Evaluation of a Bounded Plane Area by Integration

Evaluation of a Bounded Plane Area by Integration

In the beginning of this chapter, we have already introduced definite integral by a geometrical approach. In that approach, we have noted that, whenever the integrand of the definite integral is non-negative, the definite integral yields the geometrical area. In the present section, we apply the approach for finding areas of plane regions bounded by plane curves.

1. Area of the region bounded by a curve, x – axis and the lines x = a and x = b.

Case (i)

Let y = f ( x ), a ≤ x ≤ b be the equation of the portion of the continuous curve that lies above the x − axis (that is, the portion lies either in the first quadrant or in the second quadrant) between the lines x = a and x = b . See Fig.9.8. Then, y ≥ 0 for every point of the portion of the curve. Consider the region bounded by the curve, x − axis, the ordinates x = a and x = b . It is important to note that y does not change its sign in the region. Then, the area A of the region is found as follows:

Viewing in the positive direction of the y − axis, divide the region into elementary vertical strips (thin rectangles) of height y and width Δx . Then, A is the limit sum of the areas of the vertical strips.

Hence, we get

Case (ii)

Let y = f ( x ), a ≤ x ≤ b be the equation of the portion of the continuous curve that lies below the x − axis (that is, the portion lies either in the third quadrant or in the fourth quadrant). Then, y ≤ 0 for every point of the portion of the curve. It is important to note that y does not change its sign in the region. Consider the region bounded by the curve, x − axis, the ordinates x = a and x = b . See Fig.9.9. Then, the area A of the region is found as follows:

Viewing in the negative direction of the y − axis, divide the region into elementary vertical strips (thin rectangles) of height |y| = − y and width Δx . Then, A is the limit of the sum of the areas of the vertical strips. Hence, we get

A = lim ∑_{a≤x≤b} −yΔx = −^b∫_a ydx = | ^b∫_a ydx | .

Case (iii)

Let y = f ( x ), a ≤ x ≤ b be the equation of the portion of the continuous curve that lies above as well as below the x − axis (that is, the portion may lie in all quadrants).

Draw the graph of y = f ( x) in the XY − plane. The graph lies alternately above and below the x − axis and it is intercepted between the ordinates x = a and x = b.

Divide the interval[a , b] into subintervals [a , c₁ ] , [c₁ , c₂ ] , . . . , [c_k , b] such that f ( x) has the same sign on each of subintervals. Applying cases (i) and (ii), we can obtain individually, the geometrical areas of the regions corresponding to the subintervals.

Hence the geometrical area of the region bounded by the graph of y = f ( x) , the x-axis, the lines x = a and x = b is given by

For instance, consider the shaded region in Fig. 9.10. Here A₁ , A₂ , A₃ , and A₄ denote geometric areas of the individual parts. Then, the total area is given by

2. Area of the region bounded by a curve, y– axis and the lines y = c and y = d.

Case (iv)

Let x = f ( y ), c ≤ y ≤ d be the equation of the portion of the continuous curve that lies to the right side of y − axis (that is, the portion lies either in the first quadrant or in the fourth quadrant). Then, x ≥ 0 for every point of the portion of the curve. It is important to note that x does not change its sign in the region.

Consider the region bounded by the curve, y − axis, the lines y = c and y = d . The region is sketched as in Fig. 9.11. Then, the area A of the region is found as follows:

Viewing in the positive direction of the x − axis, divide the region into thin horizontal strips (thin rectangles) of length x and width Δy . Then, A is the limit of the sum of the areas of the horizontal strips. Hence, we get A = lim ∑_{c≤y≤d} xΔ y = ^d∫_c xdy. 

Case (v)

Let x = f ( y ), c ≤ y ≤ d be the equation of the portion of the continuous curve that lies to the left side of y − axis (that is, the portion lies either in the second quadrant or in the third quadrant). Then, x ≤ 0 for every point of the portion of the curve. It is important to note that x does not change its sign in  the region. Consider the region bounded by the curve, y − axis, the lines y = c and y = d . The region is sketched as in Fig. 9.12. Then, the area A of the region is found as follows:

Viewing in the positive direction of the x − axis, divide the region into thin horizontal strips (thin rectangles) of length |x| = −x and width Δy . Then, A is the limit of the sum of the areas of the horizontal strips.

Case (vi)

Let x = f ( y ), c ≤ y ≤ d be the equation of the portion of the continuous curve that lies to the right as well as to the left of the y − axis (that is, the portion may lie in all quadrants). Draw the graph of x = f ( y) in the XY − plane. The graph lies alternately to the right and to the left of the y − axis and it is intercepted between the lines y = c and y = d . Divide the interval [c , d ] into subintervals [c , a₁ ] , [a₁ , a₂ ] , . . . , [a_k , d ] such that f ( y) has the same sign on each of subintervals. Applying cases (iii) and (iv), we can obtain individually, the geometrical areas of the regions corresponding to the subintervals.

Hence the geometrical area A of the region bounded by the graph of x = f ( y) , the y-axis, the lines y = c and y = d is given by

For instance, consider the shaded region in Fig. 9.13. Here, B₁ , B₂ , B₃ and B₄ denote geometric areas of the individual parts. Then the total area B of the region bounded by the curve x = f(y), y − axis and the lines y = c and y = d is given by

B = B₁+B₂+B₃+B₄

Example 9.47

Find the area of the region bounded by the line 6x + 5y = 30 , x − axis and the lines x = −1 and x = 3 .

Solution

The region is sketched in Fig. 9.14. It lies above the x − axis. Hence, the required area is given by

Example 9.48

Find the area of the region bounded by the line 7x − 5y = 35 , x − axis and the lines x = −2 and x = 3.

Solution

The region is sketched in Fig. 9.15. It lies below the x − axis. Hence, the required area is given by

Example 9.49

Find the area of the region bounded by the ellipse

Solution

The ellipse is symmetric about both major and minor axes. It is sketched as in Fig.9.16. So, viewing in the positive direction of y -axis, the required area A is four times the area of the region bounded by the portion of the ellipse in the first quadrant

Note

Viewing in the positive direction of x -axis, the required y  area A is four times the area of the region bounded by the portion of the ellipse in the first quadrant

 y-axis, y = 0 and y = b. Hence, by taking horizontal strips (see Fig.9.17), we get

Note

Putting b = a in the above result, we get that the area of the region enclosed by the circle x²+ y² = a² is Ï€a² .

Example 9.50

Find the area of the region bounded between the parabola y² = 4ax and its latus rectum.

Solution

The equation of the latus-rectum is x = a . It intersects the parabola at  the points L ( a, 2a) and L₁ ( a, −2a) . The required area is sketched in Fig. 9.18. By symmetry, the required area A is twice the area bounded by the portion of the parabola

y = 2 √a√x , x -axis, x = 0 and x = a.

Hence, by taking vertical strips, we get

Note

Viewing in the positive direction of x-axis, and making horizontal strips (see Fig. 9.19), we get

Note

It is quite interesting to note that the above area is equal to two-thirds the base (latus-rectum) times the height (the distance between the focus and the vertex). This verifies Archimedes’ formula for areas of parabolic arches which states that the area under a parabolic arch is two-thirds the area of the rectangle having base of the arch as length and height of the arch as the breadth. It is also equal to four-thirds the area of the triangle with base (latus-rectum) and height (the distance between the focus and the vertex).

Example 9.51

Find the area of the region bounded by the y -axis and the parabola x = 5 − 4 y − y².

Solution

The equation of the parabola is ( y + 2)² = − ( x − 9) . The parabola crosses the y -axis at (0, −5) and (0,1) .The vertex is at (9, −2) and the axis of the parabola is y = −2 . The required area is sketched as in Fig. 9.20.

Viewing in the positive direction of x − axis, and making horizontal strips, the required area A is given by

Note

As in the previous problem, we again verify Archimedes’ formula that the area of the parabolic arch is equal to two-thirds the base times the height.

Example 9.52

Find the area of the region bounded by x − axis, the sine curve y = sin x , the lines x = 0 and x = 2π .

Solution

The required area is sketched in Fig. 9.21. One portion of the region lies above the x − axis between x = 0 and x = π , and the other portion lies below x − axis between x = π and x = 2π .

So, the required area is given by

Note

If we compute the definite integral ^2π∫₀sin xdx , we get

So ^2π∫₀ f (x)dx does not represent the area of the region bounded by the curve y = sin x , x − axis, the lines x = 0 and x = 2Ï€ .

Example 9.53

Find the area of the region bounded by x − axis, the curve y = |cos x|, the lines x = 0 and x = π .

Solution

It lies above the x − axis. The required area is sketched in Fig. 9.22. So, the required area is given by

3. Area of the region bounded between two curves

Case (i)

Let y = f ( x) and y = g( x) be the equations of two curves in the XOY − plane such that f (x) ≥ g ( x) for all x ∈ [a, b] . We want to find the area A of the  region bounded between the two curves, the ordinates x = a and  x = b .

The required area is sketched in Fig. 9.23. To compute A , we divide the region into thin vertical strips of width Δx and height f ( x) − g ( x) . It is important note that f (x) − g (x) ≥ 0 for all x ∈[a, b] . As before, the required area is the limit of the sum of the areas of the vertical strips. Hence, we get

 A = ^b∫_a [ f (x) - g(x)]dx

Note

Viewing in the positive direction of y − axis, the curve y = f ( x) can be termed as the upper curve (U) and the curve y = g( x) as the lower curve (L). Thus, we get A = ^b∫_a [ y_U − y_L]dx

Case (ii)

Let x = f ( y) and x = g( y) be the equations of two curves in the XOY − plane such that f ( y) ≥ g ( y) for all y ∈[c, d ] . We want to find the area A of the region bounded between the two curves, the lines y = c and y = d . The required area is sketched in Fig. 9.24. 

To compute A , we view in the positive direction of the x – axis and divide the region into thin horizontal strips of width Δy and height f ( y) − g ( y) . It is important note that f ( y) − g ( y) ≥ 0 for all y ∈[c, d ] . As before, the required area is the limit of the sum of the areas of the horizontal strips. Hence, we get

A = ^d∫_c [ f (x) - g(x)]dx

Note

Viewing in the positive direction of x − axis, the curve x = f ( y) can be termed as the right curve (R) and the curve x = f ( y) as the left curve (L). Thus, we get A = ^b∫_a [ x_R − x _L ]dy .

Example 9.54

Find the area of the region bounded between the parabolas y² = 4x and x² = 4y.

Solution

First, we get the points of intersection of the parabolas. For this, we  solve y² = 4x and x² = 4y simultaneously: Eliminating y between them, we get x⁴ = 64x and so x = 0 and x = 4 . Then the points of intersection are (0, 0) and (4, 4) . The required region is sketched in Fig.9.25.

Viewing in the direction of y -axis, the equation of the upper boundary is y = 2√x for 0 ≤ x ≤ 4 and the equation of the lower boundary is y = x²/4 for 0 ≤ x ≤ 4 . So, the required area Δ is

Note

Viewing in the positive direction of x -axis, the right bounding curve is x² = 4y and the left bounding curve is y² = 4x . See Fig. 9.26. The equation of the right boundary is x = 2√y for 0 ≤ y ≤ 4 and the equation of the left boundary is x = y²/4 for 0 ≤ y ≤ 4 . So, the required area A is

Example 9.55

Find the area of the region bounded between the parabola x² = y and the curve y = |x|.

Solution

Both the curves are symmetrical about y -axis.

It intersects the parabola x² = y at ( 1,1) and ( −1, 1).

The area of the region bounded by the curves is sketched in Fig. 9.27. It lies in the first quadrant as well as in the second quadrant. By symmetry, the required area is twice the area in the first quadrant.

In the first quadrant, the upper curve is y  = x, 0 ≤ x ≤ 1 and the lower curve is y = x², 0 ≤ x ≤ 1. Hence, the required area is given by

Example 9.56

Find the area of the region bounded by y = cos x, y = sin x, the lines x = π/4 and x = 5π/4.

Solution

The region is sketched in Fig. 9.28. The upper boundary of the region is y = sin x for π/4 ≤ x ≤ 5π/4 and the lower boundary of the region is y = cos x for π/4 ≤ x ≤ 5π/4. So the required area A is given by

Example 9.57

The region enclosed by the circle x² + y² = a² is divided into two segments by the line x = h. Find the area of the smaller segment.

Solution

The smaller segment is sketched in Fig. 9.29. Here 0 < h < a . By symmetry about the x -axis, the area of the smaller segment is given by

Example 9.58

Find the area of the region in the first quadrant bounded by the parabola y² = 4x , the line x + y = 3 and y -axis.

Solution

First, we find the points of intersection of x + y = 3 and y² − 4x :

 x + y = 3 ⇒ y = 3 − x .

 âˆ´ y ² = 4x ⇒ (3 − x )² = 4x

⇒ x ² − 10x + 9 = 0

⇒ x = 1, x = 9 .

∴ x =1 in x + y = 3 ⇒ y = 2 , and x = 9 in x + y = 3 ⇒ y = −6 .

∴(1, 2) and (9, −6) are the points of intersection.

The line x + y = 3 meets the y -axis at (0, 3) .

The required area is sketched in Fig. 9.30.

Viewing in the direction of y -axis, on the right bounding curve is given by

Example 9.59

Find, by integration, the area of the region bounded by the lines 5x − 2 y = 15, x + y + 4 = 0 and the x-axis.

Solution

The lines 5x − 2 y = 15, x + y + 4 = 0 intersect at (1, −5) . The line 5x − 2 y = 15 meets the x-axis at ( 3, 0) . The line x + y + 4 = 0 meets the x-axis at ( −4, 0) . The required area is shaded in Fig.9.31. It lies below the x-axis. It can be computed either by considering vertical strips or horizontal strips.

When we do by vertical strips, the region has to be divided into two sub-regions by the line x = 1 . Then, we get

When we do by horizontal strips, there is no need to subdivide the region. In this case, the area is bounded on the right by the line 5x − 2 y = 15 and on the left by x + y + 4 = 0 . So, we get

Note

The region is triangular with base 7 units and height 5 units. Hence its area is 35/2  without using integration.

Example 9.60

Using integration find the area of the region bounded by triangle ABC, whose vertices A, B, and C are ( −1,1), ( 3, 2) , and ( 0, 5) respectively.

Solution

See Fig. 9.32.

Example 9.61

Using integration, find the area of the region which is bounded by x-axis, the tangent and normal to the circle x² + y² = 4 drawn at (1, √3) .

Solution

We recall that the equation of the tangent to the circle  x² + y²= a² at ( x₁ , y₁ ) is xx₁ + yy₁ = a² . So, the equation of the tangent to the circle x² + y² = 4 at (1, √3) is x + y √3 = 4 ; that is, y =  âˆ’ 1/√3 (x − 4) . The tangent meets the x-axis at the point (4,0). The slope of the tangent is –1/√3. So the slope of the normal is √3 and hence equation of the normal is y − √3 = √3( x −1) ; that is y = √3x and it passes through the origin. The area to be found is shaded in the adjoining figure. It can be found by two methods.

Method 1

Viewing in the postive direction of y-axis, the required area is the area of the region bounded by x-axis, y = √3x and x + y √3 = 4 . So it can be obtained by applying the formula ^b∫_a ydx . For this, we have to split the region into sub-regions, one sub-region bounded by x-axis, the normal y = √3x and the line x = 1 ; the other sub-region bounded by x-axis, the tangent x + y√3= 4 and the line x = 1 axis.

Method 2

Viewing in the direction of x-axis, the required area is the area of the region bounded between y = √3x and x + y√3 = 4 , y = 0 and y = √3. So, it can be obtained by applying the formula

Here, c = 0 , d = √3 , x_R is the x-value on the tangent x + y√3 = 4 and x_L is the x-value on the normal y = √3x .

Working rule for finding area of the region bounded by y = f₁(x), y = f₂(x), the lines x = a and x = b , where a < b :

Draw an arbitrary line parallel to y-axis cutting the plane region. First, find the y-coordinate of the point where the line enters the region. Call it y_ENTRY . Next, find the y-coordinate of the point where the line exits the region. Call it y_EXIT . Both y_ENTRY and y_EXIT can be found from the equations of the bounding curves. Then, the required area is given by ^b∫_a [ y _EXIT − y_ENTRY ]dx.

Working rule for finding area of the region bounded by x = g₁(y), x = g₂(y), the lines y = c and y = d , where c < d :

Draw an arbitrary line parallel to x-axis cutting the plane region.

First, find the x-coordinate of the point where the line enters the region. Call it x_ENTRY .

Next, find the x-coordinate of the point where the line exits the region. Call it x_EXIT . Both x_ENTRY and x_EXIT can be found from the equations of the bounding curves. Then, the required area is given by ^d∫_c [ x _EXIT − x_ENTRY ]dy .