# Question #c5bfe

##### 1 Answer
Dec 4, 2016

$\frac{10}{3}$

#### Explanation:

Note that the leftmost point of the area is at the intersection of $x = - y$ and $x = 2 - {y}^{2}$ at $\left(- 2 , 2\right)$ and the rightmost is at the intersection of $x = 2 - {y}^{2}$ and the $x$-axis at $\left(2 , 0\right)$. Thus, we will consider integrals as $x$ goes from $- 2$ to $2$.

As the bounds of the area change at $x = 0$, we will split this problem into two integrals.

From $x = - 2$ to $x = 0$, the area is bounded above by $x = 2 - {y}^{2}$ and below by $x = - y$.

Solving for $y$ gives us the equations $y = \sqrt{2 - x}$ and $y = - x$ (note that we want the positive square root for our upper bound). Thus, we can represent this part of the area by the integral

${\int}_{- 2}^{0} \left(\sqrt{2 - x} - \left(- x\right)\right) \mathrm{dx} = {\int}_{- 2}^{0} \left(\sqrt{2 - x} + x\right) \mathrm{dx}$

$= {\left[- \frac{2}{3} {\left(2 - x\right)}^{\frac{3}{2}} + {x}^{2} / 2\right]}_{- 2}^{0}$

$= \left(- \frac{4 \sqrt{2}}{3} + 0\right) - \left(- \frac{16}{3} + 2\right)$

$= \frac{10 - 4 \sqrt{2}}{3}$

From $x = 0$ to $x = 2$, the area is still bounded above by $x = 2 - {y}^{2}$, but below by $y = 0$.

Once again rewriting $x = 2 - {y}^{2}$ as $y = \sqrt{2 - x}$, this gives us the integral

${\int}_{0}^{2} \left(\sqrt{2 - x} - 0\right) \mathrm{dx} = {\int}_{0}^{2} \sqrt{2 - x} \mathrm{dx}$

$= {\left[- \frac{2}{3} {\left(2 - x\right)}^{\frac{3}{2}}\right]}_{0}^{2}$

$= 0 - \left(- \frac{4 \sqrt{2}}{3}\right)$

$= \frac{4 \sqrt{2}}{3}$

Adding the two areas together, we get our total area:

$\frac{10 - 4 \sqrt{2}}{3} + \frac{4 \sqrt{2}}{3} = \frac{10}{3}$