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Application of Integrals

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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.

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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.

The point of intersection of the

Parabolas y² = 4x and x² = 4y are ( 0, 0 ) and ( 4, 4 )

Now the area of the region OAQBO bounded by curves y² = 4x and x² = 4y,

$\int_{0}^{4} (2 \sqrt{x} - \frac{x^{2}}{4}) dx = {[2 \frac{x^{\frac{3}{2}}}{\frac{3}{2}} - \frac{x^{3}}{12}]}_{0}^{4} = \frac{32}{3} - \frac{16}{3} = \frac{16}{3} sq . units . . . . . . . . . . . . . . (i)$

Again, the area of the region OPQAO bounded by the curve x² = 4y , x = 0,

x = 4 and the x - axis,

$\int_{0}^{4} \frac{x^{2}}{4} dx = {[\frac{x^{3}}{12}]}_{0}^{4} = (\frac{64}{12}) = \frac{16}{3} sq . units . . . . . . . . . . . . . . (ii)$

Similarly, the area of the region OBQRO bounded by the curve y² = 4x, the y-axis, y = 0 and y = 4

$\int_{0}^{4} \frac{y^{2}}{4} dy = {[\frac{y^{3}}{12}]}_{0}^{4} = \frac{16}{3} sq . units . . . . . . . . . . . . (iii)$

From (i), (ii), and (iii) it is concluded that the area of the region OAQBO =

area of the region OPQAO = area of the region OBQRO, i. e., area bounded

by parabolas y² = 4x and x² = 4y divides the area of the square 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