To find the surface area of the part of $z = x^2 + y^2$ below $z = 4$, we will use the formula for the surface area of a function $z = f(x, y)$:

$A = \iint_D \sqrt{1 + \left( \frac{\partial z}{\partial x} \right)^2 + \left( \frac{\partial z}{\partial y} \right)^2} \, dA$

Given $z = x^2 + y^2$, we need to find the partial derivatives:

$\frac{\partial z}{\partial x} = 2x$ $\frac{\partial z}{\partial y} = 2y$

The surface area integral becomes:

$A = \iint_D \sqrt{1 + (2x)^2 + (2y)^2} \, dA$ $A = \iint_D \sqrt{1 + 4x^2 + 4y^2} \, dA$

We need to determine the region $D$ over which we will integrate. Since $z = 4$ defines the boundary of our surface, we have:

$x^2 + y^2 = 4$

Thus, $D$ is the disk of radius 2 centered at the origin in the $xy$-plane. To simplify the integration, we switch to polar coordinates, where $x = r \cos \theta$ and $y = r \sin \theta$, with $r$ ranging from 0 to 2 and $\theta$ from 0 to $2\pi$.

In polar coordinates, the integrand and differential area $dA$ transform as follows:

$1 + 4x^2 + 4y^2 = 1 + 4r^2$ $dA = r \, dr \, d\theta$

Thus, the integral becomes:

$A = \int_0^{2\pi} \int_0^2 \sqrt{1 + 4r^2} \cdot r \, dr \, d\theta$

Now, evaluate the inner integral with respect to $r$:

$\int_0^2 r \sqrt{1 + 4r^2} \, dr$

To solve this, we use the substitution $u = 1 + 4r^2$. Then, $du = 8r \, dr$ or $dr = \frac{du}{8r}$. Notice $r$ can be expressed as $\sqrt{\frac{u-1}{4}}$:

$\int_0^2 r \sqrt{1 + 4r^2} \, dr = \int_1^{17} \frac{1}{8} \sqrt{u} \, du$ $= \frac{1}{8} \int_1^{17} u^{1/2} \, du$ $= \frac{1}{8} \left[ \frac{2}{3} u^{3/2} \right]_1^{17}$ $= \frac{1}{8} \cdot \frac{2}{3} \left[ 17^{3/2} - 1^{3/2} \right]$ $= \frac{1}{12} \left[ 17^{3/2} - 1 \right]$

$= \frac{1}{12} \left[ 17\sqrt{17} - 1 \right]$

Now, evaluate the outer integral with respect to $\theta$:

$A = \int_0^{2\pi} \frac{1}{12} \left[ 17\sqrt{17} - 1 \right] d\theta$ $= \frac{1}{12} \left[ 17\sqrt{17} - 1 \right] \int_0^{2\pi} d\theta$ $= \frac{1}{12} \left[ 17\sqrt{17} - 1 \right] \cdot 2\pi$ $= \frac{2\pi}{12} \left[ 17\sqrt{17} - 1 \right]$ $= \frac{\pi}{6} \left[ 17\sqrt{17} - 1 \right]$

Thus, the surface area of the part of $z = x^2 + y^2$ below $z = 4$ is:

$A = \frac{\pi}{6} \left( 17\sqrt{17} - 1 \right)$

Do you want more details or have any questions?

Here are some related questions:

1. How would the surface area change if the boundary condition was $z = 9$ instead of $z = 4$?
2. Can you find the volume of the solid bounded by $z = x^2 + y^2$ and $z = 4$?
3. What is the surface area of the part of $z = \sqrt{x^2 + y^2}$ below $z = 4$?
4. How do you calculate the surface area for an arbitrary function $z = f(x, y)$?
5. What is the significance of the Jacobian determinant in changing to polar coordinates?
6. How does changing the integrand affect the complexity of solving the integral?
7. What other methods can be used to solve surface area integrals aside from polar coordinates?
8. Can you solve the surface area of $z = x^2 + y^2$ below $z = 4$ using parametric equations?

Tip: When solving surface area problems involving double integrals, switching to polar coordinates can often simplify the integration, especially for circular or radial regions.