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Application of Integrals

Name: Zigya - For The Curious Learner
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Short Answer Type

61.

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

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Long 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.

Consider the equations
   $straight x squared plus straight y squared space equals space 1$ ...(1)
and    $straight x plus straight y space equals space 1$ ...(2)
From (2), y = 1 - x ...(3)
Putting this value of y in (1), we get,
$straight x squared plus left parenthesis 1 minus straight x right parenthesis squared space equals space 1 space space space space space space space rightwards double arrow space space space 2 space straight x squared minus 2 straight x space equals space 0$
$rightwards double arrow space space space space space straight x squared minus straight x space equals space 0 space space space space space space space rightwards double arrow space space space straight x left parenthesis straight x minus 1 right parenthesis space equals space 0 space space space space rightwards double arrow space space straight x space equals space 0 comma space space 1 therefore space space space from space left parenthesis 3 right parenthesis comma space space straight y space equals space 1 minus 0 comma space space space 1 minus 1 space equals space 1 comma space 0$

$therefore$ circle (1) and st . line x + y = 1 intersect in the points A(1, 0) and B(0, 1).
Required area = Area of shaded region = $integral subscript 0 superscript 1 square root of 1 minus straight x squared end root dx space minus integral subscript 0 superscript 1 left parenthesis 1 minus straight x right parenthesis space dx$
   $equals space open square brackets fraction numerator straight x square root of 1 minus straight x squared end root over denominator 2 end fraction plus 1 half sin to the power of negative 1 end exponent straight x close square brackets subscript 0 superscript 1 space minus space open square brackets straight x minus straight x squared over 2 close square brackets subscript 0 superscript 1 equals space open square brackets open parentheses 0 plus 1 half sin to the power of negative 1 end exponent 1 close parentheses minus open parentheses 0 plus 1 half sin to the power of negative 1 end exponent 0 close parentheses close square brackets space minus space open square brackets open parentheses 1 minus 1 half close parentheses minus left parenthesis 0 minus 0 right parenthesis close square brackets equals space 1 half open parentheses straight pi over 2 close parentheses minus 1 half space equals open parentheses straight pi over 4 minus 1 half close parentheses space sq. space units. space$

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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|.

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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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