Answer
$\displaystyle \frac{12}{5}$
Work Step by Step
Rewrite the equations as functions of y:
$\left[\begin{array}{lll}
x-y^{2/3}=0 & ... & x+y^{4}=2\\
x=y^{2/3} & & x=2-y^{4}\\
& &
\end{array}\right]$
Graphing the given equations, we find the intersections
or we solve the equation $f(y)=g(y), $ where
$ f(y)=2-y^{4} \quad$ (the curve on the right side) and
$ g(y)=y^{2/3}\quad$ (the curve on the left).
When $y\in[c,d]=[-1,1]$, the area between the graphs is
$A=\displaystyle \int_{c}^{d} [f(y)-g(y)]dy=\displaystyle \int_{-1}^{1}[2-y^{4}-y^{2/3}]dy $
$=\displaystyle \left[2y-\frac{y^{5}}{5}-\frac{y^{5/3}}{5/3} \right]_{-1}^{1}$
$=2(1+1)-\displaystyle \frac{(1+1)}{5}-\frac{3(1+1)}{5}$
$=4-\displaystyle \frac{8}{5}$
$=\displaystyle \frac{12}{5}$
