To find the volume of the solid inside the sphere $x^2 + y^2 + z^2 = 36$ and outside the cylinder $x^2 + y^2 = 9$, we'll use polar coordinates to simplify the problem.

Step 1: Convert the equations to cylindrical and polar coordinates

1. Sphere equation: In cylindrical coordinates, where $x = r \cos \theta$, $y = r \sin \theta$, and $z = z$, the equation of the sphere $x^2 + y^2 + z^2 = 36$ becomes: $r^2 + z^2 = 36$

2. Cylinder equation: The cylinder $x^2 + y^2 = 9$ in polar coordinates becomes: $r^2 = 9$ Thus, $r = 3$, which describes a cylinder of radius 3.

Step 2: Set up the integral for the volume

The solid is confined inside the sphere and outside the cylinder. Therefore, we are interested in the volume of the region between these two surfaces.

1. Bounds for $r$:

   • The radius $r$ of the cylinder is constant at $r = 3$.
   • The outer surface of the sphere gives the upper bound for $r$. From the equation $r^2 + z^2 = 36$, we can express $r$ as: $r = \sqrt{36 - z^2}$

   So, for $r$, the range will be from 3 (the cylinder) to $\sqrt{36 - z^2}$ (the sphere).

2. Bounds for $\theta$:

   • Since the cylinder is symmetric around the $z$-axis, $\theta$ will range from $0$ to $2\pi$.

3. Bounds for $z$:

   • The sphere extends from $z = -6$ to $z = 6$, because $r^2 + z^2 = 36$ means that $z$ can vary from $-6$ to $6$ (the maximum height of the sphere).

Step 3: Volume integral

Now, we can set up the volume integral in cylindrical coordinates. The volume element in cylindrical coordinates is $r \, dr \, d\theta \, dz$. The volume integral is:

$V = \int_{z=-6}^{6} \int_{\theta=0}^{2\pi} \int_{r=3}^{\sqrt{36 - z^2}} r \, dr \, d\theta \, dz$

Step 4: Solve the integral

1. Inner integral (over $r$): $\int_3^{\sqrt{36 - z^2}} r \, dr = \frac{1}{2} \left[ r^2 \right]_3^{\sqrt{36 - z^2}} = \frac{1}{2} \left( (36 - z^2) - 9 \right) = \frac{1}{2} (27 - z^2)$

2. Middle integral (over $\theta$): $\int_0^{2\pi} d\theta = 2\pi$

3. Outer integral (over $z$): $V = \int_{-6}^{6} \frac{1}{2} (27 - z^2) \cdot 2\pi \, dz = \pi \int_{-6}^{6} (27 - z^2) \, dz$

Now, we calculate the integral: $\int_{-6}^{6} (27 - z^2) \, dz = \int_{-6}^{6} 27 \, dz - \int_{-6}^{6} z^2 \, dz$ $= 27 \times (6 - (-6)) - \left[ \frac{z^3}{3} \right]_{-6}^{6}$ $= 27 \times 12 - \left( \frac{6^3}{3} - \frac{(-6)^3}{3} \right)$ $= 324 - \left( \frac{216}{3} - \frac{-216}{3} \right)$ $= 324 - (72 + 72) = 324 - 144 = 180$

Step 5: Final volume

Now, multiply by $\pi$: $V = \pi \times 180 = 180\pi$

Thus, the volume of the solid is $\boxed{180\pi}$.

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Would you like further details on any part of this process? Here are some related questions:

1. How do cylindrical coordinates simplify volume integrals compared to Cartesian coordinates?
2. What would the volume be if the cylinder were inside the sphere, instead of outside?
3. How does the presence of the cylinder modify the volume of the solid?
4. Can this method be extended to solids with other boundary conditions, like ellipsoids or more complex shapes?
5. What is the significance of using polar coordinates for these types of 3D volume integrals?

Tip: For volume integrals, always carefully consider the geometry of the problem to set up the limits correctly in your coordinate system.