Find the area of the region enclosed by the parabola y2 = 4 a x

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 Multiple Choice QuestionsLong Answer Type

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31. Find the area of the region enclosed by the parabola y2 = 4 a x and the line y = mx.


The equation of parabola is y2 = 4 a x    ... (1)
The line y = mx meets (1) where straight m squared straight x squared space equals space 4 space straight a space straight x
or
   straight x space left parenthesis straight m squared straight x space minus space 4 straight a right parenthesis space equals space 0 space space or space straight x space equals space 0 comma space space fraction numerator 4 straight a over denominator straight m squared end fraction
therefore space space straight y space equals space 0 comma space space fraction numerator 4 straight a over denominator straight m end fraction
rightwards double arrow space space space straight O space is space left parenthesis 0 comma space 0 right parenthesis space space and space straight A comma space is space open parentheses fraction numerator 4 straight a over denominator straight m squared end fraction comma space fraction numerator 4 straight a over denominator straight m end fraction close parentheses
Area bounded by curve (1) above the x - axis ordinates straight x space equals space 0 comma space space straight x space equals space fraction numerator 4 straight a over denominator straight m squared end fraction is
equals space integral subscript 0 superscript 4 straight a divided by straight m squared end superscript straight y space dx space equals space integral subscript 0 superscript 4 straight a divided by straight m squared end superscript square root of straight x space dx space equals 2 square root of straight a open square brackets fraction numerator straight x to the power of begin display style 3 over 2 end style end exponent over denominator begin display style 3 over 2 end style end fraction close square brackets subscript 0 superscript fraction numerator 4 straight a over denominator straight m squared end fraction end superscript
equals space fraction numerator 4 square root of straight a over denominator 3 end fraction open square brackets open parentheses fraction numerator 4 straight a over denominator straight m squared end fraction close parentheses to the power of 3 over 2 end exponent minus 0 close square brackets space space equals space fraction numerator 4 square root of straight a over denominator 3 end fraction cross times fraction numerator 8 straight a to the power of begin display style 3 over 2 end style end exponent over denominator straight m cubed end fraction space equals space 32 over 3 straight a squared over straight m cubed

Area bounded by line y = m x, above the x-axis and ordinates x = 0,  straight x space equals space fraction numerator 4 straight a over denominator straight m squared end fraction space is
                     equals space integral subscript 0 superscript 4 straight a divided by straight m squared end superscript mx space dx space equals space straight m open square brackets straight x squared over 2 close square brackets subscript 0 superscript fraction numerator 4 straight a over denominator straight m squared end fraction end superscript space equals space straight m over 2 open square brackets straight x squared close square brackets subscript 0 superscript fraction numerator 4 straight a over denominator straight m squared end fraction end superscript
equals space straight m over 2 open square brackets fraction numerator 16 straight a squared over denominator straight m to the power of 4 end fraction minus 0 close square brackets space equals space straight m over 2 cross times fraction numerator 16 straight a squared over denominator straight m to the power of 4 end fraction space equals space fraction numerator 8 straight a squared over denominator straight m cubed end fraction space sq. space units
therefore space space space required space area space equals space fraction numerator 32 space straight a squared over denominator 3 space straight m cubed end fraction minus fraction numerator 8 space straight a squared over denominator straight m cubed end fraction space equals space fraction numerator 32 straight a squared minus 24 straight a squared over denominator 3 straight m cubed end fraction space equals space fraction numerator 8 straight a squared over denominator 3 straight m cubed end fraction space sq. space units.

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32.

Find the area bounded by the curve  y = x2 and the line y = x.
OR
Find the area of the region {(x. y): x2 ≤ y ≤ x}.

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33. Using the method of integration find the area bounded by the curve |x| + |y| = 1.
[Hint: The required region is bounded by lines x + y = 1, x– y = 1, – x + y = 1 and
– x – y = 1].


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34.

Find the area of the region bounded by the line y = 3 x + 2, the x-axis and the ordinates x = - 1 and x = 1.

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35. Find the area of the region included between the parabola straight y space equals 3 over 4 straight x squared space and space the space line space 3 straight x space minus space 2 straight y space plus space 12 space equals space 0
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36. Find the area bounded by the parabola x2 = 4 y and the straight line x = 4 y - 2.
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37. Find the area of the region included between the parabola y2 = x and the line x + y = 2.
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38.

Find the area of the region enclosed by the parabola x2 = y, the line y = x + 2 and the x-axis.
OR
Draw the rough sketch and find the area of the region:
{(x, y): x2 < y < x + 2}

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 Multiple Choice QuestionsShort Answer Type

39.

Draw a rough sketch of the curves y = sin x and y = cos x as x varies from 0 to straight pi over 2 and find the area of the region enclosed by them and the x-axis.

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40.

Find the area bounded by the curve y = cos x between x = 0 and x = 2 straight pi.

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