The area enclosed by |4-x^2| ≤ y ≤ x^2 ; y ≤ 4, x ≤ 0 equals to (in square units)

Question

The area enclosed by \(\left|4-x^{2}\right| \leq y \leq x^{2} ; y \leq 4, x \leq 0\) equals to (in square units)

(1) \(\frac{4}{3}(20 \sqrt{2}-24)\)

(2) \(\frac{2}{3}(20 \sqrt{2}+24)\)

(3) \(\frac{4}{3}(10 \sqrt{2}+24)\)

(4) \(\frac{2}{3}(20 \sqrt{2}-24)\)

Answer 1

Correct option is: (4) \(\frac{2}{3}(20 \sqrt{2}-24)\) 

Shaded region is the required area.

Area \(=\int_\limits{0}^{2}(-\sqrt{4-y}+\sqrt{4+y}) d y+\int_\limits{2}^{4}(-\sqrt{y}+\sqrt{4+y}) d y\)

\(\left.\left.=\frac{2(4+y)^{3 / 2}}{3}+\frac{2(4-y)^{3 / 2}}{3}\right]_{0}^{2}+\frac{2(4+y)^{3 / 2}}{3}-\frac{2 y^{3 / 2}}{3}\right]_{0}^{4}\)

\(=\frac{2}{3}\left[\left(6^{3 / 2}+2^{3 / 2}\right)-(8+8)+\left(8^{3 / 2}-8\right)-\left(6^{3 / 2}-2^{3 / 2}\right]\right.\)

\(=\frac{2}{3}\left[2^{3 / 2}-16+8^{3 / 2}-8+2^{3 / 2}\right]\)

\(=\frac{2}{3}[4 \sqrt{2}+16 \sqrt{2}-24]\)

\(=\frac{2}{3}[20 \sqrt{2}-24]\) sq. unit