Find the volume between the regions $x^2 + y^2+ z^2 = 4$ and $x = 4-y^2$ I want the volume of the sphere $x^2 + y^2 + z^2 = 4$ from $x = 0$ to $x = 4-y^2$.  The integral that gives this volume is
$$\int\limits_{-2}^2 \int\limits_0^{4-y^2} \int\limits_{-\sqrt{4-x^2-y^2}}^{\sqrt{4-x^2-y^2}}\ 1\ dz\ dx\ dy$$
I don't find $T$ such that $T\bar{u} = \bar{x}$ for every $\bar{x}\in D$ where $D$ is our desired volume.
I mean, I'm trying to find $D^*$ such that for every $\bar{u} \in D^*$, $T$ is a change of variables for our problem.
So, I'm trying the find the volume between the regions $x^2 + y^2 + z^2 = 4$ and $x = 4-y^2$ but I can neither find that scalar.
Any ideas please.
 A: Nobody commented on my scalar in the comments section so far.
The projection of the solid on the $xy$-plane is determined by curves, $x^2+y^2=4$, $x=4-y^2$ and $x=0$. The common solution of the circle and the parabola gives $(1,\pm \sqrt{3})$. Hence the quarter volume integral due to symmetries w.r.t $xy$-plane and $x$-axis, is
$$\frac{1}{4}V=\int_0^{\sqrt{3}}\int_0^{\sqrt{4-y^2}}\int_0^{\sqrt{4-x^2-y^2}}dzdxdy+\int_{\sqrt{3}}^2\int_0^{4-y^2}\int_0^{\sqrt{4-x^2-y^2}}dzdxdy$$
and
$$\frac{1}{4}V=\int_0^{\sqrt{3}}\int_0^{\sqrt{4-y^2}}\sqrt{4-x^2-y^2}dxdy+\int_{\sqrt{3}}^2\int_0^{4-y^2}\sqrt{4-x^2-y^2}dxdy$$
and
$$\frac{1}{4}V=\int_0^{\sqrt{3}}\frac{\pi}{4}(4-y^2)dy+\int_{\sqrt{3}}^2\left(\frac{x\sqrt{4-x^2-y^2}}{2}+\frac{(4-y^2)\arcsin(\frac{x}{\sqrt{4-y^2}})}{2}\right)\vert_0^{4-y^2}dy$$
and hence $V$ is
$$V=3\sqrt{3}\pi+2\int_{\sqrt{3}}^{2}\sqrt{y^2-3}\sqrt{4-y^2}+(4-y^2)\arcsin(\sqrt{4-y^2})dy$$
By WolframAlpha, $V\approx 16.324+0.482=16.806$. The second integral which gives the tiny volume is not nice.
