Using integration, find the area of the region enclosed between

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 Multiple Choice QuestionsLong Answer Type

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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.


Given equations of the circles are:
straight x squared plus straight y squared space equals space 4 space space space... left parenthesis 1 right parenthesis
left parenthesis straight x minus 2 right parenthesis squared plus straight y squared space equals space 4 space... left parenthesis 2 right parenthesis
Equation (1) is a circle with centre O at the origin and radius 2. Equation (2) is a circle with centre C(2, 0) and radius 2.
Solving (1) and (2), we have:
left parenthesis straight x minus 2 right parenthesis squared plus straight y squared space equals space straight x squared plus straight y squared
straight x squared minus 4 straight x plus 4 plus straight y squared space equals space straight x squared plus straight y squared
straight x space equals space 1
This gives straight y equals plus-or-minus square root of 3
Thus, the points of intersection of the given circles are straight A open parentheses 1 comma space square root of 3 close parentheses space and space straight A apostrophe left parenthesis 1 comma space minus square root of 3 right parenthesis space as space shown space in space the space figure. space

Required area
 = Area of the region OACA'O
 = 2[area of the region ODCAO] 
 =2[area of the region ODAO + area of the region DCAD]
equals 2 open square brackets integral subscript 0 superscript 1 ydx plus integral subscript 1 superscript 2 ydx close square brackets
equals 2 open square brackets integral subscript 0 superscript 1 square root of 4 minus left parenthesis straight x minus 2 right parenthesis squared end root dx plus integral subscript 1 superscript 2 square root of 4 minus straight x squared end root dx close square brackets
equals 2 open square brackets 1 half left parenthesis straight x minus 2 right parenthesis square root of 4 minus left parenthesis straight x minus 2 right parenthesis squared end root plus 1 half cross times 4 sin to the power of negative 1 end exponent open parentheses fraction numerator straight x minus 2 over denominator 2 end fraction close parentheses close square brackets subscript 0 superscript 1 plus 2 open square brackets 1 half straight x square root of 4 minus straight x squared end root plus 1 half cross times 4 sin to the power of negative 1 end exponent straight x over 2 close square brackets subscript 1 superscript 2
equals open square brackets left parenthesis straight x minus 2 right parenthesis square root of 4 minus left parenthesis straight x minus 2 right parenthesis squared end root plus 4 sin to the power of negative 1 end exponent open parentheses fraction numerator straight x minus 2 over denominator 2 end fraction close parentheses close square brackets subscript 0 superscript 1 plus open square brackets straight x square root of 4 minus straight x squared end root plus 4 sin to the power of negative 1 end exponent straight x over 2 close square brackets subscript 1 superscript 2
equals open square brackets negative square root of 3 plus 4 sin to the power of negative 1 end exponent open parentheses fraction numerator negative 1 over denominator 2 end fraction close parentheses minus 4 sin to the power of negative 1 end exponent left parenthesis negative 1 right parenthesis close square brackets plus open square brackets 4 sin to the power of negative 1 end exponent 1 minus square root of 3 minus 4 sin to the power of negative 1 end exponent 1 half close square brackets
equals open square brackets open parentheses negative square root of 3 minus 4 cross times straight pi over 6 close parentheses plus 4 cross times straight pi over 2 close square brackets plus open square brackets 4 cross times straight pi over 2 minus square root of 3 minus 4 cross times straight pi over 6 close square brackets
equals fraction numerator 8 straight pi over denominator 3 end fraction minus 2 square root of 3


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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


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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 :  x29 + y24  1  x3 + y2 


98.

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


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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 QuestionsMultiple Choice Questions

100.

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

  • 23

  • 16

  • 13

  • 1


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