This question asks us to find the surface area of a specific portion of the plane defined by the equation $x + 2y + 2z = 12$. This portion is enclosed by the coordinate planes $x = 0$, $y = 0$, and the cylinder $x^2 + y^2 = 16$. Let's break down the steps to calculate this surface area.

Understanding the Plane Surface

The given plane is $x + 2y + 2z = 12$. To work with surface area calculations, it's often easier to express one variable in terms of the others. We can rewrite the equation to solve for $z$:

$2z = 12 - x - 2y$

$z = \frac{12 - x - 2y}{2}$

$z = 6 - \frac{1}{2}x - y$

This equation represents the plane as a function $z = f(x, y)$, where $f(x, y) = 6 - \frac{1}{2}x - y$.

Calculating the Surface Area Element

The formula for the surface area $S$ of a surface $z = f(x, y)$ over a region $D$ in the xy-plane is given by:

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

First, we need to find the partial derivatives of $z$ with respect to $x$ and $y$:

• $\frac{\partial z}{\partial x} = \frac{\partial}{\partial x} \left(6 - \frac{1}{2}x - y\right) = -\frac{1}{2}$
• $\frac{\partial z}{\partial y} = \frac{\partial}{\partial y} \left(6 - \frac{1}{2}x - y\right) = -1$

Now, we substitute these into the square root term:

$\sqrt{1 + \left(-\frac{1}{2}\right)^2 + (-1)^2} = \sqrt{1 + \frac{1}{4} + 1} = \sqrt{2 + \frac{1}{4}} = \sqrt{\frac{8}{4} + \frac{1}{4}} = \sqrt{\frac{9}{4}} = \frac{3}{2}$

So, the surface area integral becomes:

$S = \iint_D \frac{3}{2} \, dA$

Defining the Region of Integration D

The region $D$ in the xy-plane is determined by the boundaries given: $x = 0$, $y = 0$, and $x^2 + y^2 = 16$.

• $x = 0$ is the yz-plane.
• $y = 0$ is the xz-plane.
• $x^2 + y^2 = 16$ is a cylinder of radius 4 centered around the z-axis.

Since we are bounded by $x=0$ and $y=0$, and the cylinder $x^2+y^2=16$, the region $D$ is the part of the circular disk $x^2 + y^2 \le 16$ that lies in the first quadrant (where both $x \ge 0$ and $y \ge 0$). This shape is a quarter-disk of radius 4.

Calculating the Area of Region D

The integral $\iint_D dA$ represents the area of the region $D$. Since $D$ is a quarter of a circle with radius $r = 4$, its area is:

$\text{Area}(D) = \frac{1}{4} \times (\text{Area of a circle with radius 4})$

$\text{Area}(D) = \frac{1}{4} \times \pi r^2 = \frac{1}{4} \times \pi (4^2)$

$\text{Area}(D) = \frac{1}{4} \times 16\pi = 4\pi$

Final Surface Area Calculation

Now we can calculate the total surface area $S$ using the result from the integral setup:

$S = \frac{3}{2} \iint_D dA = \frac{3}{2} \times \text{Area}(D)$

$S = \frac{3}{2} \times 4\pi$

$S = 6\pi$

Therefore, the surface area of the plane cut off by the given boundaries is $6\pi$.