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**shawsend** That's just a double integral. You're integrating from the x-y plane up to the surface that is $\displaystyle z=f(x,y)=y^2$. That's the z-direction. And we're integrating that over the foot-print in the x-y plane that's just the two unit squares underneath the surface which because it's symmetric, just multiply by two, the volume underneath the unit square as x goes from 0 to 1 and y goes from 0 to one or:

$\displaystyle V=2\int_0^1\int_0^1 y^2 dydx$

or if you wanna' be a purist, a triple integral:

$\displaystyle V=2\int_0^1\int_0^1\int_0^{y^2} dzdydx$

but same dif.