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 $x=0$


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

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,$$\sqrt2\cdot2^{5/2} = 2^{1/2}2^{5/2} = 2^{3} = 8 \implies 4- \frac{8}{5} = \frac{12}{5}$$

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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 $A=4-\int{f(x)dx}=4-\frac{8}{5}=\frac{12}{5}$


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