Area enclosed by plane curves

 

AREA ENCLOSED BY PLANE CURVES

 

Area in Cartesian System:

Let R be the plane region, the area which is required. Let us divided the area into a large number of elemental areas like PQRS (shaded) by drawing lines parallel to the y‒axis at intervals of ∆x and line parallel to x axis at the intervals of ∆y above figure

Area of the elemental rectangle PQRS = ∆x. ∆y.

Required area A of the region R is the sum of elemental area like PQRS.

viz., A = lim_{∆x→0, ∆y→0} (ΣΣ ∆x ∆y)

Area = ∫∫_R dxdy

The area A enclosed by the curves y = f₁(x) and y = f₂(x) and the ordinates x = a and x = b is given by

Similarly, the area enclosed by x = ϕ₁(y) and x = ϕ₂(y) and y = c, y= d is given by

 

Remark 1.

If the region is similar to the region given in Figure 5.3a then the horizontal strip is not possible so we take the vertical strip PQ as in Figure 5.3b In this case

 

Remark 2.

If the region is similar to the region given in Figure 5.3c then the vertical strip is not possible so we take the horizontal strip PQ as in Figure 5.3d In this case

 

Example 42. Find the area of the circle of radius a.

Solution:

Let us assume a circle with center (0,0) and radius a

Equation of the circle is x² + y² = a²

To plot the region:

x² + y² = a² is a circle with centre at the orgin and radius 'a'

To evaluate the area:

Area of the circle = 4 x Area of region in the first quadrant

= 4 ∫∫_R dydx

Since the order of integration is dydx, draw a vertical strip PQ

At P, y = 0;

At 0, x = 0;

At Q, y = √[a² ‒ x²]

At A, x = a

 

Example 43. Find the area of the ellipse x²/a² + y²/b² = 1.

Solution:

To plot the region:

 x²/a² + y²/b² = 1 is an ellipse with vertex at origin.

To evaluate the area:

Area of the ellipse = 4 × Area of region in the first quadrant

  = 4 ∫∫_R dydx

Since the order of integration is dydx, draw a vertical strip PQ

At P, y = 0;

At Q, y = b/a √[a² ‒ x²]

At O, x = 0;

At A, x = a

Area = πab sq.units.

 

Example 44. Find the smaller of the areas bounded by the ellipse 4x² + 9y² = 36 and the straight line 2x + 3y = 6.

Solution:

To plot the region:

The equation of the ellipse is x²/9 + y²/4 = 1      ........ (1)

and the line is x/3 + y/2= 1       ..:(2)

Both meet x‒axis at A(3,0) and y‒axis at B(0,2)

The smaller of the area is shown in the figure.

To evaluate the area:

Area = ∫∫ dx dy

Since the order of integration is dx dy, draw horizontal strip PQO

Using horizontal strips, the required area lies between

At P, x= 3/2(2− y),

At Q, x = 3/2√[4‒ y²],

At A, y = 0,

At B, y = 2

Required area is

Area = 3/2 (π‒2) sq. units

 

Example 45. Find the area of the region bounded by the parabolas y² = 4ax and x² = 4ay.

Solution:

To plot the region:

y² = 4ax is a parabola symmetrical about x‒axis and passing through the origin

 x² = 4ay is a parabola symmetrical about y‒axis and passing through the origin

The region of integration is OAO (shaded region)

To find the points of intersection:

 y² = 4ax  

 x = y² / 4a   ..... (1)

 x² = 4ay   ……(2)

(2) (y² / 4a)² = 4ay Using (1)

y⁴ / 16a² = 4ay

⇒ y² = 64a³y.

y⁴‒64a³y = 0

y(y³ ‒ 64a³)=0

y= 0, y³ = 64a³

y = 4a

The points of intersection are (0,0), (4a, 4a)

To evaluate the area:

Area = ∫∫_R dydx

Since the order of integration is dydx,

draw a vertical strip PQ

At P, y = x² / 4a

At Q, y = 2√α√x

At 0, x = 0;

At A, x = 4a

Area = 16a² / 3 sq. units.

 

Example 46. Find the area between the curves y² = 4x and x² = 4y.

Solution:

To plot the region:

 y² = 4x is a parabola symmetrical about x‒axis and passing through the origin

 x² = 4y is a parabola symmetrical about y‒axis and passing through the origin

The region of integration is OAO (shaded region)

To find the points of intersection:

y²=4x

x = y² / 4 ………(1)

x² = 4y   …….. (2)

(2) => (y²/4)² = 4y      Using (1)

 y⁴/16 = 4y

y⁴ = 64y

y⁴ – 64y =0

y(y³ – 64) = 0

y = 0, y³ = 64 ⇒ y = 4

The points of intersection are (0,0), (4, 4)

To evaluate the area:

Area = ∫∫_R dydx

Since the order of integration is dydx,

draw a vertical strip PQ

At P, y = x²/4

At Q, y = 2√x

At 0, x=0;

At A, x=4

Area = 16/3 sq. units.

 

Example 47. Find the area bounded by y² = 4 ‒ x and y² = x.

Solution:

To plot the region:

y² = 4‒x = ‒(x‒4) is a parabola symmetrical about x‒axis (left side) with vertex at (4, 0) and passing through the origin

y² = x is a parabola symmetrical about x‒axis and passing through the origin

The region of integration is OBACO (shaded region)

To find the points of intersection

y² = 4‒x  ………(1)

y² = x  ………(2)

From (1) and (2),

x = 4‒x ⇒ 2x= 4 ⇒ x=2

Put x = 2 in (2) ⇒ y²=2 ⇒ y = ±√2

 ∴ The points of intersection are (2, √2), (2, −√2)

To evaluate the integral:

Area = 2 × Area of OABO

= 2 ∫∫ dxdy

Since the order of integration is dxdy, draw a horizontal strip PQ

At P, x = y²;

At Q, x = 4‒y²

At O, y = 0;

At B, y = √2

 

Example 48. Find the area enclosed by the parabola y² = 4ax, the x‒axis and the latus rectum of the parabola.

Solution:

To plot the region:

 y² = 4ax is a parabola symmetrical about x‒axis with vertex at origin.

The equation of latus rectum is x = a, a line perpendicular to x‒axis.

To find the points of intersection

y² = 4ax        ……..(1)

Equation of latus rectum is x = a

Put x = a in (1) ⇒ y² = 4a²

 ⇒ y = 2a

[Take only positive value since the region is first quadrant.]

The point of intersection is (a, 2a)

The region of integration is OABO (shaded region)

To evaluate the area:

Area = ∫∫_R dydx

Since the order of integration is dydx, draw a vertical strip PQ

At P, y = 0;

At Q, y = 2√a√x

At O, x = 0;

At A, x = a

 

Example 49. Find the area of the region R enclosed by the parabola y = x² and the line y=x+2.

Solution: We need to describe the region between the graph of the parabola y = x² and the straight line y = x + 2

To find the points of intersection

y = x²              ... (1)

y = x + 2           ... (2)

From (1) and (2),

x²=x+2 ⇒ x²‒x‒2=0

(x‒2)(x+1)=0 ⇒ x=2,‒1

Put x = 2 in (2) ⇒ y = 4

Put x = ‒1 in (2) ⇒ y = 1

∴ The points of intersection are (2,4), (−1,1)

The area is given by

A = ∫∫_R dy dx

Since the order of integration is dydx, draw vertical strip PQ

At P, y = x²;

At Q, y=x+2

At A, x= ‒1;

At B, x = 2

 

Example 51. Find by double integration, the area lying between the parabola y = 4x‒x² and the line y = x.

Solution:

To plot the region:

y = x       …….(1) a straight line intersect the XOY plane

y = 4x‒x²            …………(2)

= ‒( x²‒4x)

y = ‒[ (x‒2)²‒4 ]

 (x‒4) = (x‒2)² is a parabola with vertex (2, 4) which is symmetrical about y‒axis (downwards) and passing through the origin

The region of integration is the shaded region OAO. To find the point of intersection from (1) and (2)

x = 4x − x²

x² ‒ 3x = 0 ⇒ x(x − 3) = 0 ⇒ x = 0 or x = 3

The points of intersections are (0, 0) and (3, 3)

To evaluate the area:

Area = ∫∫_R dydx

Since the order of integration is dydx, draw a vertical strip PQ.

At P, y = x;

At Q, y = 4x‒x²

At O, x = 0;

At A, x = 3

 

Example 52. Evaluate ∫∫_R dx dy where R is the shaded region in the figure.

Solution:

Given region is a semi‒circular part of the circle x² + y² = 4.

∫∫_R dx dy = Area of the shaded region (Semi – circle)

 = ½ Area of the circle (x² + y² = 4) = ½ π (4)

Area = 2π sq. units

 

AREA IN POLAR SYSTEM

 

We divided the area A of the given region R into a large number of elemental curvilinear rectangular areas like PQRS (shaded) by drawing radial lines and concentric circular arcs, where P and R have a polar coordinates (r, θ) and (r+ Δr, θ + Δθ) above figure

Area of the element PQRS = r Δr Δθ

(PS = r Δθ and PQ = Δr)

Required area A = lim_{∆r→0, ∆θ→0} ( ΣΣ r Δr Δθ)

Area = ∫∫_R r dr dθ

 

Example 53. Find the area of a circle x²+ y² = a² using polar coordinates in double integrals.

Solution:

 

Example 54. Find the area between the two circles x² + y² = a² and x² + y² = b² where a > b.

Solution:

To plot the region:

The region of integration is the annular region between x² + y² = a² and x² + y² = b²

Since the region is circular,

we use polar co‒ordinates.

x = r cosθ, y = r sinθ

x² + y² = r²

dx dy = r dr dθ

To evaluate the area:

 x² + y² = a² ⇒ r² = a²

⇒ r = a

x² + y² = b² ⇒ r² = b²

⇒ r = b

 r varies from r = a to r = b

Since the region lies in all the four quadrants,

 θ varies from θ = 0 to θ = 2π

Area = π(a² ‒ b²) sq. units.

 

Example 55. Find the area between the circles r = 2 sinθ and r = 4 sinθ.

Solution:

To plot the region

 r = 2 sinθ is a circle with diameter 2 passing through the origin and symmetrical about θ = π/2

 r = 4 sinθ is a circle with diameter 4 passing through the origin and symmetrical about θ = π/2

∴ Total area = 2 (Area of region in the first quadrant)

To evaluate the area:

Limits of r: At P, r = 2 sinθ; At Q, r = 4 sinθ

Limits of θ: θ = 0 to θ = π/2

 

Example 56. Find the area of the region outside the inner circle r = 2 cos θ and inside the outer circle r = 4 cos θ by double integration.

Solution:

 r = 2 cosθ is the circle with diameter 2 passing through the origin and symmetrical about the initial line θ = 0.

r = 4 cosθ is the circle with diameter 4 passing through the origin and symmetrical about the initial line θ=0.

To evaluate the area:

Area = 2 × Area of the region above the initial line.

= 2 ∫∫_OABO r dr dθ

Limits of r: At P, r = 2 cos θ,

Limits of θ: θ = 0,

Area = 3π sq units.

 

Example 57. Find the area of the cardioid r = a(1 + cosθ).

Solution:

To plot the region:

 r = a(1 + cosθ) is a cardioid passes through the origin and symmetrical about the initial line.

To evaluate the area:

 ∴ Area = 2 × Area of the region above the initial line.

= 2 ∫∫_OABO r dr dθ

Limits of r: At 0, r = 0 to r = a(1 + cos θ)

Since the region lies in the first and second quadrant, θ varies from θ = 0 to θ = π

 I = 3/2 πa² sq. units.

 

Example 58. Find the area outside to the circle r = 1 and inside to the cardioid r = 1+ cosθ

Solution:

To plot the region:

r = 1, a circle with centre (0,0) and radius 1

 r = 1+ cosθ a cardioid, symmetrical about the initial line θ = 0 and passes through the origin (0,0)

The region of integration is ABCDA. (shaded region).

The region is symmetric about the line θ = 0.

To evaluate the area:

Area = 2 × Area of the region above the initial line

= 2 ∫∫_ABDA r dr dθ

The point of intersection of the circle r = 1 and r = 1 + cosθ are obtained as

 1 = 1 + cosθ

 cos θ = 0

 θ = ± π/2

Since the region ABDA lies in the first quadrant and touches the x‒axis & y‒axis,

 θ varies from θ = 0 to θ = π/2

 

 

EXERCISE

 

26. Find the area enclosed by y = x² and x = y². Ans: 16/3

 

27. Find the area of the circle x² + y² = 16. Ans: 16π

 

28. Find the area of the ellipse x²/16 + y²/9 = 1. Ans: 12π

 

29. Find the area of the region bounded by x = 0, y = 0 and x + y = a. Ans: 1/2 a²

 

30. Find the area of the region enclosed by the curves y = x and y = x². Ans: 1/6 sq.units.

 

31. Find by double integration, the area between the parabola y² = 4ax and the line y = x. Ans: 8a² / 3 sq.units.

 

32. Find the area bounded by the parabolas y = x² and the straight line 2x − y + 3=0. Ans: 32/3 sq.units.

 

33. Find the area of the region bounded by y = x ‒ 1 and y² = 2x + 6. Ans: 18 sq.units

 

34. Find the area bounded by x² + y² = 4, x + y = 2 lying in the first quadrant. Ans: π‒2

 

35. Find the area of the region bounded by x = 0, y = 0 and x + y = 1. Ans: 1/2.

 

36. Find the area enclosed by the parabola x² = 4ay and the latus rectum of the parabola.

Ans: 4a² / 3

 

37. Find the area enclosed by the curve y² = 4ax and the lines x + y = 3a, y = 0. Ans: 10a² / 3 sq.units

 

38. Find the area between the circle x² + y² = a² and the line x + y = a lying in the first quadrant. Ans: a²/4 (π‒2) sq. units

 

39. Find the area of the region R enclosed by the parabola y = x² and the line y = x + 2. Ans: 9/2

 

40. Find the area between the two parabolas 3y² = 25x and 5x² = 9y. Ans: 5 sq. units

 

41. Find the area bounded by the hypocycloid (x/a)^2/3 +(y/b)^2/3 = 1. Ans: 3/8 πab sq. units

 

42. Find the area outside to the circle r = a and inside to the cardioid r = a( 1+cos θ ). Ans: a²(π+8) sq. units.