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Application of Integrals

Name: Zigya - For The Curious Learner
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Long Answer Type

11. Find the area of the region bounded by the curve

straight y space equals space 2 square root of 1 minus straight x squared end root

and x-axis. Draw a rough sketch..

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Short Answer Type

12. Calculate the area bounded by the parabola y² = 4 a x and its latus rectum.

824 Views

13.

Give the rough sketch of the curve y² = x and the line x = 4 and find the area between the curve and the line.

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14.

Find the area bounded between the curve y² = x and the line x = 3. Draw the rough sketch also.

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15. The area between x = y² and x = 4 is divided into two equal parts by the line x = a, find the value of a.

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16.

Find the area of the region bounded by the curve y = x² and the line y = 4.

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17.

Find the area of the region bounded by the curve y² = 4x and the line x = 3.

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18.
Using definite integrals, find the area of the circle x² + y² = a².

The equation of circle is
x² + y² = a² ...(1)
Its centre is a and radius a. We know that circle x² + y² = a² is symmetrical about both axes.
required area = $4 integral subscript 0 superscript straight a straight y space dx space equals space 4 space integral subscript 0 superscript straight a square root of straight a squared minus straight x squared end root dx$
$space equals space 4 space open square brackets fraction numerator straight x square root of straight a squared minus straight x squared end root over denominator 2 end fraction plus straight a squared over 2 sin to the power of negative 1 end exponent straight x over straight a close square brackets subscript 0 superscript straight a space equals space 4 open square brackets open parentheses 0 plus straight a squared over 2 sin to the power of negative 1 end exponent 1 close parentheses minus open parentheses 0 plus straight a squared over 2 sin to the power of negative 1 end exponent 0 close parentheses close square brackets space equals space 4 open square brackets straight a squared over 2 cross times straight pi over 2 close square brackets space space space left square bracket because space sin to the power of negative 1 end exponent 0 space equals space 0 right square bracket equals space straight pi space straight a squared.$