<s>[INST] 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.
 [/INST] 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}
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