JEE PYQ: Area Under Curves Question 6
Question 6 - 2021 (24 Feb Shift 1)
The area (in sq. units) of the part of the circle $x^2 + y^2 = 36$, which is outside the parabola $y^2 = 9x$, is:
(1) $24\pi + 3\sqrt{3}$
(2) $12\pi + 3\sqrt{3}$
(3) $12\pi - 3\sqrt{3}$
(4) $24\pi - 3\sqrt{3}$
Show Answer
Answer: (4) $24\pi - 3\sqrt{3}$
Solution
Curves intersect at $(3, \pm 3\sqrt{3})$. Required area $= 36\pi - 2\left[\int_0^3 \sqrt{9x},dx + \int_3^6 \sqrt{36-x^2},dx\right] = 24\pi - 3\sqrt{3}$.
BETA
