Find the volume of the solid restricted to $x^2+y^2+z^2=1$ and $x^2+(y-1)^2+z^2=1$
I think the volume is given by $$\int_{0}^{\pi}\int_{0}^{\pi}\int_{0}^{1}\rho^{2}\sin\left(\varphi\right)d\rho d\theta d \varphi$$
Is that right? What if I wanted to use Cartesian coordinates? I would say it's something like $$\iint_{D}^{ }\int_{\sqrt{1-x^{2}-z^{2}}+1}^{\sqrt{1-x^{2}-z^{2}}}dydzdx$$ Though I can't find $D$.
Evaluate the volume as follows
\begin{align} V &= \iint_{x^2+z^2<\frac34} \int_{1-\sqrt{1-x^{2}-z^{2}}}^{\sqrt{1-x^{2}-z^{2}}}dy \> dxdz\\ &=\iint_{x^2+z^2<\frac34} \left(2\sqrt{1-x^{2}-z^{2}}-1\right)\> dxdz\\ &=2\pi\int_0^{\sqrt3/2} \left(2\sqrt{1-r^2}-1\right)\> rdr \\ &= \frac{5\pi}{12} \end{align}