Hi -
I attach a drawing of the problem as I understand it, with three axes: $\displaystyle Ox$, $\displaystyle Oy$ and $\displaystyle Oz$.
The values of $\displaystyle x$ go from $\displaystyle 0$ to $\displaystyle 10$. At $\displaystyle x = a$, the parabola (which lies in a plane parallel to the $\displaystyle y-z$ plane) has equation $\displaystyle z=a-y^2$. This parabola will have values of $\displaystyle y$ from $\displaystyle -\sqrt a$ to $\displaystyle \sqrt a$, and values of $\displaystyle z$ from $\displaystyle 0$ to $\displaystyle a$.
What you need to do, then, to find the volume of the solid enclosed by all these parabolas, the $\displaystyle x-y $ plane and the plane $\displaystyle x = 10$, is:
- Find the area $\displaystyle A$ of the typical parabola shown. (Do this in the usual way with an integral, whose limits are $\displaystyle y=-\sqrt a$ to $\displaystyle y=\sqrt a$.)
- Replace $\displaystyle a$ by $\displaystyle x$ in your formula for $\displaystyle A$.
- Now imagine $\displaystyle x$ increasing by an amount $\displaystyle \delta x$. As it does so, the volume 'swept out' is approximately $\displaystyle A\delta x$. So the total volume will be $\displaystyle \int_0^{10} A dx$. So, replace $\displaystyle A$ by your formula in terms of $\displaystyle x$, and then work out the integral.
Have I given you enough to go on? Let me know if you want me to check your working.
Grandad