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

The equation of ellipse is
$straight x squared over 9 plus straight y squared over 4 space equals space 1$ ...(1)
The equation of line is
$straight x over 3 plus straight y over 2 space equals space 1$ ...(2)
Let line (2) meet ellipse (1) in A (3, 0) and B(0, 2).
Area of shaded region = Area of region OAB - area of $increment OAB$

$equals space 2 over 3 integral subscript 0 superscript 3 square root of 9 minus straight x squared end root space dx minus space 2 over 3 integral subscript 0 superscript 3 left parenthesis 3 minus straight x right parenthesis space dx$ $open square brackets because space of space left parenthesis 1 right parenthesis comma space left parenthesis 2 right parenthesis close square brackets$
$equals space 2 over 3 open square brackets fraction numerator straight x square root of 9 minus straight x squared end root over denominator 2 end fraction plus 9 over 2 sin to the power of negative 1 end exponent straight x over 3 close square brackets subscript 0 superscript 3 space minus space 2 over 3 open square brackets fraction numerator left parenthesis 2 minus straight x right parenthesis squared over denominator left parenthesis negative 1 right parenthesis thin space left parenthesis 2 right parenthesis end fraction close square brackets subscript 0 superscript 3 equals space 2 over 3 open square brackets open parentheses fraction numerator 3 square root of 9 minus 9 end root over denominator 2 end fraction plus 9 over 2 sin to the power of negative 1 end exponent 1 close parentheses minus open parentheses 0 plus 9 over 2 sin to the power of negative 1 end exponent 0 close parentheses close square brackets plus 1 third left square bracket 0 minus 9 right square bracket space equals space 2 over 3 open square brackets open parentheses 0 plus 9 over 2 cross times straight pi over 2 close parentheses minus left parenthesis 0 plus 0 right parenthesis close square brackets plus 1 third left parenthesis 0 minus 9 right parenthesis equals space 2 over 3 cross times 9 over 2 cross times straight pi over 2 minus 1 third space equals space fraction numerator 3 straight pi over denominator 2 end fraction minus 3 equals space fraction numerator 3 space straight pi minus 6 over denominator 2 end fraction space equals 3 over 2 left parenthesis straight pi minus 2 right parenthesis space sq. space units.$

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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.

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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}.