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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 $straight x equals fraction numerator straight a over denominator square root of 2 end fraction$

The equation of circle is
$straight x squared minus straight y squared space equals straight a squared$ ...(1)
The equation of line is
$straight x space equals space fraction numerator straight a over denominator square root of 2 end fraction$ ...(2)
From (1) and (2), we get,
$straight a squared over 2 minus straight y squared space equals space straight a squared space space space space or space space space straight y squared space equals space straight a squared over 2$
$therefore space space space space space straight y space equals space fraction numerator straight a over denominator square root of 2 end fraction$
$because space space space straight C$ is $open parentheses fraction numerator straight a over denominator square root of 2 end fraction comma space fraction numerator straight a over denominator square root of 2 end fraction close parentheses$

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

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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.

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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}.