I guess I am kind of lost here. I have a sphere $x^2+y^2+z^2=25$
Some region $E = \{ \ \ (x,y,z) \ \ | \ \ a\le x \le b, \ \ c\le y \le d, \ \ e\le z \le f \ \ \}$
I have found:
$$ \begin{align} & a = -5 \\ & b = -5 \\ & c = -\sqrt{25-x^2-z^2} \\ & d = \sqrt{25-x^2-z^2} \\ & e = 0 \\ & f = \sqrt{25-x^2-y^2} \end{align}$$
However, $c$ and $d$ are wrong. I thought it was just a matter of algebra but I am not getting it for nothing.
So you're given the sphere $S:=\{(x,y,z)\in\Bbb R^3:x^2+y^2+z^2=25\}$. By a simple rearrangement this is described by the relation $z=\pm\sqrt{25-x^2-y^2}$, where the $-$ portion describes the lower half of the sphere, and the $+$ portion is the upper half. So if we integrate with respect to $z$ first, then we want to go from the bottom of the sphere to the top of the sphere, which is $-\sqrt{25-x^2-y^2}\leq z\leq \sqrt{25-x^2-y^2}$. Now, if we integrate with respect to $y$ next, then we can forget about $z$, because we already took care of it. So the $y$ value will range along the center circle in the sphere (i.e. $x^2+y^2=25$) because this circle produces the largest $y$ values in terms of magnitude on the sphere. So this circle is described by the relation $y=\pm\sqrt{25-x^2}$. So $-\sqrt{25-x^2}\leq y\leq\sqrt{25-x^2}$. Finally, $x$ ranges from the endpoints of the sphere along the $x$ axis, which is simply $-5\leq x\leq 5$. Thus, we have $$\iiint_S dV=\int_{-5}^5\int_{-\sqrt{25-x^2}}^\sqrt{25-x^2}\int_{-\sqrt{25-x^2-y^2}}^\sqrt{25-x^2-y^2}dzdydx$$
As I mentioned, spherical coordinates (as suggested by the name) are very useful for problems like these (integrating over spheres). Using them, we have $$\iiint_S dV=\int_0^{2\pi}\int_0^\pi\int_0^5\rho^2\sin(\phi)d\rho d\phi d\theta$$where $\rho$ is the radius of the sphere so $-5\leq\rho\leq 5$, $\phi$ is the vertical angle (inclination) so $0\leq\phi\leq \pi$, and $\theta$ is the radial angle (so in the $xy$-plane) so $0\leq\theta\leq 2\pi$. Note, the factor $\rho^2\sin(\phi)$ is the Jacobian of the coordinate transformation.