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Vector Algebra

Question
CBSEENMA12032807
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Evaluate integral subscript 0 superscript straight r square root of straight r squared minus straight x squared end root dx,  where x is a fixed positive number, Hence, prove that the area of a circle of radius r is straight pi space straight r squared.

Solution

Let I = integral subscript 0 superscript straight r square root of straight r squared minus straight x squared end root space dx
          equals space open square brackets fraction numerator straight x square root of straight r squared minus straight x squared end root over denominator 2 end fraction plus straight r squared over 2. sin to the power of negative 1 end exponent open parentheses straight x over straight r close parentheses close square brackets subscript 0 superscript straight r equals space open square brackets fraction numerator straight r square root of straight r squared minus straight r squared end root over denominator 2 end fraction plus straight r squared over 2 sin to the power of negative 1 end exponent open parentheses straight r over straight r close parentheses close square brackets space minus space open square brackets 0 plus straight r squared over 2 sin to the power of negative 1 end exponent left parenthesis 0 right parenthesis close square brackets space equals space open square brackets 0 plus straight r squared over 2 sin to the power of negative 1 end exponent left parenthesis 1 right parenthesis close square brackets space minus space open square brackets 0 plus straight r squared over 2 sin to the power of negative 1 end exponent left parenthesis 0 right parenthesis close square brackets space equals space 0 plus straight r squared over 2 cross times straight pi over 2 minus 0 space equals space πr squared over 4

Let straight f left parenthesis straight x right parenthesis space equals space straight y space equals space square root of straight r squared minus straight x squared end root
Now straight y equals space square root of straight r squared minus straight x squared end root   rightwards double arrow space space space straight y squared space equals space straight r squared minus straight x squared
rightwards double arrow space space space space space straight x squared plus straight y squared space equals space straight r squared comma space space which space is space straight a space circle space at space the space origin space and space radius space straight r.
therefore space space space space integral subscript 0 superscript straight r square root of straight r squared minus straight x squared end root space space dx  represents the area bounded  by the circle straight x squared plus straight y squared space equals space straight r squared comma the x-axis and the ordinate x = 0 and the ordinate x = r, i.e., it represents the area of the region in the first quadrant. 
therefore space space space space area space of space circle space equals space 4. space open parentheses πr squared over 4 close parentheses space equals space straight pi space straight r squared.

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