## 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 *x*^{2} = *y*^{3}*,* the line *y* = 4, and the *Y*-axis.

*Solution.*

EXAMPLE 2. Find the area included between the curves *x*^{2} = 4*y* and *x*^{2} = 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

and area *ABCOA* equals

An alternative analysis would give:

**EXERCISE 15—1**

**1.** Find the area bounded by the curve *x*^{½} *+ y*^{½} = *a*^{½} 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* = 2*π,* 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* = *x*^{3}, the *X*-axis, and the lines *x* = *a* and *x* = *b*.

**5.** Find the area between the curve *y* = *x*^{3} — *ax*^{2} and the *X*-axis.

**6.** Find the area between the parabola *y*^{2} = *ax* and the circle *y*^{2} = 2*ax* — *x*^{2}.

**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* = *t*_{1} when *x* = *a* and *t* = *t*_{2} 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 *t*_{1} = ; when *x* = *r*, cos *t* = 1, and *t*_{2} = 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

EXAMPLE 2. Find the area under the curve

from *x* = 0 to *x* = 8.

*Solution.* When *x* = 0, t_{1} = 0; when *x* = 8, *t*_{2} = 2. Here

*ø*(*t*) = *t*^{2} + 4,

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

*Alternative method:* By eliminating the parameter *t,* the equation of the curve is found to be ; 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 *ρ*_{1}, *ρ*_{2}, … , and let the respective angles of the sectors be Δ*θ*_{1}, Δ*θ*_{2}, ... .

We shall now apply the Fundamental Theorem. With pole *0* as center, and with *ρ*_{0}, *ρ*_{1}, … *ρ _{n}*

_{–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 Δ

*θ*, the closer the approximation; as

*n*→ ∞, or Δ

*θ*→ 0, the limit of this sum will be the area sought. Now, the area of a circular sector is equal to radius × arc; hence the area of the first sector the area of the second sector etc. Hence, the entire area

*OR*

_{1}

*S*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 cos^{2} *θ* *dθ* therefore,

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

*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 hence

area of loop (sin *2θ*)^{2} *dθ.*

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

Therefore,

**EXERCISE 15—2**

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

**2.** Find the area under the curve *x* = *t +* 4, *y* = *t*^{2}, 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 2*θ*.

**6.** Find the area of the cardioid *ρ* = *a*(l — cos*θ*).

**7.** Find the area under the curve *ρ*^{2} = *a*^{2} cos *2θ.*

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

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

**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.