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

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Class 12 Chapter 7 - Application of Integrals

Area under Simple Curves and The area bounded by a curve and a line

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
about y-axis only, therefore

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
quadrant (Fig 8.11). Draw perpendicular
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 ]`