Area under the curve $y= -(x+1)^2$ and lines $y=y_1$ and $x=x_1$

Question

Find the area bounded by region $x= x_1$ , $y= y_1$ and $y= -(x+1)^2 $ where $x_1$ and $y_1$ are the values of $x , y$ from the equation $\sin^{-1}(x) + \sin^{-1}(y) = -\pi$ will be (nearer to origin).refer image

I have got the value of $x = y = -1$ by solving the equation $\sin^{-1}(x) + \sin^{-1}(y) = -\pi$

Then integrating gives $-\frac13$ , but answer seems to be $\frac23$ what I am missing?

I have also doubt that should i consider area above the curve or below the curve.

Answer 1

You should consider the area below the curve $y = -(x+1)^2$ which is $(1- \text{Area above the curve} )= 1-\frac{1}{3}=\frac{2}{3}$

Area above the curve $= \int_{-1}^{0} (-y)dx = \int_{-1}^{0} -(-(1+x)^2)dx = \int_{0}^{1} u^2du =\frac{1}{3}$ where $u = (1+x)$

Answer 2

You've actually just found the area shown below

The area you want is given by ;

$A = \int_{-1}^0(-1+(x+1)^2)\,dx$

$A = -x +\frac{(x+1)^3}{3}\bigg|_{-1}^0$

$A = \frac13 - 1$

$A =-\frac23$

The area is negative as the curve is below the X axis . The area is $|-\frac23|= \frac23 $sq. units