To evaluate the given surface integral:

$\iint_S f(x, y, z) \, dS, \quad \text{where } f(x, y, z) = x^2 + y^2 + z^2,$ and $S$ is the surface defined by $z = x + y$, constrained by $x^2 + y^2 \leq 1$.

Step 1: Surface Parameterization

The surface $S$ can be parameterized in terms of $x$ and $y$: $\vec{r}(x, y) = \langle x, y, x + y \rangle.$ Here, $x^2 + y^2 \leq 1$ defines a circular region in the $x\text{-}y$ plane.

Step 2: Compute the Surface Element $dS$

The surface element $dS$ is obtained as: $dS = \|\vec{r}_x \times \vec{r}_y\| \, dx \, dy.$

1. Compute partial derivatives: $\vec{r}_x = \langle 1, 0, 1 \rangle, \quad \vec{r}_y = \langle 0, 1, 1 \rangle.$

2. Compute the cross product $\vec{r}_x \times \vec{r}_y$:

   \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & 0 & 1 \\ 0 & 1 & 1 \end{vmatrix} = \langle -1, -1, 1 \rangle.$$

3. Compute the magnitude: $\|\vec{r}_x \times \vec{r}_y\| = \sqrt{(-1)^2 + (-1)^2 + (1)^2} = \sqrt{3}.$

Thus, $dS = \sqrt{3} \, dx \, dy$.

Step 3: Express the Integral

Substitute $z = x + y$ into $f(x, y, z)$: $f(x, y, z) = x^2 + y^2 + z^2 = x^2 + y^2 + (x + y)^2 = x^2 + y^2 + x^2 + 2xy + y^2 = 2x^2 + 2y^2 + 2xy.$

The integral becomes: $\iint_S f(x, y, z) \, dS = \iint_{x^2 + y^2 \leq 1} \left( 2x^2 + 2y^2 + 2xy \right) \sqrt{3} \, dx \, dy.$

Step 4: Switch to Polar Coordinates

Convert to polar coordinates: $x = r\cos\theta, \quad y = r\sin\theta, \quad x^2 + y^2 = r^2, \quad dx \, dy = r \, dr \, d\theta.$

Substitute into the integral: $2x^2 + 2y^2 + 2xy = 2r^2 + 2r^2\cos\theta\sin\theta = 2r^2 + r^2\sin(2\theta).$

The integral becomes: $\sqrt{3} \iint_{x^2 + y^2 \leq 1} \left( 2r^2 + r^2\sin(2\theta) \right) r \, dr \, d\theta.$

Split the integral: $\sqrt{3} \int_0^{2\pi} \int_0^1 \left( 2r^3 + r^3\sin(2\theta) \right) \, dr \, d\theta = \sqrt{3} \left( \int_0^{2\pi} \int_0^1 2r^3 \, dr \, d\theta + \int_0^{2\pi} \int_0^1 r^3\sin(2\theta) \, dr \, d\theta \right).$

Step 5: Evaluate the Integrals

1. First term: $\int_0^1 2r^3 \, dr = 2 \cdot \frac{r^4}{4} \Big|_0^1 = \frac{1}{2}.$ The $\theta$-integral is: $\int_0^{2\pi} \frac{1}{2} \, d\theta = \frac{1}{2} \cdot 2\pi = \pi.$

2. Second term: The $r$-integral is the same: $\int_0^1 r^3 \, dr = \frac{1}{4}.$ For the $\theta$-integral: $\int_0^{2\pi} \sin(2\theta) \, d\theta = 0 \quad \text{(since sine integrates to 0 over a full period)}.$

Thus, the second term vanishes.

Step 6: Final Answer

The surface integral evaluates to: $\boxed{\sqrt{3} \pi}.$

Would you like further details or have any questions?

Here are 5 related questions for exploration:

1. What is the physical significance of surface integrals in physics?
2. How do we handle surface integrals for parametric surfaces not aligned with coordinate planes?
3. Can you derive the formula for $\|\vec{r}_u \times \vec{r}_v\|$?
4. What are the conditions under which a surface integral can be evaluated using Stokes' Theorem?
5. How would the computation change if $f(x, y, z)$ were a vector field?

Tip: When switching to polar coordinates, always adjust $dx\,dy$ to $r \, dr \, d\theta$ carefully to avoid errors in the Jacobian transformation.