Volume of 3 intersecting cylinders in Cartesian coordinates

Question

I'm trying to find the volume of 3 intersecting cylinders:

$y^2 + z^2 = a^2$

$x^2 + z^2 = a^2$

$x^2 + y^2 = a^2$

However, I need to find the volumes using a triple integral in Cartesian coordinates, $dx,dy,dz$. I cannot use cylindrical coordinates.

Edit: I need to find what tu put in my 3 integrals, I already know the anwser.

Answer 1

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