Answer
The area of the region is $12/5$.
Work Step by Step
$x-y^{2/3}=0$ and $x+y^4=2$
- For $x-y^{2/3}=0$, we can rewrite $$x=y^{2/3}$$
- For $x+y^4=2$, we can rewrite $$x=2-y^4$$
1) Draft the graph:
The graph is drafted below. We would use integration with respect to y here.
2) Find the limits of integration:
To find the limits of integration, we find intersection points with respect to y.
$$y^{2/3}=2-y^4$$ $$y^{2/3}+y^4-2=0$$ $$y=\pm1$$
So the upper limit is $1$ and the lower limit is $-1$.
3) Find the area:
The region from $y=-1$ to $y=1$ is bounded on the right by $x=2-y^4$ and on the left by $x=y^{2/3}$. Therefore, the area of the region is
$$A=\int^1_{-1}[2-y^4-y^{2/3}]dy$$ $$A=2y-\frac{y^5}{5}-\frac{3y^{5/3}}{5}\Big]^1_{-1}$$ $$A=\Big(2-\frac{1}{5}-\frac{3}{5}\Big)-\Big(-2+\frac{1}{5}+\frac{3}{5}\Big)$$ $$A=2\Big(2-\frac{1}{5}-\frac{3}{5}\Big)$$ $$A=2\times\frac{6}{5}=\frac{12}{5}$$
