# 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= ab y dx= ab 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 = cd x dy = cd 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 = 24 y dx

= 24 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/√2a Area under the curve (y2 = a2 - x2) dx]

= 2 a/√2a√(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 04 √x dx+ 48 √(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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