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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.

Given region is
$open curly brackets left parenthesis straight x comma space straight y right parenthesis space colon space straight y squared less or equal than 3 straight x comma space 3 straight x squared plus 3 straight y squared less or equal than 16 close curly brackets$
Consider the equations
   $straight y squared equals 3 space straight x comma space$ ...(1)
   $3 straight x squared plus 3 straight y squared space equals space 16$
i.e.,    $straight x squared plus straight y squared space equals space 16 over 3$ ...(2)
From (1) and (2), we get,
$straight x squared plus 3 straight x space equals space 16 over 3 space space space or space space 3 straight x squared plus 9 straight x minus 16 space equals space 0$
$therefore space space straight x space equals space fraction numerator negative 9 plus-or-minus square root of 81 plus 192 end root over denominator 6 end fraction space equals fraction numerator negative 9 plus-or-minus square root of 273 over denominator 6 end fraction$
Rejecting negative values of x, we get
$straight x equals fraction numerator negative 9 plus square root of 273 over denominator 6 end fraction equals space straight a$

which gives the abscissa of the points of intersection P and Q.
Required area = Area of shaded region
= 2 [area of region APOA]
= 2 [area of region AMPA + area of region MOPM ]
= $2 open square brackets integral subscript 0 superscript straight a square root of 3 space square root of straight x dx plus space integral subscript straight a superscript 4 divided by square root of 3 end superscript square root of 16 over 3 minus straight x squared end root space dx close square brackets$
= $2 square root of 3 open square brackets fraction numerator straight x to the power of 3 divided by 2 end exponent over denominator begin display style 3 over 2 end style end fraction close square brackets subscript 0 superscript straight a plus 2 open square brackets fraction numerator straight x square root of begin display style 16 over 3 end style minus straight x squared end root over denominator 2 end fraction plus fraction numerator begin display style 16 over 3 end style over denominator 2 end fraction sin to the power of negative 1 end exponent open parentheses fraction numerator straight x over denominator 4 divided by square root of 3 end fraction close parentheses close square brackets subscript 0 superscript fraction numerator 4 over denominator square root of 3 end fraction end superscript$
$equals space fraction numerator 4 over denominator square root of 3 end fraction left parenthesis straight a to the power of 3 divided by 2 end exponent minus 0 right parenthesis space plus space open square brackets open curly brackets 0 plus 16 over 3 sin to the power of negative 1 end exponent left parenthesis 1 right parenthesis close curly brackets minus open curly brackets straight a square root of 16 over 3 minus straight a squared end root plus 16 over 3 sin to the power of negative 1 end exponent open parentheses fraction numerator straight a square root of 3 over denominator 4 end fraction close parentheses close curly brackets close square brackets equals space fraction numerator 4 over denominator square root of 3 end fraction straight a to the power of 3 divided by 2 end exponent plus 8 over 3 straight pi minus straight a square root of 16 over 3 minus straight a squared end root minus 16 over 3 sin to the power of negative 1 end exponent open parentheses fraction numerator straight a square root of 3 over denominator 4 end fraction close parentheses$

111 Views

72.

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

135 Views

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

386 Views

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 $straight x over 4 plus straight y over 3 equals 1.$

138 Views

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 $straight x over straight a plus straight y over straight b equals 1$

110 Views

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.

97 Views

Short Answer Type

77.

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

350 Views

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

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

221 Views

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

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

107 Views

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}.