To convert the given integral to polar coordinates in the $xz$-plane, let's analyze each part of the original Cartesian double integral:

Step 1: Set up polar coordinates

In the $xz$-plane, we can define polar coordinates as: $x = r \cos \theta \quad \text{and} \quad z = r \sin \theta$ The Jacobian determinant for converting to polar coordinates is $r$, meaning $dx \, dz$ becomes $r \, dr \, d\theta$ in the polar coordinates.

Step 2: Determine the limits in polar coordinates

The $x$-limits are from $-3$ to $3$, and for a given $x$, $z$ ranges from $-\sqrt{9 - x^2}$ to $\sqrt{9 - x^2}$. This describes a circle of radius 3 in the $xz$-plane, which suggests that in polar coordinates: $0 \leq r \leq 3 \quad \text{and} \quad 0 \leq \theta \leq 2\pi$

Step 3: Convert the integrand

The integrand is: $63 - 2x^2 - 2z^2 - (5x^2 + 5z^2)$ We can simplify this by substituting $x = r \cos \theta$ and $z = r \sin \theta$ to express $x^2 + z^2 = r^2$: $63 - 2(r^2) - 5(r^2) = 63 - 7r^2$

Step 4: Set up the polar integral

Now we can write the integral in polar coordinates: $\text{Volume} = \int_0^{2\pi} \int_0^3 \left(63 - 7r^2\right) r \, dr \, d\theta$

Final Polar Form of the Integral

The complete expression in polar coordinates is: $\text{Volume} = \int_0^{2\pi} \int_0^3 \left(63 - 7r^2\right) r \, dr \, d\theta$

Would you like a detailed solution for evaluating this integral, or do you have any other questions?

Here are five related questions to consider:

1. How is the Jacobian determinant derived for polar coordinates in the $xz$-plane?
2. Why does $dx \, dz$ become $r \, dr \, d\theta$ in polar coordinates?
3. What are the conditions for choosing $\theta$ and $r$ limits in polar coordinates?
4. How would this conversion differ if the integral were in the $xy$-plane instead?
5. How would we approach evaluating this polar integral for the volume?

Tip: When converting to polar coordinates, ensure you correctly apply the Jacobian and adjust all limits and expressions in the integrand accordingly.