I am trying to understand how is the cylindrical conversion of the rectangular triple integral is true given that the integral calculates the volume inside a paraboloid bounded by $z=25$ (height-wise) and the radius is not constant throughout the height of $z=x^2+y^2$. Basically, I am trying to understand why I don't need to write the limits for $r$ as a function of theta since the radius is not constant throughout the paraboloid?
$$\int_{-\color{red}5}^{\color{red}5} \int_{-\sqrt{\color{red}{25}-z^2}}^{\sqrt{\color{red}{25}-z^2}} \int_{x^2+y^2}^{\color{red}{25}}x \, dz \, dy \, dx$$
$$\int_0^{2\pi} \int_0^5 \int_{r^2}^{25} r^2\cos(\theta) \, dz \, dr \, d\theta$$
As a function of $\theta$, the bounds on $r$ wouldn't change. Think about a slice $d \theta$ of the paraboloid. This cross-section would be a parabola ($z > r^2$). No matter what $\theta$ is, $r$ would go from $0$ to $5$. This picture shows the region of integration for a fixed $\theta$. As you can see, $r$ goes from $0$ to $5$, while $z$ goes from $r^2$ to $25$.