**NCERT Solutions of class 12 maths exercise 8.2-Application of Integrals **

**NCERT Solutions of class 12 maths exercise 8.2 of chapter 8-Application of Integrals** are the **solutions of class 12 NCERT maths chapter 8 exercise 8.2- Applications of integrals** that are created by** future study point** for the purpose of clearing the concept of the **chapter-Application of integrals**. All these **NCERT** **solutions of exercise 8.1** are well explained by a step by step ways in order to be assessable for every student of **class 12.**

Applications of integrations are used in calculating the area, volume, mass, and density of the different types of shapes of the figure but exercise 8.2 is based on the area of curves covered by different kinds of curves.

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**Exercise 8.1-Application of Integrals**

**Exercise 8.3 – Application of Integrals**

** Miscellaneous exercise -Application of Integrals**

**NCERT Solutions of class 12 maths exercise 8.2-Application of Integrals **

**Q1. Find the area of circle 4x² + 4y² = 9 which is interior to the parabola x²= 4y.**

Ans.We are given the equation of circle

4x² + 4y² = 9

Equation of parabola is given to us

x²= 4y……..(ii)

Equation of circle shows that its centre is at (0,0) radius is 3/2

Solving (i) and (ii) equations we get that parabola and circle intersect each other in first quadrant at (√2,1/2) and second quadrant at (-√2, 1/2)

Area interior to parabola=2×area of OBC

Area of OBC= area of OPBC -area of OPB

Area of OPBC

From (i) equation, we have

Area of OPB

Where y is equation of a parabola

x² = 4y

Therefore

Hence the area of OBC

Area interior to parabola (OBCD)

**NCERT Solutions of class 12 maths exercise 8.2-Application of Integrals **

Q2. Find the area bounded by the curves (x-1)² + y² = 1 and x² + y² =1.

Ans.We are given the equations of circle (x-1)² + y² = 1 and x² + y² =1,solving both of these equation for getting point of intersecting

Both equations of the circle show that their centre lies at (1,0) and (0,0)

(x-1)² + y² = 1

x² +1 -2x +y² =1

x² = 2x – y², putting in equation x² + y² =1

2x – y² + y² = 1

2x = 1⇒ x=1/2,putting in x² + y² =1

y² = 1-1/4

y= ±√3/2

Hence the points of intersection of the both circle are (1/2, √3/2) and (1/2, -√3/2)

Solving both equation for the value of y

y = √[1-(x²-1)]….(i) and y =√( 1 -x²)……(ii)

Required area = ar ABDO= ar ABDC + arAODC

ar AODC= 2×ar AOC ( Area of AODC is symmetric about x-axis)

ar AOC

As seen in the figure

Applying the integral formula

Therefore ar AODC

ar ABDC= 2×ar ABC ( Area of ABDC is symmetric about x-axis)

From the figure

ar ABC

Applying the integral formula

ar ABDC

Required area = ar ABDO

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Chapter 2-Inverse Trigonometric Functions | Chapter 10-Vector Algebra |

Chapter 3-Matrices | Chapter 11 – Three Dimensional Geometry |

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