The problem asks for the surface area of the portion of the paraboloid $z = x^2 + y^2$ bounded by the planes $z = 0$ and $z = \frac{1}{4}$.

Surface Area Integral Setup

The surface area $A$ for a function $z = f(x, y)$ over a region $D$ in the xy-plane is calculated using the formula:

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

For $z = x^2 + y^2$, we find the partial derivatives:

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

The integrand becomes:

$ \sqrt{1 + (2x)^2 + (2y)^2} = \sqrt{1 + 4(x^2 + y^2)} $

Defining the Region D

The paraboloid portion is between $z=0$ and $z=\frac{1}{4}$. This corresponds to $x^2+y^2 = 0$ and $x^2+y^2 = \frac{1}{4}$.

The projection $D$ onto the xy-plane is the disk $x^2 + y^2 \le \frac{1}{4}$.

Switching to Polar Coordinates

Using polar coordinates ($r^2 = x^2 + y^2$, $dA = r \, dr \, d\theta$), the integral becomes easier.

• The integrand is $\sqrt{1 + 4r^2}$.
• The region $D$ is defined by $0 \le r \le \frac{1}{2}$ and $0 \le \theta \le 2\pi$.

The integral is:

$ A = \int_{0}^{2\pi} \int_{0}^{1/2} \sqrt{1 + 4r^2} \cdot r \, dr \, d\theta $

Evaluating the Integral

First, evaluate the inner integral $\int_{0}^{1/2} r\sqrt{1 + 4r^2} \, dr$.

Use substitution: let $u = 1 + 4r^2$, then $du = 8r \, dr$, so $r \, dr = \frac{1}{8} du$. Change limits:

• When $r=0$, $u=1$.
• When $r=\frac{1}{2}$, $u=1 + 4(\frac{1}{4}) = 2$.

The inner integral evaluates to:

$ \int_{1}^{2} \sqrt{u} \cdot \frac{1}{8} du = \frac{1}{8} \left[ \frac{u^{3/2}}{3/2} \right]_{1}^{2} = \frac{1}{12} \left[ u^{3/2} \right]_{1}^{2} = \frac{1}{12} (2^{3/2} - 1^{3/2}) = \frac{1}{12} (2\sqrt{2} - 1) $

Now, integrate with respect to $\theta$:

$ A = \int_{0}^{2\pi} \frac{1}{12} (2\sqrt{2} - 1) \, d\theta = \frac{1}{12} (2\sqrt{2} - 1) [\theta]_{0}^{2\pi} $

$ A = \frac{1}{12} (2\sqrt{2} - 1) (2\pi) = \frac{2\pi}{12} (2\sqrt{2} - 1) = \frac{\pi}{6} (2\sqrt{2} - 1) $

Conclusion

The surface area is $\frac{\pi}{6}(2\sqrt{2}-1)$.