The problem asks to calculate the surface area of the surface $z = 4 - x^2 - y^2$ within the boundaries of the cylinder $x^2 + y^2 \leq 4$.

To find the surface area of a given surface $z = f(x, y)$, we use the formula for surface area in three dimensions:

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

Here, $D$ is the projection of the surface onto the $xy$-plane, which is given by $x^2 + y^2 \leq 4$, a disk of radius 2 centered at the origin.

Step 1: Partial derivatives

The surface equation is $z = 4 - x^2 - y^2$. Compute the partial derivatives: $\frac{\partial z}{\partial x} = -2x, \quad \frac{\partial z}{\partial y} = -2y$

Square these derivatives: $\left( \frac{\partial z}{\partial x} \right)^2 = 4x^2, \quad \left( \frac{\partial z}{\partial y} \right)^2 = 4y^2$

Substitute into the surface area formula: $A = \iint\limits_{D} \sqrt{1 + 4x^2 + 4y^2} \, dA$

Step 2: Convert to polar coordinates

In polar coordinates, $x = r \cos\theta$ and $y = r \sin\theta$, with $x^2 + y^2 = r^2$ and $dA = r \, dr \, d\theta$.

The limits for $r$ are from 0 to 2 (radius of the cylinder), and for $\theta$, from 0 to $2\pi$. The integral becomes: $A = \int_{0}^{2\pi} \int_{0}^{2} \sqrt{1 + 4r^2} \, r \, dr \, d\theta$

Step 3: Simplify and solve the integral

Separate the integral: $A = \int_{0}^{2\pi} d\theta \int_{0}^{2} r \sqrt{1 + 4r^2} \, dr$

The angular integral evaluates to: $\int_{0}^{2\pi} d\theta = 2\pi$

Now, compute the radial integral: $I = \int_{0}^{2} r \sqrt{1 + 4r^2} \, dr$

Use the substitution $u = 1 + 4r^2$, so $du = 8r \, dr$ and $r \, dr = \frac{1}{8} du$. When $r = 0$, $u = 1$, and when $r = 2$, $u = 17$. The integral becomes: $I = \frac{1}{8} \int_{1}^{17} \sqrt{u} \, du$

Evaluate: $\int \sqrt{u} \, du = \frac{2}{3} u^{3/2}$

Substitute back: [ I = \frac{1}{8} \left[ \frac{2}{3} u^{3/2} \right]_{1}^{17} = \frac{1}{12} \left[ (17)^{3/2} - (1)^{3/2} \right] ]

Simplify: $I = \frac{1}{12} \left( 17\sqrt{17} - 1 \right)$

Finally, substitute $I$ into the area formula: $A = 2\pi \cdot I = \frac{\pi}{6} \left( 17\sqrt{17} - 1 \right)$

This is the surface area of the given surface.

Let me know if you need a deeper explanation or further assistance! Here are five related questions:

1. How would the surface area change if the cylinder's radius increased?
2. What is the geometric interpretation of the surface integral in this context?
3. How does converting to polar coordinates simplify the computation?
4. What role does the gradient of $z(x, y)$ play in the surface area calculation?
5. How can this method be extended to surfaces not defined explicitly as $z = f(x, y)$?

Tip: When working with circular regions, always check if polar coordinates can simplify the problem.