To find the volume of the solid under the paraboloid $z = x^2 + y^2$ and above the disk defined by $x^2 + y^2 \leq 9$, we can convert to polar coordinates.

In polar coordinates, we have:

• $x = r \cos \theta$
• $y = r \sin \theta$
• $z = r^2$ (since $z = x^2 + y^2 = r^2$)

The region of integration is defined by the disk $r^2 \leq 9$, which translates to $0 \leq r \leq 3$ and $0 \leq \theta < 2\pi$.

The volume $V$ can be expressed as:

$V = \int_0^{2\pi} \int_0^3 (r^2) \, r \, dr \, d\theta$

Here, $r^2$ is the height of the paraboloid, and $r \, dr \, d\theta$ accounts for the area element in polar coordinates.

Now we can set the integral parameters:

• $A = 2\pi$
• $B = 3$

Integral Setup

The integral for the volume $V$ becomes:

$V = \int_0^{2\pi} \int_0^3 r^3 \, dr \, d\theta$

Finding the Volume

1. Evaluate the inner integral:

$\int_0^3 r^3 \, dr = \left[ \frac{r^4}{4} \right]_0^3 = \frac{3^4}{4} - 0 = \frac{81}{4}$

2. Evaluate the outer integral:

$V = \int_0^{2\pi} \frac{81}{4} \, d\theta = \frac{81}{4} \cdot \left[ \theta \right]_0^{2\pi} = \frac{81}{4} \cdot 2\pi = \frac{162\pi}{4} = \frac{81\pi}{2}$

Final Result

Thus, the volume of the solid is:

$V = \frac{81\pi}{2}$

Summary

• $A = 2\pi$
• $B = 3$
• Volume $V = \frac{81\pi}{2}$

Would you like more details on any step, or do you have any questions? Here are some related questions to consider:

1. What is the geometric interpretation of the solid bounded by the paraboloid and the disk?
2. How would the volume change if the disk had a different radius?
3. Can you derive the volume using Cartesian coordinates for comparison?
4. What is the significance of changing to polar coordinates for this problem?
5. How would you set up the integral if the paraboloid opened downward?

Tip: When working with polar coordinates, always remember to include the Jacobian $r$ when converting area elements.