QuizSolver
  • Bank PO
  • CBSE
  • IIT JEE
 
 

Solved Subjective Question on Area Set 3

Posted on - 04-01-2017

Math

IIT JEE

Example.1

Find the area enclosed between the circle x2 + y2 – 2x + 4y –11 = 0 and the parabola, y= -x2 + 2x + 1 -2.

Solution

The given curves are

(x -1)2 + (y + 2)2 = 16 ….(1).

y =-(x –1)2 -2+ 2 ….(2)

Solving (1) and (2) we get x = -1, 3

Hence required area,

A =

=

=

=sq. units

Example.2

Let f(x) = max then determine the area of the region bounded by the curves y = f(x), x-axis, y axis and x = 2p

Solution

Given f(x) = max

Hence required area =

=

=

= sq. units

Example.3

Find the area enclosed by |x| +|y| = 1.

Solution

Due to the modulus sign we consider four different cases :

Case-I:

x ≥ 0 , y ≥ 0, then |x| = x and |y| = y,

and the given equation becomes x + y =1 . . . . . . (1).

Case-II:

x ≥ 0 , y < 0 then |x| = x, |y| = – y and the given equation becomes

x - y = 1 . . . . . (2).

Case-III:

x < 0 , y ≥ 0 then |x| = -x , |y| = y and the given equation becomes

– x + y = 1 . . . . . (3).

Case-IV:

x < 0 , y < 0 then |x| = – x , |y| = – y

and the given equation becomes – x – y = 1 . . . . . (4).

Taking into the considerations of these (1)-(4) equations, the graph of |x|+|y| =1is as shown :

Obviously the quadrilateral ABCD is a square, whose side is of length . Hence required area is 2 sq. units.

Example.4

Find the area enclosed between the curves y = ln(x + e), x = ln(1/y) and the x-axis . .

Solution

The area enclosed between the curves is as shown in the fig. .

Required area
=

=

=

=

= (0 + 1 – e + e) – (0 + 0 –1 + 0) = 2 .

Example.5

ABCD is a square of side 2a, circles of radius 2a are drawn with centers at A, B, C, D then find the common area of these circles.

Solution

Given AB = BC = CD = DA = 2a

Co-ordinates of A, B, C, D are (a, a): (-a, a); (-a, -a); (a, -a) respectively

Equations of circles if centres are at A, B, C, D are

(x -a)2 + (y -a)2 = 4a2 …..(1).

(x +a)2 + (y -a)2 = 4a2 …..(2).

(x +a)2 + (y +a)2 = 4a2 …..(3).

(x -a)2 + (y +a)2 = 4a2 …..(4).

Solving these we get

P (0, a(-1)); Q (-a(-1), 0); R(0, -a(-1)); S(a(-1),0)

Required area PQRSP = 4 area of OSPO

=

=

=

=

=sq. units

 
Quadratic Equations - Solved Objective Questions Part 2 for Conceptual Clarity
Quadratic Equations - Solved Objective Questions Part 1 for Conceptual Clarity
Solved Objective Question on Probability Set 2
Solved Objective Question on Probability Set 1
Solved Objective Question on Progression and Series Set 2
Solved Objective Question on Permutations and Combinations Set 3
Solved Objective Question on Permutations and Combinations Set 2
Solved Objective Question on Progression and Series Set 1
Solved Objective Question on Permutations and Combinations Set 1
Quadratic Equations - Solved Subjective Questions Part 4
Quadratic Equations - Solved Subjective Questions Part 2
Quadratic Equations - Solved Subjective Questions Part 3
Quadratic Equations - Solved Subjective Questions Part 1
Solved Subjective Questions on Circle Set 9
Solving Equations Reducible to Quadratic Equations
Theory of Polynomial Equations and Remainder Theorem
Solved Subjective Questions on Circle Set 8
Solving Quadratic Inequalities Using Wavy Curve Methods
Division and Distribution of Objects - Permutation and Combination
Basics of Quadratic Inequality or Inequations

Comments