Question

Area lying in the first quadrant and bounded by the circle x² + y² = 4 and the lines x = 0 and x = 2 is
$straight pi$
$straight pi over 2$
$straight pi over 3$
$straight pi over 4$

Solution

straight pi

The equation of circle is
$straight x squared plus straight y squared space equals space 4$ ...(1)
We are to find the area of the circle lying between the lines x = 0 and x = 2 in the first quadrant.
Required area = $integral subscript 0 superscript 2 straight y space dx$
$equals space integral subscript 0 superscript 2 square root of 4 minus straight x squared space end root dx$ $open square brackets because space space space of space left parenthesis 1 right parenthesis close square brackets$

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

For Daily Free Study Material Join wiredfaculty

Vector Algebra

Area lying in the first quadrant and bounded by the circle x² + y² = 4 and the lines x = 0 and x = 2 is
$straight pi$
$straight pi over 2$
$straight pi over 3$
$straight pi over 4$

Some More Questions From Vector Algebra Chapter

Mock Test Series

NCERT Sample Papers

Entrance Exams Preparation

Phone Number

Email Address

Loaction

For Daily Free Study Material Join wiredfaculty

Vector Algebra

Area lying in the first quadrant and bounded by the circle x2 + y2 = 4 and the lines x = 0 and x = 2 is

Some More Questions From Vector Algebra Chapter

Mock Test Series

NCERT Sample Papers

Entrance Exams Preparation

Area lying in the first quadrant and bounded by the circle x² + y² = 4 and the lines x = 0 and x = 2 is
$straight pi$
$straight pi over 2$
$straight pi over 3$
$straight pi over 4$