The problem asks to find the volume of a solid bounded by the surfaces:

• $z = \sqrt{x^2 + y^2}$
• $z = 0$
• $x^2 + y^2 = 25$

This suggests the solid is a cone with a circular base, centered at the origin, and capped by the circle $x^2 + y^2 = 25$. We are tasked to find the correct double integral in polar coordinates.

The volume can be computed using a double integral in polar coordinates. In polar coordinates, we have the following transformations:

• $x = r \cos \theta$
• $y = r \sin \theta$
• $x^2 + y^2 = r^2$

The given solid has a radius of 5 (since $x^2 + y^2 = 25 \Rightarrow r = 5$), and the height at any point is $z = \sqrt{x^2 + y^2} = r$.

The volume $V$ is given by:

$V = \int_0^{2\pi} \int_0^5 z(r) \, r \, dr \, d\theta$

Substituting $z(r) = r$ (since the height of the cone is $z = r$), the integral becomes:

$V = \int_0^{2\pi} \int_0^5 r^2 \, dr \, d\theta$

Now looking at the given options, the correct answer corresponds to:

$V = \int_0^{2\pi} \int_0^5 r^2 \, dr \, d\theta$

which matches the third option.

Do you want further details or have any questions?

Here are 5 related questions:

1. How would the limits change if the height of the cone was modified?
2. What is the geometric interpretation of this double integral?
3. How would this integral change if we had an ellipsoid instead of a cone?
4. What happens to the volume if the radius increases?
5. How can we calculate the surface area of this cone?

Tip: When using polar coordinates, always express the variables in terms of $r$ and $\theta$ to simplify integrals over circular regions.