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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