Find the volume of the solid bounded by $z^2=xy$; $x+y=a$; $x+y=b$ $(0Find the volume of the solid bounded by $z^2=xy$; $x+y=a$; $x+y=b$ $(0<a<b)$ by applying variable substitution. $xy>0$ and $x,y$ can't be negative at the same time so $x>0;y>0.$ Now, $a - x < y < b - x$ and we get that $a < x < b$. So the integral to compute is $\int_{a}^{b}\int_{a-x}^{b-x}{\sqrt{xy}\,dxdy}$. Now this integral, as requested by the problem, should be solved by applying variable substitution. But the answer in the textbook is $\dfrac{\pi(b^3-a^3)}{12}$. I don't get where this $\pi$ comes from. The only way I know to obtain such a result is by changing to polar coordinate system, but I don't see any way to obtain relations between $a,b$, $a-x$, $b-x$ and the new coordinate system especially because in none of these equations are $x$ or $y$ squared. Am I doing some mistake or how does this $\pi$ appear in the answer? Thanks in advance.
 A: The trasformation $$(x,y,z)\mapsto(x+y,{y \over x},z^2)=(u,v,w)$$ maps the region $$\big\{(x,y,z)\in \mathbb{R}^3:x>0,y>0,a<x+y<b,0<z<\sqrt{xy}\big\}$$ bijectively into the the region $$\Bigg\{(u,v,w)\in \mathbb{R}^3:a<u<b,v>0,0<w<\frac{u^2 v}{(v+1)^2}\Bigg\}$$ Furthermore, $$\frac{\partial(u,v,w)}{\partial(x,y,z)}=2z\left(\frac{1}{x}+\frac{y}{x^2}\right)=\frac{2(v+1)^2\sqrt{w}}{u}$$ So using symmetry volume of your region is $$\begin{eqnarray*}2\int_a^b \int_0^{\infty} \int_0^{\frac{u^2v}{(v+1)^2}}\frac{u}{2(v+1)^2\sqrt{w}}\mathrm{d}w\mathrm{d}v\mathrm{d}u &=& \left(\int_a^b 2u^2 \mathrm{d}u\right)\cdot \left(\int_0^{\infty}\frac{\sqrt{v}}{(v+1)^3}\mathrm{d}v\right) \end{eqnarray*}$$ Can you finish?
A: If you do $X=x+y$, $Y=x-y$, and $Z=z$, then $z^2=xy$ becomes $Z^2=\frac14(X^2-Y^2)$, $x+y=a$ becomes $X=a$, and $x+y=b$ becomes $X=b$. Since $Z^2\geqslant0$, $Y^2\leqslant X^2$, which means that $-X\leqslant Y\leqslant X$. Finally, you have $(x,y,z)=\left(\frac{X+Y}2,\frac{X-Y}2,Z\right)$, and the absolute value of the Jacobian of$$(X,Y,Z)\mapsto\left(\frac{X+Y}2,\frac{X-Y}2,Z\right)$$is $\frac12$. So, the volume that you're after is\begin{align}\int_a^b\int_{-X}^X\int_{-\frac12\sqrt{X^2-Y^2}}^{\frac12\sqrt{X^2-Y^2}}\frac12\,\mathrm dZ\,\mathrm dY\,\mathrm dX&=\frac12\int_a^b\int_{-X}^X\sqrt{X^2-Y^2}\,\mathrm dY\,\mathrm dX\\&=\int_a^b\frac14\pi X^2\,\mathrm dX\\&=\frac{\pi(b^3-a^3)}{12}.\end{align}
A: The quadric $z^2 = x y $ corresponds to the quadratic form
$ r^T Q r = 0 $
where $r = [x, y, z]^T$ and
$Q = \begin{bmatrix} 0 && -1/2 && 0 \\ -1/2 && 0 && 0 \\ 0 && 0 && 1 \end{bmatrix} $
Diagonalizing $Q$ it becomes $Q = R D R^T $ where
$ D = \begin{bmatrix} -0.5 && 0 && 0 \\ 0 && 0.5 && 0 \\ 0 && 0 && 1 \end{bmatrix} $
$ R = \begin{bmatrix} \dfrac{1}{\sqrt{2}} && - \dfrac{1}{\sqrt{2}} && 0 \\ \dfrac{1}{\sqrt{2}} && \dfrac{1}{\sqrt{2}} && 0 \\ 0 && 0 && 1 \end{bmatrix} $
Define the change of variable as follows:
$ w = R^T r $ i.e. $r = R w $.  So $x + y = [1, 1, 0] r = [1, 1, 0] R w = [\sqrt{2}, 0, 0 ] w $
The equation of the quadric is
$ 0.5 w_1^2 = 0.5 w_2^2 + w_3^2 $
which simplifies to
$ w_1^2 = w_2^2 + 2 w_3^2 $
Now the equation of the boundary $r^T Q r = 0 $ is a cone whose axis is along the the first column of $R$, so in the changed variable it is a cone whose axis is the $w_1$ axis.  The limits of $w_1$ are from $\dfrac{a}{\sqrt{2}} $ to $\dfrac{b}{\sqrt{2}} $.  If we take the volume between $w_1 = 0$ and $w_1=c$, then the volume is just one third of the area of base (in the $w_2, w_3$ plane) times the height $c$; the area is $ \pi \left( \dfrac{c^2}{\sqrt{2}} \right) $ so that volume is
$ \dfrac{1}{3 \sqrt{2}} \pi c ^ 3 $
Now we want to take the difference of volumes between $c = \dfrac{b}{\sqrt{2}}$ and $c = \dfrac{a}{\sqrt{2}} $ , and this is
$ V = \dfrac{\pi}{3\sqrt{2}} \left( \dfrac{b^3 }{2 \sqrt{2}} - \dfrac{a^2}{2 \sqrt{2}} \right) = \dfrac{ \pi (b^3 - a^3) }{12}$
A: The integral is $\int_T 2\sqrt{xy}\,dx\,dy$ where $T$ is the trapezoidal region defined by the $x$ and $y$ axes and the lines $x + y = a$, $x + y = b$. (You have a factor of 2 in the integrand because it goes from $z = -\sqrt{xy}$ to $z = \sqrt{xy}$.)
So if you let $x = u^2$ and $y = v^2$, you get the integral
$\int_U 8u^2v^2 \,du\,dv$ where $U$ is the part of the annulus $a < u^2 + v^2 < b$ for which $u, v > 0$. This is easily done in polar coordinates; it just becomes $$\int_0^{\pi \over 2} 8\sin^2\theta\cos^2\theta\,d\theta \times \int_{\sqrt{a}}^{\sqrt{b}} r^5\,dr$$
I'll leave it to you to verify the answer ${\pi \over 12} (b^3 - a^3)$ in your solution manual.
