<s>[INST] 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"
 [/INST] The area should be 
$$\int^2_0[f(x)-g(x)]dx$$ as $f(x)\ge g(x)$ in the given interval.
So,$$\sqrt2\cdot2^{5/2} = 2^{1/2}2^{5/2} = 2^{3} = 8 \implies 4- \frac{8}{5} = \frac{12}{5}$$

</s>