We are tasked with finding the volume of the solid under the paraboloid $z = x^2 + y^2$ and above the disk $x^2 + y^2 \leq 1$. This problem is well-suited for polar coordinates, where $x = r\cos\theta$ and $y = r\sin\theta$.

Step-by-Step Solution:

1. Expressing the paraboloid in polar coordinates: In polar coordinates, we have the following conversions: $x^2 + y^2 = r^2$ So, the equation of the paraboloid becomes: $z = r^2$

2. Setting up the bounds:

   • The disk $x^2 + y^2 \leq 1$ translates to $r \leq 1$ in polar coordinates.
   • The bounds for $r$ range from 0 to 1.
   • The bounds for $\theta$ are from $0$ to $2\pi$ since the disk is circular.

3. Volume integral: The volume of the solid can be expressed as: $V = \int \int_D z \, dA$ Where $z = r^2$ and $dA = r \, dr \, d\theta$ is the differential area element in polar coordinates.

   Therefore, the volume integral becomes: $V = \int_0^{2\pi} \int_0^1 r^2 \cdot r \, dr \, d\theta$ Simplifying the integrand: $V = \int_0^{2\pi} \int_0^1 r^3 \, dr \, d\theta$

4. Evaluate the integral:

   First, we evaluate the inner integral with respect to $r$: $\int_0^1 r^3 \, dr = \left[\frac{r^4}{4}\right]_0^1 = \frac{1^4}{4} - \frac{0^4}{4} = \frac{1}{4}$

   Now, integrate with respect to $\theta$: $\int_0^{2\pi} \frac{1}{4} \, d\theta = \frac{1}{4} \cdot \theta \Big|_0^{2\pi} = \frac{1}{4} \cdot 2\pi = \frac{\pi}{2}$

Thus, the volume of the solid is: $V = \frac{\pi}{2}$

Conclusion:

The volume of the solid under the paraboloid $z = x^2 + y^2$ and above the disk $x^2 + y^2 \leq 1$ is $\frac{\pi}{2}$.

Would you like further details or have any questions?

Related Questions:

1. How would the volume change if the paraboloid was $z = 2(x^2 + y^2)$?
2. What would be the volume of the solid if the disk was $x^2 + y^2 \leq 4$?
3. How can we compute the surface area of the paraboloid above the disk?
4. How would you compute the volume using Cartesian coordinates instead of polar?
5. How does the result change if the paraboloid is rotated about a different axis?

Tip:

When solving problems involving circular symmetry, switching to polar coordinates often simplifies the process!