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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.

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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.

Given region is
   $open curly brackets left parenthesis straight x comma space straight y right parenthesis space colon space straight y squared less or equal than 5 straight x comma space 5 straight x squared plus 5 straight y squared space less or equal than 36 close curly brackets$
Consider the equations
$straight y squared space equals space 5 straight x$ ...(1)
$5 straight x squared plus 5 straight y squared space equals space 36$
i.e.    $straight x squared plus straight y squared space equals space 36 over 5$ ...(2)
From (1) and (2), we get,
   $5 straight x squared plus 25 straight x space equals space 36 space space space or space space 5 straight x squared plus 25 straight x minus 36 space equals space 0$
$therefore space space space straight x space equals space fraction numerator negative 25 plus-or-minus square root of 625 plus 720 end root over denominator 10 end fraction equals space fraction numerator negative 25 plus-or-minus square root of 1345 over denominator 10 end fraction$
Rejecting negative value of x, we get,
$straight x equals fraction numerator negative 25 plus square root of 1345 over denominator 10 end fraction equals straight a space left parenthesis say right parenthesis$

which gives the abscissa of the points of intersection P and Q.
Required area = Area of shaded region = 2 (area of region OAPO)
= 2[area of region OMPO + area of region MAPM]
$equals space 2 open square brackets integral subscript 0 superscript straight a square root of 5 square root of straight x dx plus integral subscript straight a superscript 6 divided by square root of 5 end superscript square root of 36 over 5 minus straight x squared end root close square brackets dx$
$equals space 2 square root of 5 open square brackets fraction numerator straight x to the power of begin display style 3 over 2 end style end exponent over denominator begin display style 3 over 2 end style end fraction close square brackets subscript 0 superscript straight a plus 2 open square brackets fraction numerator straight x square root of begin display style 36 over 5 end style minus straight x squared end root over denominator 2 end fraction plus fraction numerator begin display style 36 over 5 end style over denominator 2 end fraction sin to the power of negative 1 end exponent open parentheses fraction numerator straight x over denominator 6 divided by square root of 5 end fraction close parentheses close square brackets subscript straight a superscript fraction numerator 6 over denominator square root of 5 end fraction end superscript equals space fraction numerator 4 square root of 5 over denominator 3 end fraction left parenthesis straight a to the power of 3 over 2 end exponent minus 0 right parenthesis space space plus open square brackets open parentheses 0 plus 36 over 5 sin to the power of negative 1 end exponent 1 close parentheses space minus space straight a square root of 36 over 5 minus straight a squared end root sin to the power of negative 1 end exponent open parentheses fraction numerator straight a square root of 5 over denominator 6 end fraction close parentheses close square brackets equals space fraction numerator 4 square root of 5 over denominator 3 end fraction straight a to the power of 3 divided by 2 end exponent plus 36 over 10 straight pi minus straight a square root of 36 over 5 minus straight a squared end root minus 36 over 5 sin to the power of negative 1 end exponent open parentheses fraction numerator straight a square root of 5 over denominator 6 end fraction close parentheses$
where $straight a equals fraction numerator negative 25 plus square root of 1345 over denominator 10 end fraction$