Two variations of the calculations with $a = 1$. You can generalize easily.
$$\eqalign{\frac{V}{16}
&= \int_0^{\sqrt2/2}\!\!\int_y^{\sqrt{1-y^2}}\!\!\sqrt{1-x^2}\,dxdy =
\int_0^{\sqrt2/2}\frac12\left(\arcsin x + x\,\sqrt{1-x^2}\right)\Bigg|_{x=y}^{x=\sqrt{1-y^2}}dy\cr
&= \int_0^{\sqrt2/2}\frac12\left(\arcsin\sqrt{1-y^2} + \sqrt{1-y^2}\,y - \arcsin y - y\,\sqrt{1-y^2}\right)dy\cr
&= \int_0^{\sqrt2/2}\frac12\left(\arccos y - \arcsin y\right)dy = \int_0^{\sqrt2/2}\!\!\left(\frac\pi4 - \arcsin y\right)dy\cr
&= \left(\frac{\pi y}4 -y\arcsin y - \sqrt{1-y^2}\right)\Bigg|_{y=0}^{y=\sqrt2/2} = 1 - \frac{\sqrt2}2.
}$$
$$\eqalign{\frac{V}{16}
&= \int_0^{\sqrt2/2}\!\!\int_0^x\!\sqrt{1-x^2}\,dydx\ + \int_{\sqrt2/2}^1\int_0^{\sqrt{1-x^2}}\!\!\sqrt{1-x^2}\,dydx\cr
&= \int_0^{\sqrt2/2}\!\!x\,\sqrt{1-x^2}\,dydx\ + \int_{\sqrt2/2}^1(1-x^2)\,dydx\cr
&= -\frac13(1-x)^{3/2}\Bigg|_{x=0}^{x=\sqrt2/2} +\ \left(x - \frac13 x^3\right)\Bigg|_{x=\sqrt2/2}^{x=1}\cr
& = \left(\frac13 - \frac{\sqrt2}{12}\right) - \left(\frac23 - \frac{5\sqrt2}{12}\right) = 1 - \frac{\sqrt2}2.
}$$