## 1. Area under Simple Curves especially lines, circles/ parabolas/ellipses (in standard form only)

- Books Name
- Mathmatics Book Based on NCERT

- Publication
- KRISHNA PUBLICATIONS

- Course
- CBSE Class 12

- Subject
- Mathmatics

**Chapter 8**

**Area under Simple Curves especially lines, circles/ parabolas/ellipses (in standard form**

One of the major application of integrals is in determining the area under the curves.

Consider a function y = f(x), then the area is given as

Consider the two curves having equation of f(x) and g(x), the area between the region a,b of the two curves is given as-

dA = f(x) – g(x)]dx, and the total area A can be taken as-

**Application of Integrals Examples**

**Example 1: **

Determine the area enclosed by the circle x^{2} + y^{2 }= a^{2}

**Solution:**

Given, circle equation is x^{2} + y^{2 }= a^{2}

From the given figure, we can say that the whole area enclosed by the given circle is as

= 4(Area of the region AOBA bounded by the curve, coordinates x=0 and x=a, and the x-axis)

As the circle is symmetric about both x-axis and y-axis, the equation can be written as

= 4_{ 0}∫^{a }y dx (By taking the vertical strips) ….(1)

From the given circle equation, y can be written as

y = ±√(a^{2}-x^{2})

As the region, AOBA lies in the first quadrant of the circle, we can take y as positive, so the value of y becomes √(a^{2}-x^{2})

Now, substitute y = √(a^{2}-x^{2}) in equation (1), we get

= 4_{ 0}∫^{a }√(a^{2}-x^{2}) dx

Integrate the above function, we get

= 4 [(x/2)√(a^{2}-x^{2}) +(a^{2}/2)sin^{-1}(x/a)]_{0} ^{a}

Now, substitute the upper and lower limit, we get

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

= 4(a^{2}/2)(π/2)

= πa^{2}.

Hence, the area enclosed by the circle x^{2} +y^{2} =a^{2 } is πa^{2}.

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

- Books Name
- Mathmatics Book Based on NCERT

- Publication
- KRISHNA PUBLICATIONS

- Course
- CBSE Class 12

- Subject
- Mathmatics

**To find the area of the region bounded by a curve and a line**

Let’s have a look at the example to understand how to find the area of the region bounded by a curve and a line.

**Example 2:**

Find the area of the region bounded between the line x = 2 and the parabola y^{2} = 8x.

**Solution:**

Given equation of parabola is y^{2} = 8x.

Equation of line is x = 2.

Here, y^{2} = 8x as a right handed parabola having its vertex at the origin and x = 2 is the line which is parallel to y-axis at x = 2 units distance

Similarly,

y^{2} = 8x has only even power of y and is symmetrical about x-axis.

So, the required area = Area of OAC + Area of OAB

= 2 (Area of OAB)

= 2 ∫_{0}^{2} y dx

Substituting the value of y, i.e. y2 = 8x and y = √(8x) = 2 √2 √x, we get;

= 2 ∫_{0}^{2} (2 √2 √x) dx

= 4√2 ∫_{0}^{2} (√x) dx

= 4√2 [x^{3/2}/ (3/2)]_{0}^{2}

By applying the limits,

= 4√2 {[2^{3/2}/ (3/2)] – 0}

= (8√2/3) × 2√2

= (16 × √2 × √2)

= 32/3

Go through the example given below to learn how to find the area between two curves.

**Example 3:**

Determine the area which lies above the x-axis and included between the circle and parabola, where the circle equation is given as x^{2}+y^{2} = 8x, and parabola equation is y^{2} = 4x.

**Solution:**

The circle equation x^{2}+y^{2} = 8x can be written as (x-4)^{2}+y^{2}=16. Hence, the centre of the circle is (4, 0), and the radius is 4 units. The intersection of the circle with the parabola y^{2} = 4x is as follows:

Now, substitute y^{2} = 4x in the given circle equation,

x^{2}+4x = 8x

x^{2}– 4x = 0

On solving the above equation, we get

x=0 and x=4

Therefore, the point of intersection of the circle and the parabola above the x-axis is obtained as O(0,0) and P(4,4).

Hence, from the above figure, the area of the region OPQCO included between these two curves above the x-axis is written as

= Area of OCPO + Area of PCQP

= _{0}∫^{4} y dx + _{4}∫^{8} y dx

= 2 _{0}∫^{4} √x dx + _{4}∫^{8} √[4^{2}– (x-4)^{2}]dx

Now take x-4 = t, then the above equation is written in the form

= 2 _{0}∫^{4} √x dx + _{0}∫^{4 } √[4^{2}– t^{2}]dx …. (1)

Now, integrate the functions.

2 _{0}∫^{4} √x dx = (2)(⅔) (x^{3}/2)_{0}^{4}

2 _{0}∫^{4} √x dx = 32/3 …..(2)

_{0}∫^{4} √[4^{2}– t^{2}]dx = [(t/2)(√[4^{2}-t^{2}] + (½)(4^{2})(sin^{-1}(t/4)]_{0}^{4}

_{0}∫^{4} √[4^{2}– t^{2}]dx = 4π …..(3)

Now, substitute (2) and (3) in (1), we get

= (32/3) + 4π

= (4/3) (8+3π)

Therefore, the area of the region that lies above the x-axis, and included between the circle and parabola is (4/3) (8+3π).

## 3. Area between Two Curves

- Books Name
- Mathmatics Book Based on NCERT

- Publication
- KRISHNA PUBLICATIONS

- Course
- CBSE Class 12

- Subject
- Mathmatics

**Area Between Two Curves**

**Type-1:**

**Area between two curves = ∫ _{a}^{b }[f(x) – g(x)] dx**

**Type-2:**

**Example: **

Find the area of the region bounded by the parabolas y = x^{2} and x = y^{2}.

**Solution:**

When the graph of both the parabolas is sketched we see that the points of intersection of the curves are (0, 0) and (1, 1) as shown in the figure below.

So, we need to find the area enclosed between these points which would give us the area between two curves. Also, in the given region as we can see,

y = x^{2 }= g(x)

and

x = y^{2}

or, y = √x = f(x).

As we can see in the given region,

The area enclosed will be given as,

**Example :**

Find the area bounded by curves (x – 1)^{2} + y^{2} = 1 and x^{2} + y^{2} = 1.

**Solution:**

Given equations of curves:

x^{2} + y^{2} = 1 ….(i)

(x – 1)^{2} + y^{2} = 1 ….(ii)

From (i),

y^{2} = 1 – x^{2}

By substituting it in equation (2), we get;

(x – 1)^{2} + 1 – x^{2} = 1

On further simplification

(x – 1)^{2} – x^{2} = 0

Using the identity a^{2} – b^{2} = (a – b)(a + b),

(x – 1 – x) (x – 1 + x) = 0

-1(2x – 1) = 0

– 2x + 1 = 0

2x = 1

x = 1/2

Using this in equation (1) we get;

y = ± √3/2

Thus, both the equations intersect at point A (1/2, √3/2) and B (1/2, -√3/2).

Also, (0, 0) is the centre of first circle and radius 1

Similarly, (1, 0) is the centre of second circle and radius is 1.

Here, both the circles are symmetrical about x-axis and the required area is shaded here.

So, the required area = area OACB

= 2 (area OAC)

= 2 [area of OAD + area DCA]