# How should this volume integral be set up?

I would like to find the (four-dimensional) volume of the region given by

$$xy>zw \quad\wedge \quad x>-y \quad\wedge \quad x^2+y^2+z^2+w^2<1,$$

for $x,y,z,w\in\mathbb{R}$ and where the last condition means that the whole thing is bounded by the unit $4$-ball. The boundary of the region given by the two first conditions would be (I think)

which can be seen in the figure below.

$\quad\quad\quad$

But how do I go from this to actually setting up the integrals? I believe the answer should be one fourth of the volume of the unit $4$-ball.

Thanks.

The 4D ball condition implies that all $4$ variables be constrained within $(-1,1)$. So we can rewrite the bounds as $$\bbox[lightyellow] { \left\{ \begin{gathered} - 1 < x,y,z,w < 1 \hfill \\ zw < xy \hfill \\ 0 < x + y \hfill \\ x^{\,2} + y^{\,2} < 1 - \left( {z^{\,2} + w^{\,2} } \right) \hfill \\ \end{gathered} \right. } \tag {1}$$ Note the overall symmetry of the problem, which allows:
- the exchange of $x$ with $y$, and $w$ with $z$;
- that inverting the 2nd and/ or 3rd inequality will not change the result;
- that exchanging the $(x,y)$ couple with the $(w,z)$ one will not change the result.
and finally note that, dealing with volumes, the $<$ sign can be replaced by the $\leqslant$.

The integral to be found is $$\bbox[lightyellow] { \begin{gathered} \int\limits_{G(x,y,w,z)} {1\,dx\,dy\,dw\,dz} = \hfill \\ = \int\limits_{ - 1\, < w,z\, < \,1} {\left( {\int\limits_{G\left( {x(w,z),y(w,z)} \right)} {1\,dx\,dy\,} } \right)dw\,dz} = \hfill \\ = \int\limits_{ - 1\, < w,z\, < \,1} {A(w,z)dw\,dz} \hfill \\ \end{gathered} }$$ where $G(x,y,w,z)$ is the region bounded by the given conditions, and $A(w,z)$ is the area of the projection of that region on the $x,y$ plane, given the values of $z$ and $w$.
Clearly $A(w,z)$ will be null when the values of $w$ and $z$ are incompatible with the bounds, e.g. when $1 < z^2+w^2$, although each variable is within the $\pm 1$ bound.

Let's take a view about the projection of the integration region onto the $x,y$ plane, with one sketch for the case $0 < z \cdot w$ and one for the case $z \cdot w < 0$.
(those illustrated correspond to $w=0.5, \; z= \pm 0.2$)

The various inequalities in play are evident.
The region satisfying all the bounds is shown in deep purple, i.e. the "lens-shaped" region $AD$ in the first sketch, and the semicircle minus the two half-lenses (insisting on $AC$) for the second.

While it is not difficult to compute the area of the lens region, this will lead to an expression not simple to handle, as it involves an hyperbolic arc.

We can by-pass this calculation, by availing ourselves of the complementary symmetry clearly evidentiated by the two sketches, i.e.
that the two hyperbolas $xy= \pm zw$ , have the diagonal symmetry as shown.
Thus the sum of the two areas, for the same $|z \cdot w|$ value, totals that of the semicircle, and that keeps valid even when the hyperbola and the circle do not intersecate.

We can therefore split the integral as follows $$\bbox[lightyellow] { \begin{gathered} \int\limits_{G(x,y,w,z)} {1\,dx\,dy\,dw\,dz} = \int\limits_{ \begin{subarray}{l} - 1\, < w\, < \,1 \\ - 1\, < \,z\, < \,1 \end{subarray} } {A(w,z)dw\,dz} = \hfill \\ = \int\limits_{ \begin{subarray}{l} - 1\, < w\, < \,1 \\ - 1\, < \,z\, < \,0 \end{subarray} } {A(w,z)dw\,dz} + \int\limits_{ \begin{subarray}{l} - 1\, < w\, < \,1 \\ 0\, \leqslant \,z\, < \,1 \end{subarray} } {A(w,z)dw\,dz} = \hfill \\ = \int\limits_{ \begin{subarray}{l} - 1\, < w\, < \,1 \\ 0\, \leqslant \,z\, < \,1 \end{subarray} } {A(w, - z)dw\,dz} ^{\;\left( \diamondsuit \right)} + \int\limits_{ \begin{subarray}{l} - 1\, < w\, < \,1 \\ 0\, \leqslant \,z\, < \,1 \end{subarray} } {A(w,z)dw\,dz} = \hfill \\ = \int\limits_{ \begin{subarray}{l} - 1\, < w\, < \,1 \\ 0\, \leqslant \,z\, < \,1 \end{subarray} } {\left( {A(w, - z) + A(w,z)} \right)dw\,dz} = \hfill \\ = \int\limits_{ \begin{subarray}{l} - 1\, < w\, < \,1 \\ 0\, \leqslant \,z\, < \,1 \\ 0\, \leqslant \,z^{\,2} + w^{\,2} \, \leqslant \,1 \end{subarray} } {\frac{1} {2}\pi \left( {1 - \left( {z^{\,2} + w^{\,2} } \right)} \right)dw\,dz} = \hfill \\ = \frac{1} {4}\pi \int\limits_{0\, \leqslant \,z^{\,2} + w^{\,2} \, \leqslant \,1} {\left( {z^{\,2} + w^{\,2} } \right)dw\,dz} = \hfill \\ = \frac{1} {4}2\pi ^{\,2} \int\limits_{0\, \leqslant \,r\, \leqslant \,1} {r^{\,3} dr} = \hfill \\ = \frac{1} {4}\left( {2\pi ^{\,2} \frac{1} {4}} \right) = \frac{1} {4}\left( {\frac{{\pi ^{\,4/2} 1^{\,4} }} {{\Gamma \left( {4/2 + 1} \right)}}} \right) = \frac{1} {4}V_{\text{ball}} \hfill \\ \end{gathered} } \tag {2}$$

Note($\diamondsuit$): working with areas we shall use the absolute sign of $dz$ (absolute Jacobian).

The above intuitively summarizes into
- the circle $A,B,C,D$ is the projection on the $x,y$ plane of the whole ball, its integral in $dz\,dw$ gives back the volume of the ball;
- the condition $0<x+y$ divides the circle in half (on the $x,y$ plane);
- folding the negative $z$ half plane onto the positive one, to get the complementary superposition, divides the volume in half (on the $z,w$ plane);
- the total result is to divide the whole ball by $4$.

• @Lovsovs: fully re-edited: hope it is enough clear now. – G Cab Feb 13 '17 at 2:19
• Beautiful! I will read it thoroughly later, but it looks very much clearer! (I have deleted my old comments, as they are no longer relevant.) – Bobson Dugnutt Feb 13 '17 at 10:09
• @Lovsovs: good that it is clearer (although less concise). I deleted my previous comments as well. – G Cab Feb 13 '17 at 15:00