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Application of Integrals

Name: Zigya - For The Curious Learner
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Long Answer Type

91.

Using integration, find the area of the region enclosed between the two circles:
$straight x squared plus straight y squared space equals space 4 space and space left parenthesis straight x minus 2 right parenthesis squared plus straight y squared space equals space 4.$

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92.
Find the area bounded by the circle x² + y² = 16 and the line √3y=x in the first quadrant, using integration.

The area of the region bounded by the circle, x²+y²=16, x=√3y, and the x-axis is the area OAB.

Solving x²+y²=16 and x=√3y, we have

(√3y)²+y²=16
⇒3y²+y²=16
⇒4y²=16
⇒y²=4
⇒y=2
(In the first quadrant, y is positive)

When y = 2, x = 2√3

So, the point of intersection of the given line and circle in the first quadrant is (2√3,2).

The graph of the given line and circle is shown below:

Required area = Area of the shaded region = Area OABO = Area OCAO + Area ACB
Area OCAO = 12×2√3×2=2√3 sq units
$Area space ABC space equals space integral subscript 2 square root of 3 end subscript superscript 4 space straight y space dx space equals space integral subscript 2 square root of 3 end subscript superscript 4 square root of 16 minus straight x squared dx end root space equals space open square brackets straight x over 2 square root of 16 minus straight x squared end root space plus 16 over 2 space sin to the power of negative 1 end exponent space straight x over 4 close square brackets subscript 2 square root of 3 end subscript superscript 4 space equals space open square brackets open parentheses 0 plus space 8 space sin to the power of negative 1 end exponent 1 close parentheses minus open parentheses fraction numerator 2 square root of 3 over denominator 2 end fraction space straight x space 2 space plus 8 space straight x space sin to the power of negative 1 end exponent fraction numerator square root of 3 over denominator 2 end fraction close parentheses close square brackets space equals space 8 space straight x space straight pi over 2 minus 2 square root of 3 space minus 8 space straight x space straight pi over 3 space equals open parentheses fraction numerator 4 straight pi over denominator 3 end fraction minus 2 square root of 3 space close parentheses space sq space unit therefore space Required space area space equals space open parentheses fraction numerator 4 straight pi over denominator 3 end fraction minus 2 square root of 3 close parentheses space plus 2 square root of 3 space equals space fraction numerator 4 straight pi over denominator 3 end fraction space sq space units$

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93.

Using integration, find the area of region bounded by the triangle whose vertices are (–2, 1), (0, 4) and (2, 3).

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94.

Using integration, find the area of the region in the first quadrant enclosed by the x-axis, the line y = x and the circle x² + y² =32

95.

Using integration find the area of the region bounded by the parabola y² = 4x and the circle 4x² + 4y² = 9.

96.

Prove that the curves y²= 4x and x²= 4y divide the area of the square bonded by x = 0, x = 4, y = 4, and y = 0 into three equal parts.

97.

Using integration, find the area of the following region:

$\{(x, y) : \frac{x^{2}}{9} + \frac{y^{2}}{4} \leq 1 \leq \frac{x}{3} + \frac{y}{2}\}$

98.

Using integration find the area of the triangular region whose sides have equations y=2x+1, y=3x+1 and x=4.

99.

Using the method of method of integration, find the area of the region bounded by the following lines:

3x – y – 3 = 0,

2x + y – 12 = 0,

x – 2y – 1 = 0

Multiple Choice Questions

100.

The area (in sq. units) enclosed between the curves y = x² and y = $|x|$ is

$\frac{2}{3}$
$\frac{1}{6}$
$\frac{1}{3}$
1