Changing the order of integration to find the volume My goal is to set up the triple integral that will solve the volume $\iiint \ xyz \ dV $ if S if the region bounded by the cylinders $x^2 + y^2 = 25$ and $ x^2+ z^2 = 25$ and the 1st octant with dV = dxdydz.
With order dV = dzdydx, the value of the volume is $\frac{15625}{24}$. My attempt is to set up the triple integral with order dV = dxdydz in which it should have the same answer with the triple integral with order dV = dzdydx.
The S I have computed is S $\lbrace (x, y, z) \in \mathbb{R}^3: 0 \leq x \leq 5, 0 \leq y \leq \sqrt{25 - x^2}, 0 \leq z \leq \sqrt{25 - x^2} \rbrace$ this is for triple integral with order dV = dzdydx
The S I have computed is S $\lbrace (x, y, z) \in \mathbb{R}^3: 0 \leq x \leq \sqrt{25 - \frac{y^2}{2} - \frac{z^2}{2}}, 0 \leq y \leq \sqrt{50 - z^2}, 0 \leq z \leq \sqrt{50} \rbrace$ this is for triple integral with order dV = dxdydz
When I set up the triple integral, I found out that it should be $\frac{1}{2} \int_0^\sqrt{50} \int_0^\sqrt{50-z^2} \int_0^\sqrt{25 - \frac{y^2}{2} - \frac{z^2}{2}} xyz \ dxdydz$ and not simply $\int_0^\sqrt{50} \int_0^\sqrt{50-z^2} \int_0^\sqrt{25 - \frac{y^2}{2} - \frac{z^2}{2}} xyz \ dxdydz$. What is the principle behind this? Why I should divide the volume of this triple integral to 2?
 A: The second $S$ is incorrect. On one surface we have $0\le x\le\sqrt{25-y^2}$ and on the other $0\le x\le\sqrt{25-z^2}$, which one do you choose? Since we have to take the intersection of the regions enclosed by these surfaces, we take the intersection of these inequalities as well, i.e.$$\begin{align*}0\le x\le\sqrt{25-y^2}\text{ and }0\le x\le\sqrt{25-z^2}&\iff0\le x\le\min\{\sqrt{25-y^2},\sqrt{25-z^2}\}\\&
\iff0\le x\le\begin{cases}\sqrt{25-y^2},&0\le z<y\le5\\
\sqrt{25-z^2},&0\le y\le z\le5\end{cases}
\end{align*}$$
I have attached a visual of the region. The horizontal plane is the $xy$ plane and the vertical axis is $z$ axis. I have tried to show its cross-sections lying in $xy,yz,xz$ planes. You may see the cross-section in the $yz$ plane is the square $0\le y,z\le5$.

Thus we will have to split the integral according as $y>z$ or $y\le z$:$$I=\left[\int_{z=0}^{z=5}\int_{y=z}^{y=5}\int_{x=0}^{x=\sqrt{25-y^2}}+\int_{z=0}^{z=5}\int_{y=0}^{y=z}\int_{x=0}^{x=\sqrt{25-z^2}}\right]xyz~dx~dy~dz$$
