Find volume using double integral but symmetry to xy plane

Question

$V=\{(x,y,z)\in R^3\mid x^2+y^2\le1, x^2+z^2\le1\} $ find the volume
Here i was confused why i need to multiply the double integral with 2? It is said that because it is symmetry to $xy$ plane? But i dont understand when i find problem like this whether i have to multiply it by 2 or not? Should i draw? And also it is a bit hard to draw in $xyz$ plane, is there another way to know whether i should multiply by two or not? $z=\sqrt{1-x^2}$ $2 \int \int_D \sqrt{1-x^2}\; dx \; dy$, where i need to multiply it by two???

Answer 1

Hint: First draw what the first condition says: $x^2+y^2 \leq 1$ is a disk, and $z$ is arbitrary, so this is a cylinder that runs along the $z$-axis, whereas the latter condition gives a cylinder along the $y$ axis. The two conditions together then provide the volume of two intersected cylinders.

There is a vizualization here

The solid is known as the steinmetz solid.

One can re-write the coniditions as saying that $x^2+y^2 \leq 1$ while $-\sqrt{1-x^2} \leq z \leq \sqrt{1-x^2}$ so we get

$$\int_{-1}^{1} \int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}}\int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}}1 dz \, dy\,dx=2 \int_{-1}^{1} \int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}}\sqrt{1-x^2} dy\,dx=4\int_{-1}^{1}1-x^2 dx.$$