Mathematics Area under Simple Curves and The area bounded by a curve and a line
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### Topic Covered

color{red}** Introduction
color{red}** Area under Simple Curves
color{red}** The area of the region bounded by a curve and a line

### Introduction

● Here we shall study a specific application of integrals to find the area under simple curves, area between lines and arcs of circles, parabolas and ellipses (standard forms only).

### Area under Simple Curves

● We can think of area under the curve as composed of large number of very thin vertical strips.

● Consider an arbitrary strip of height y and width color{blue}{dx},

color{green} ✍️ color{red}{dA "(area of the elementary strip)"= ydx, } where, y = f (x).

● Total area A of the region between x-axis, ordinates x = a, x = b and the curve y = f (x) as the result of adding up the elementary areas of thin strips across the region PQRSP. Symbolically, we express

● Here, we consider vertical strips as shown in the Fig

color{blue} {A = int_a^b d A = int_a^b ydx = int_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, y = d is given by

● Here, we consider horizontal strips as shown in the Fig

color{blue} {A = int_c^d x dy = int_c^d g(y) dy}

### color{red} text{Important Note } :

"Case 1 :" If the position of the curve under consideration is below the y-axis, then since f (x) < 0 from x = a  to x = b, as shown in Fig

● The area bounded by the curve, x-axis and the ordinates x = a, x = b come out to be negative. But, it is only the numerical value of the area which is taken into consideration.

● Thus, if the area is negative, we take its absolute value, i.e ,color{red}{| int_a^b f(x) dx |}

"Case 2 : " If some portion of the curve is above x-axis and some is below the x-axis as shown in the Fig Here, A_1 < 0 and A_2 > 0.

● The area A bounded by the curve y = f (x), x-axis and the ordinates x = a and x = b is given by color{red}{A = |A_1| + A_2}
Q 3146534473

Find the area enclosed by the circle x^2 + y^2 =a^2
Class 12 Chapter 8 Example 1
Solution:

From Fig, the whole area enclosed by the given circle
= 4 (area of the region AOBA bounded by the curve, x-axis and the ordinates x = 0 and x = a) [as the circle is symmetrical about both
x-axis and y-axis]

= 4 int_0^a ydx (taking vertical strips)

= 4 int_0^a sqrt(a^2 -x^2 ) dx

Since x^2 + y^2 = a^2 gives y = pm sqrt(a^2 - x^2)

As the region AOBA lies in the first quadrant, y is taken as positive. Integrating, we get the whole area enclosed by the given circle

= 4 [ x/2 sqrt(a^2 -x^2 + a^2/2sin^-1 \ \ x/a]_0^a

= 4 [ (a/2 xx a + a^2/2 sin^-1 1 ) - 0 ]

= 4 (a^2/2) ( pi/2 ) = pia^2
Q 3176534476

Find the area enclosed by the ellipse x^2/a^2 + y^2/b^2 = 1
Class 12 Chapter 8 Example 2
Solution:

From Fig , the area of the region ABA′B′A bounded by the ellipse

= 4 ( Area of the region AOBA the first quadrant bounded by the curve x axis and the ordinates x =0 ,x = a)
(as the ellipse is symmetrical about both x-axis and y-axis)

= 4 int _0^a y dx  ( taking vertical strips)

Now x^/a^2 + y^2/b^2 =1 gives  y = pm b/a sqrt (a^2 - x^2), but as the region AOBA lies in the first quadrant, y is taken as positive. So, the required area is

= 4 int_0^a b/a sqrt (a^2 -x^2) dx

= (4b)/a [ x/2 sqrt (a^2 - x^2) + a^2/2 sin^-1(x/a) ]_0^a (Why ?)

= (4b)/a [ (a/2 xx a + a^2/2 sin^-1 ) - 0]

= (4b)/a a^2/2 pi/2 = piab

### The area of the region bounded by a curve and a line

● Here , we will find the area of the region bounded by a line and a circle, a line and a parabola, a line and an ellipse.

color{red} text{Remarks} :
● It is inferred that we can consider either vertical strips or horizontal strips for calculating the area of the region. Henceforth, we shall consider either of these two, most preferably vertical strips.
Q 3176145076

Find the area of the region bounded by the curve y = x^2 and the line y = 4.
Class 12 Chapter 8 Example 3
Solution:

Since the given curve represented by
the equation y = x^2 is a parabola symmetrical

the required area of the region AOBA is given by

 2 int xdy =

 2 ( area of the region BONB bounded by curve , y - axis and the lines  y = 0 and  y= 4 )

 = 2 int_0^4 sqrt( y)dy = 2 xx 2/3 [y^(3/2) ]_0^4 = 4/3 xx 8 = (32)/3  (Why?)

Here, we have taken horizontal strips as indicated in the

Q 3186145077

Find the area of the region in the first quadrant enclosed by the x-axis,
the line y = x, and the circle x^2 + y^2 = 32.
Class 12 Chapter 8 Example 4
Solution:

The given equations are

y = x ... (1)

and x^2 + y^2 = 32 ... (2)
Solving (1) and (2), we find that the line
and the circle meet at  B(4, 4) in the first
BM to the x-axis.
Therefore, the required area = area of
the region OBMO + area of the region
BMAB.
Now, the area of the region OBMO

 = ∫_0^4 ydx = ∫_0^4 xdx  ... (3)

= 1/2 [ x^2]_0^4 = 8

Again, the area of the region BMAB

= int_4^(4sqrt2) y dx = ∫_4^(4sqrt2) sqrt( 32 - x^ 2 ) dx

 = [ 1/2 .4 sqrt2 xx0 + 1/2 xx 32 xx sin^(-1) 1) - ( 4/2 sqrt(32 -16) + 1/2 xx 32 xx sin^(-1) \ (1/sqrt2))

= 8 π – (8 + 4π) = 4π – 8 ... (4)
Adding (3) and (4), we get, the required area = 4π.
Q 3136245172

Find the area bounded by the ellipse x^2/a^2 + y^2/b^ 2 =1 and the ordinates x = 0
and x = a e, where, b^2 = a^2 (1 – e^2) and e < 1.
Class 12 Chapter 8 Example 5
Solution:

The required area (Fig 8.12) of the region BOB′RFSB is enclosed by the
ellipse and the lines x = 0 and x = ae.
Note that the area of the region BOB′RFSB

= 2 ∫_0^(ae) ydx = 2 b/a ∫_0^(ae) sqrt(a^2 - x^2) dx

 = (2b)/a [ x/2 sqrt(a^2 - x^2) + a^2/2 sin^(-1) \ x/a ]_0^(ae)

 = (2b)/(2a) [ ae sqrt(a^2 - a^2 e^2) + a^2 sin^(-1) e]

 = ab [ e sqrt(1 - e^2) + sin^(-1) e ]