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Application of Integrals

Name: Zigya - For The Curious Learner
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Long Answer Type

71. Draw a rough sketch of the region {(x, y): y² ≤ 3 x, 3 x² + 3 y² ≤ 16} and find the area enclosed by the region using method of integration.

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72.

Find the area of the smaller part of the circle x² + y² = a cut oil by the line $straight x equals fraction numerator straight a over denominator square root of 2 end fraction$

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73. Find the area of the smaller region bounded by the ellipse

straight x squared over 9 plus straight y squared over 4 space equals space 1

and the straight line

straight x over 3 plus straight y over 2 space equals space 1

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74.

Using integration, find the area of the smaller region bounded by the curve $straight x squared over 16 plus straight y squared over 9 space equals space 1$ and the straight line $straight x over 4 plus straight y over 3 equals 1.$

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75.

Find the area of smaller region bounded by the ellipse $straight x squared over straight a squared plus straight y squared over straight b squared space equals 1$ and the straight line $straight x over straight a plus straight y over straight b equals 1$

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76. Evaluate $integral subscript 0 superscript straight r square root of straight r squared minus straight x squared end root dx$ , where x is a fixed positive number, Hence, prove that the area of a circle of radius r is $straight pi space straight r squared.$

Let I = $integral subscript 0 superscript straight r square root of straight r squared minus straight x squared end root space dx$
$equals space open square brackets fraction numerator straight x square root of straight r squared minus straight x squared end root over denominator 2 end fraction plus straight r squared over 2. sin to the power of negative 1 end exponent open parentheses straight x over straight r close parentheses close square brackets subscript 0 superscript straight r equals space open square brackets fraction numerator straight r square root of straight r squared minus straight r squared end root over denominator 2 end fraction plus straight r squared over 2 sin to the power of negative 1 end exponent open parentheses straight r over straight r close parentheses close square brackets space minus space open square brackets 0 plus straight r squared over 2 sin to the power of negative 1 end exponent left parenthesis 0 right parenthesis close square brackets space equals space open square brackets 0 plus straight r squared over 2 sin to the power of negative 1 end exponent left parenthesis 1 right parenthesis close square brackets space minus space open square brackets 0 plus straight r squared over 2 sin to the power of negative 1 end exponent left parenthesis 0 right parenthesis close square brackets space equals space 0 plus straight r squared over 2 cross times straight pi over 2 minus 0 space equals space πr squared over 4$

Let $straight f left parenthesis straight x right parenthesis space equals space straight y space equals space square root of straight r squared minus straight x squared end root$

Now $straight y equals space square root of straight r squared minus straight x squared end root$ $rightwards double arrow space space space straight y squared space equals space straight r squared minus straight x squared$
$rightwards double arrow space space space space space straight x squared plus straight y squared space equals space straight r squared comma space space which space is space straight a space circle space at space the space origin space and space radius space straight r.$
$therefore space space space space integral subscript 0 superscript straight r square root of straight r squared minus straight x squared end root space space dx$ represents the area bounded by the circle $straight x squared plus straight y squared space equals space straight r squared comma$ the x-axis and the ordinate x = 0 and the ordinate x = r, i.e., it represents the area of the region in the first quadrant.
$therefore space space space space area space of space circle space equals space 4. space open parentheses πr squared over 4 close parentheses space equals space straight pi space straight r squared.$

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Short Answer Type

77.

Find the whole area of the circle x² + y² = a².

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Multiple Choice Questions

78. Area lying in the first quadrant and bounded by the circle x² + y² = 4 and the lines x = 0 and x = 2 is

$straight pi$
$straight pi over 2$
$straight pi over 3$
$straight pi over 3$

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79. Area of the region bounded by the curve y² = 4 x, y-axis and the line y = 3 is

2
$9 over 4$
$9 over 3$
$9 over 3$

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80. Smaller area enclosed by the circle x² + y² = 4 and the line x + y = 2 is
Choose the correct answer.
or
Draw the rough sketch and find the area of the region:
{(x, y): x² + y² < 4, x + y > 2}.