Question

the area of region bounded by curve y = (16-x²)/4 and y = sec⁻¹[-sin²x]​

Answer 1

Question:-

find the area of region bounded by curve $\sf y = \dfrac{16-x^2}{4}$ and $\sf y = sec^{-1} [ - sin^2 x]$

Answer:-

$\sf 0 \leqslant sin^2 x \leqslant 1$

$\sf \implies -1 \leqslant - sin^2 x \leqslant 0$

$\sf \therefore [-sin^2 x] = 0 \ or \ -1$

let $\sf sec^{-1}$ is not defined

hence, $\sf y = sec^{-1} [-sin^2x] = sec^{-1}(-1) = \pi$

now, $\sf \pi = \dfrac{16-x^2}{4}$

$\sf \implies x^2 = 16-4\pi = 4(4-\pi)$

$\sf \implies x = \pm \:2 \sqrt{4-\pi}$

Area $\displaystyle\sf = \int\limits_{-2\sqrt{4-x}}^{2\sqrt{4-x}} \left( \dfrac{16-x^2}{4} - \pi \right) dx$

$\boxed{\sf = \dfrac{8}{3} (4-\pi)^{3/2} \ sq.\ units}$