Find the area of the region {(x, y): x2 ≤ y ≤ |x|}.OrFind

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 Multiple Choice QuestionsShort Answer Type

61.

Find the area of the region {(x, y): x2 + y2 ≤ 1 ≤ x + y}.

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 Multiple Choice QuestionsLong Answer Type

62.

Find the area of the region bounded by the circle x2 + y2 = 1 and x + y = 1. Also draw a rough sketch.

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63.

Find the area of the region {(x, y): x2 ≤ y ≤ |x|}.
Or
Find the area of the region bounded by the parabola y = x2 and y = |x|.


The given region is
      open curly brackets left parenthesis straight x comma space straight y right parenthesis space colon space straight x squared space less or equal than straight y less or equal than open vertical bar straight x close vertical bar close curly brackets
This region is the intersection of the following regions
                        straight R subscript 1 space equals space open curly brackets left parenthesis straight x comma space straight y right parenthesis space colon space straight x squared space less or equal than space straight y close curly brackets
straight R subscript 2 space equals open curly brackets open parentheses straight x comma space straight y close parentheses colon space straight y space less or equal than open vertical bar straight x close vertical bar close curly brackets
Consider the equations
                          straight y equals straight x squared                                            ...(1)
                          y = x                                               ...(2)
                   and  y = -x                                            ...(3)

From (1) and (2), we get
                        straight x equals straight x squared space space space space space space space space space rightwards double arrow space space space straight x squared minus straight x space equals space 0 space space space space space space space space rightwards double arrow space space space straight x left parenthesis straight x minus 1 right parenthesis space space space space space space rightwards double arrow space space space space space straight x space equals space 0 comma space space 1
  from (2), y = 0,  1
∴   curve (1) and (2) intersect in the points O (0, 0) and A (1, 1).
Similarly, curves (1) and (3) intersect in the points O (0, 0) and B (-1, 1)
Required area = area of shaded region = 2 (area of region OAO)
equals 2 open square brackets integral subscript 0 superscript 1 straight x space dx minus integral subscript 0 superscript 1 straight x squared space dx close square brackets space equals space 2 space open curly brackets open square brackets straight x squared over 2 close square brackets subscript 0 superscript 1 minus open square brackets straight x cubed over 3 close square brackets subscript 0 superscript 1 close curly brackets
equals space 2 open curly brackets open parentheses 1 half minus 0 close parentheses minus open parentheses 1 third minus 0 close parentheses close curly brackets space equals 2 open parentheses 1 half minus 1 third close parentheses space equals space 2 cross times 1 over 6 equals 1 third space sq space. units.

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64. Draw the rough sketch and find the area of the region:
{(x, y) : y2 ≤ 8 x, x2 + x2 ≤ 9} 
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65. Find the area of the region {(x, y): y2 ≤ 4 x, 4x2 + 4 y2 ≤ 9}
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66. Find the area lying above x-axis and included between the circle x2 + y2 = 8 x and inside of the parabola y2 = 4 x.
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67. Calculate the area enclosed in the region:
open curly brackets left parenthesis straight x comma space straight y right parenthesis space semicolon space space straight x squared plus straight y squared space less or equal than space 1 space less than space straight x plus 1 half straight y close curly brackets
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68. Find the area of the circle x2 + y2 = 16 which is exterior to the parabola y2 = 6x. 

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69. Find the area of the circle 4x2 + 4y2 = 9 which is interior to the parabola y2= 4 x. 
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70. Draw a rough sketch of the region {(x, y): y2 ≤ 5 x, 5 x2 + 5 y2 ≤ 36} and find the area enclosed by the region using method of integration.
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