle image

EXAMPLE 2. Find the area under the curve

from x = 0 to x = 8.

Solution. When x = 0, t1 = 0; when x = 8, t2 = 2. Here

ø(t) = t2 + 4,

and f(t) = 4$. From equation [1]:

Alternative method: By eliminating the parameter t, the equation of the curve is found to be image; hence, area equals

15—7. Area under a Curve with Equation in Polar Coordinates. Consider the curve whose equation is ρ = f(θ), where we wish to find the area bounded by the curve and any two radius vectors OR and OS corresponding to θ= α and θ = β. Let the radius vectors be ρ1, ρ2, … , and let the respective angles of the sectors be Δθ1, Δθ2, ... .

We shall now apply the Fundamental Theorem. With pole 0 as center, and with ρ0, ρ1, … ρn–1 as radii, draw circular arcs; the sum of these sectors is an approximation to the desired area. The greater the value of n, or the smaller the value of Δθ, the closer the approximation; as n → ∞, or Δθ → 0, the limit of this sum will be the area sought. Now, the area of a circular sector is equal to image radius × arc; hence the area of the first sector image the area of the second sector image etc. Hence, the entire area OR1S is given by

Applying the Fundamental Theorem:

From (1), we see that the area under a polar curve equals

where ρ = f(θ), and the value of ρ is expressed in terms of θ as given by the equation ρ = f(θ).

EXAMPLE 1. Find the area of the circle ρ = a cos θ.

Solution. When the angle θ varies from θ = 0 to θ = π/2, the radius vector OP sweeps over the area ABO; from θ = π/2 to θ = π, the radius vector sweeps over the area OCA. Therefore the limits of integration are 0 and π; hence

But image cos2 θ dθ image therefore,

EXAMPLE 2. Find the entire area enclosed by the curve ρ = a sin 2θ.

Solution. Since the curve is symmetrical with respect to both OA and OR, we shall first find the area of one loop. The loop in the first quadrant is formed as the radius vector sweeps out from image hence

area of loop image (sin 2θ)2 dθ.

To find image (sin 2θ)2 dθ, we use formula [89], page 377, making the substitution z = 2θ; thus

Therefore,

EXERCISE 15—2

1. Find the area, from x = 2 to x = 6, under the curve x = 2t, y = 6/t.

2. Find the area under the curve x = t + 4, y = image t2, from x = 0 to x = 6.

3. Find the area enclosed by the ellipse

where the parameter is the eccentric angle θ.

4. Find the area under one arch of the cycloid

where θ is the parameter. Hint: Remember that, by definition of the cycloid, x varies from 0 to 2πa, where a is the radius of the generating circle; hence angle θ varies from 0 to 2π.

5. Find the area of one loop of the four-leaved rose ρ = a cos 2θ.

6. Find the area of the cardioid ρ = a(l — cosθ).

7. Find the area under the curve ρ2 = a2 cos 2θ.

8. Find the area of the three loops of the curve ρ = a sin 3θ.

9. Find the area under the curve ρ = a (cos 2θ + sin 2θ).

10. Find the area swept over as the radius vector of the spiral of Archimedes, ρ = aθ, makes one revolution from θ = 0 to θ = 2π.