# Find the area between functions $x^3=2y^2;x=0;y=-2$

Take the element of area parallel to the y axis: $$x^3=2y^2;x=0;y=-2$$

First, I isolated in terms of $$y$$,

$$y= \pm \sqrt{x^3\over2} \\ = \pm{\sqrt2\over 2}x^{3 \over 2}$$

Since bounded by $$y=-2$$, consider the negative portion:

$$= -{\sqrt2\over 2}x^{3 \over 2}$$

and let

$$f(x)= -{\sqrt2\over 2}x^{3 \over 2}$$

$$g(x) = -2$$

since $$f(x)$$ and $$g(x)$$ intersect at $$x=2$$ and bound by $$x=0$$

So,

$$\int_0^2 [-g(x) -f(x)]dx \\ = \int_0^2 (2 + {\sqrt2\over 2}x^{3 \over 2} ) dx \\ = 2x + {\sqrt2\over 5}x^{5 \over 2}\Bigg]^2_0 \\ = 4 + {\sqrt2\over 5}2^{5 \over 2}$$

However the answer i supposed to be $$12\over 5$$

And I do not understand what the question means by "The element of area parallel to y axis"

The area should be $$\int^2_0[f(x)-g(x)]dx$$ as $$f(x)\ge g(x)$$ in the given interval.
So,$$\sqrt2\cdot2^{5/2} = 2^{1/2}2^{5/2} = 2^{3} = 8 \implies 4- \frac{8}{5} = \frac{12}{5}$$
$$f(x)=-\frac{\sqrt2}{2}x^{\frac{3}{2}}$$
When intersected with $$g(x)=-2$$ you get the point $$x=2$$ as you showed.
So the area between $$x=0$$ , $$y=-2$$ and $$f(x)$$ is the area of $$x=0$$, $$y=2$$ and $$-f(x)$$.
This area is the same as the area of $$x=0$$ , $$x=2$$ and $$y=2$$ minus the area of $$-f(x)$$ between $$0$$ and $$2$$.
So $$A=4-\int{f(x)dx}=4-\frac{8}{5}=\frac{12}{5}$$