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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.

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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.

The equation of circle is
   $straight x squared plus straight y squared space equals 16$ ...(1)
The equation of parabola is
   $straight y squared space equals 6 straight x$ ...(2)
From (1) and (2),
$straight x squared plus 6 straight x space equals space 16$
or $straight x squared plus 6 straight x minus 16 space equals space 0$
$therefore space space space left parenthesis straight x plus 8 right parenthesis space left parenthesis straight x minus 2 right parenthesis space equals space 0$
$rightwards double arrow$    $straight x equals negative 8 comma space 2$

Rejecting x = -8 as parabola lies in 1st or 4th quadrant , we get x =2

When $straight x equals 2 comma space space straight y equals square root of 12 space equals space 2 square root of 3$

$therefore space space left parenthesis 1 right parenthesis space and space left parenthesis 2 right parenthesis space intersect space in space straight P left parenthesis 2 comma space 2 square root of 3 right parenthesis$

Required Area = Area of the circle - area of circle interior to the parabola
$equals 16 straight pi minus 2 integral subscript 0 superscript 2 straight y space dx space minus space 2 integral subscript 2 superscript 4 straight y space dx space equals space 16 straight pi minus 2 integral subscript 0 superscript 2 square root of 6 straight x to the power of 1 half end exponent dx minus 2 integral subscript 0 superscript 4 square root of 16 minus straight x squared end root dx$
$equals 16 straight pi minus 2 square root of 6 open square brackets 2 over 3 straight x to the power of 3 over 2 end exponent close square brackets subscript 0 superscript 2 minus space 2 open square brackets straight x over 2 square root of 16 minus straight x squared end root plus 16 over 2 sin to the power of negative 1 end exponent straight x over 4 close square brackets subscript 2 superscript 4 equals space 16 straight pi minus 4 over 3 square root of 6. space 2 square root of 2 minus 2 open square brackets 4 over 2 square root of 16 minus 16 end root plus 16 over 2 sin to the power of negative 1 end exponent 1 minus 2 over 2 square root of 16 minus 4 end root minus 16 over 2 sin to the power of negative 1 end exponent 2 over 4 close square brackets equals 16 straight pi minus fraction numerator 16 over denominator square root of 3 end fraction minus 16. space straight pi over 2 plus 4 square root of 3 plus fraction numerator 8 straight pi over denominator 3 end fraction equals negative fraction numerator 4 over denominator square root of 3 end fraction plus fraction numerator 32 space straight pi over denominator 3 end fraction space equals 4 over 3 left parenthesis 8 straight pi minus square root of 3 right parenthesis$