Area lying in the first quadrant between the curves x2+y2=π 2 and y=sinx , is equal to

Question

Area lying in the first quadrant between the curves x2+y2=π 2 and y=sinx , is equal to (A) (((π )2 - 8)/2) (B) (((π )3 - 8)/3) (C) (((π )2 - 8)/4) (D) (((π )3 - 8)/4)

Tardigrade Admin · Accepted Answer

Area the circle in 1st quadrate. =π .(π 2/4) Area of one loop of sinx=2 (π 3/4)-2=(π 3 - 8/4) <img class="img-fluid question-image" alt="Solution" src="https://cdn.tardigrade.in/q/nta/m-9zvoj88pvpim4mga.jpg" />

Q. Area lying in the first quadrant between the curves $x^{2} + y^{2} = π^{2}$ and $y = s in x$ , is equal to

$\frac{( ( π ) ^{2} - 8 )}{2}$

$\frac{( ( π ) ^{3} - 8 )}{3}$

$\frac{( ( π ) ^{2} - 8 )}{4}$

$\frac{( ( π ) ^{3} - 8 )}{4}$

Area the circle in $1^{s t}$ quadrate. $= π . \frac{π ^{2}}{4}$
Area of one loop of $s in x = 2$
$\frac{π ^{3}}{4} - 2 = \frac{π ^{3} - 8}{4}$

Q. Area lying in the first quadrant between the curves x2+y2=π2 and y=sinx , is equal to

2((π)2−8)​

3((π)3−8)​

4((π)2−8)​

4((π)3−8)​