What is the relation? The function $x^2 = y\quad$ limits two areas $A$ and $B$: 
$A$ is further limited with the line $x= a$, $a\gt 0$.
$A$ rotates around the $x$-axis, which gives Volume $A = Va$.
$B$ is limited with the line $y=b$, $b\gt 0$.
$B$ rotates around the $y$-axis, which gives Volume $B = Vb$.
What are the relations between $a$ and $b$, when $Vb = Va$?
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I have come to the solution that: 
$Vb = (\pi b^2 )/ 2$
$Va = (\pi a^5) / 5$
so the relation between them is:
$2.5b^2 = a^5$ 
Is that the final solution or is it more? 
 A: If your calculations are correct this is what you should have found.
A: Without a drawing or a more detailed description, I cannot be certain.  But under the reasonable interpretation of what you wrote, your conclusion is absolutely correct.  Maybe, since $a$ and $b$ are positive, it might be slightly better to say that
$$b=a^2\sqrt{\frac{2a}{5}}$$
A: If we take the region bounded by the $y$-axis, the $x$-axis, the line $x=a$ (with $a\gt 0$), and the parabola $y=x^2$, and rotate it about the $x$-axis, the volume of the resulting solid of revolution is easily computed (using, for example, discs perpendicular to the $x$-axis) to be
$$\text{Volume A} = \int_0^a \pi(x^2)^2\,dx = \frac{\pi}{5}x^5\Bigm|_0^a = \frac{\pi a^5}{5}.$$
If the region bounded by the $y$-axis, the $x$ axis, the line $y=b$ (with $b\gt 0$), and the parabola $y=x^2$ is revolved around the $y$-axis, then using discs perpendicular to the $y$-axis we obtain the volume to be:
$$\text{Volume B} = \int_0^b \pi (\sqrt{y})^2\,dy = \frac{\pi}{2}y^2\Bigm|_0^b = \frac{\pi b^2}{2}.$$
So your computations are correct there.
If the two volumes are the same, then we must have
$$\text{Volume A} = \frac{\pi a^5}{5} = \frac{\pi b^2}{2} = \text{Volume B};$$
there are many ways to express this: you can solve for one of $a$ or $b$ in terms of the other:
$$b = \sqrt{\frac{2a^5}{5}} = a^{5/2}\sqrt{\frac{2}{5}},$$
or, if you want to express $a$ in terms of $b$ instead,
$$ a = \sqrt[5]{\frac{5}{2}b^2} = b^{2/5}\sqrt[5]{\frac{5}{2}}.$$
Or you can simply express this relation by saying, say
$$2a^5 = 5b^2.$$
Note. If $a\lt 0$, then the volume of $A$ can be computed the same way, but the integral would go from $a$ to $0$, so that the volume would be $-\frac{\pi a^5}{5}$; to account for both possibilities, both $a\gt 0$ and $a\lt 0$, you can simply write that the volume is $\frac{\pi|a|^5}{5}$. For solid $B$, however, it makes no sense to talk about $b\lt 0$, because then we don't have a finite area "enclosed" by the curves in question. 
