Triple integration problem Prove that:
$$\iiint\limits_B \frac{1}{x} dx\,dy\,dz=\frac{8-4\sqrt2}{3}$$
where
$$B=\{(x,y,z):1\leqslant x \leqslant e^z, y\geqslant z, y^2+x^2\leqslant 4\}.$$
I used Mathematica's regionplot3D to visualize $B$. However I could use some help setting the correct integration intervals.
 A: If you put $x$ and $y$ on the outside, you've got
$$
\iint\limits_{\begin{smallmatrix} x^2+y^2\le 4 \\[2pt]  x\ge 1 \\[2pt] y\ge\log x \end{smallmatrix}} \left( \int_{\log x}^y \frac 1 x \, dz \right)\,dx\,dy.
$$
The inner integral is now routine:
$$
\iint\limits_{\begin{smallmatrix} x^2+y^2\le 4 \\[2pt]  x\ge 1 \\[2pt] y\ge\log x \end{smallmatrix}} \frac{y-\log x}{x} \,dx\,dy.
$$

The constraint $x\ge 1$ means that if we put $\int\cdots\cdots\,dx$ on the outside, then the bounds work out:
$$
\int_1^2 \left( \int_{-\sqrt{4-x^2}}^{\sqrt{4-x^2}} \frac{y-\log x}{x} dy \right)\,dx
$$
The integral of $y\,dy$ over an interval symmetric about $0$ is $0$, and the integral of $-(\log x)/x$ with respect to $y$ is easy because $-(\log x)/x$ does not change as $y$-changes.  We get
$$
\int_1^2  2\sqrt{4-x^2} \left(\frac{-\log x}{x}\right) \,dx.
$$
That takes care of the bounds.
Postscript: The bounds don't work out simply this way, and I'm not sure they do any other way either.  The inner integral with respect to $y$ would need to be $\displaystyle\int_{\log x}^{\sqrt{4-x^2}}$, and the outer one could not go all the way up to $2$ since once $x$ is too big, we get $\log x>\sqrt{4-x^2}$.  You can tell that will happen because $\log x$ increases from $0$ to a positive number as $x$ goes from $0$ to $2$ while $\sqrt{4-x^2}$ decreases from a positive number to $0$.  Perhaps to be continued [ . . . . ]
A: @Michael Hardy has already elaborated much on the integral and there remains only the last part to solve. So I will try to do that. The integral now that we need to solve, after the simplification by @Michael Hardy, is $$I=\iint\limits_{\begin{smallmatrix} x^2+y^2\le 4,\\ x\ge 1,\\ y\ge \ln x
\end{smallmatrix}}\frac{y-\ln x}{x}dx$$ Now the curve $y=\ln x$ intersects with the circle $x^2+y^2=4$ at two points, one at $x<1,y<0$ and one at $x>1,y>0$. Let the later on has coordinates $(x_0,y_0)$. Note that since the function $\phi(x)=\sqrt{4-x^2}-\ln x$ is monotonic decreasing for $x>1$, and $\phi(2)=-\ln 2<0\Rightarrow x_0<2$. Then our integral becomes $$I=\iint\limits_{\begin{smallmatrix}  1\le x\le x_0,\\ \ln x\le y\le \sqrt{4-x^2}
\end{smallmatrix}}\frac{y-\ln x}{x}dx=\int_{x=1}^{x_0}\int_{y=\ln x}^{\sqrt{4-x^2}}\frac{y-\ln x}{x}dy \ dx=\int_{x=1}^{x_0}\frac{\frac{1}{2}(4-x^2-(\ln x)^2)-\ln x(\sqrt{4-x^2}-\ln x)}{x}dx\\=\int_{x=1}^{x_0}\left(\frac{2}{x}-\frac{x}{2}\right)dx+\int_{x=1}^{x_0}\frac{(\ln x)^2}{2x}dx-\int_{x=1}^{x_0}\frac{\ln x\sqrt{4-x^2}}{x}dx\\=2\ln x_0-\frac{x_0^2-1}{4}+\frac{(\ln x_0)^3}{6}-J$$ where $$J=\int_{x=1}^{x_0}\frac{\ln x\sqrt{4-x^2}}{x}dx$$
