Introduction to Bounded Regions
In this article we will study a specific application of integrals to find the area under simple curves, area between lines and arcs of circles, parabolas and ellipses.
Area under simple curves
The area bounded by the curve y = f(x), x-axis and the ordinates x = a and x = b is given by A= a∫b y dx= a∫b f(x) dx
The area A of the region bounded by the curve x = g(y), y-axis and the lines y = c and y = d is given by
A = c∫d x dy = c∫d g(y) dy
Example: Find the area of the region bounded by the curve y2 = 9x, x = 2, x = 4 and the x-axis in the first quadrant.
Solution: Area = 2∫4 y dx
= 2∫4 3√x dx
= 3 [x3/2 / (3/2)]24
= (16 - 4√2) sq. units
Area between a curve and a line
Example: Find the area of the smaller part of the circle x2 + y2 = a2 cut off by the line x = a/√2
Solution: x2 + y2 = a2 ………….(1)
x = a/√2 …………..(2)
Solving (1) and (2) we will get the point of intersection
We have to find the area of shaded region which is given by
A=2*[ a/√2∫a Area under the curve (y2 = a2 - x2) dx]
= 2 a/√2∫a√(a2 - x2) dx
= 2 [x/2 (√a2 - x2) – (a2/2) sin-1 (x/a)]a/√2a
= a2/2 [(π/2) – 1] sq. units
Area between two curves
Example: Find the area lying above x-axis and included between the circle x2 + y2 = 8x and inside the parabola y2 = 4x
Solution: The given equation of the circle x2 + y2 = 8x can be expressed as (x - 4)2 + y2 = 16, which is a circle with center (4, 0) and radius 4.
The point of intersection gives x = 0, 4
Hence the curves intersect at O (0, 0) and P (4, 4) above the x-axis.
Required area=2 0∫4 √x dx+ 4∫8 √(42 - (x - 4)2) dx
=2(2/3) [x3/2]04 + [((x-4)/2) √(42-(x-2)2) + (42/2)sin-1(x-2)/2]48
= 32/3 + [4/2 *0+1/2 *16*sin-1(1)]
= (32/3) + 4π
= (4/3) (8 + 3π) sq. units
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