Question

The area enclosed by the curve $y^2+x^4=x^2$ is:

• A. $\frac{2}{3}$
• B. $\frac{4}{3}$
• C. $\frac{8}{3}$
• D. $\frac{10}{3}$

Answer 1

The correct option is B $\frac{4}{3}$

$y=\pm x\sqrt{1-x^2}$
Clearly given cirve is symmetrical about both x and y axis.
Hence required area is,
A = $4{\int }_0^{1}x\sqrt{1-x^2}dx$
$=-2{\left[\frac{{(1-x^2)}^{3/2}}{3/2}\right]}_0^1=\frac{4}{3}$