find the area of the shaded region $x=y^2-1, y=1, x=\sqrt y$

Question

find the area of the shaded region $$x=y^2-1, y=1, x=\sqrt y$$ The shaded region contains the point $(0,1)$

$$y=\sqrt{x+1}, y=x^2, y=1$$

$$\int _{-1}^{1}\left(\sqrt{x+1}-1-x^2\right)dx$$

I'm getting something different than the book

Answer 1

Hint:

$$\int_0^1 \sqrt{y}-(y^2-1) \, dy$$

Answer 2

First, notice there's a fourth boundary in the picture you forgot to mention: $y=0$. If you really wanted to integrate over $x$, you would need to split up the integral. Left of the $y$-axis, the top curve is $y=\sqrt{x+1}$ and the bottom curve is $y=0$. Right of the $y$-axis, you have $y=1$ on top and $y=x^2$ on the bottom. Therefore, your integral would be

$$\int_{-1}^0\sqrt{x+1}dx+\int_0^1(1-x^2)dx$$

Looks like the first integral is $\frac23$ while the second is $1-\frac13$. Add those and you have your book's answer. The problem looks like it was made to be integrated over $y$ though, especially since you have $x$ given as a function of $y$.