Question

The area (in square units) bounded by the curves $x=-2y^2$ and $x=1-3y^2$ is

• A. $\frac{2}{3}$
• B. $1$
• C. $\frac{4}{3}$
• D. $\frac{5}{3}$

Answer 1

Answer: (C)

$\frac{4}{3}$

$x=-2y^2$

$x=1-3y^2$

$-2y^2=1-3y^2\Rightarrow y^2=1\Rightarrow y=\pm 1$

Area between the curves $=A_1-A_2$

$={\int }_{-1}^1(1-3y^2+2y^2)dy$

$={\int }_{-1}^1(1-y^2)dy$

$=[y-\frac{y^3}{3}{]}_{-1}^1=\frac{4}{3}$