We are asked to find the surface area of the portion of the plane $z = 2x + 2y - 100$ that lies inside the cylinder $x^2 + y^2 = 1$. The surface area is given as $\alpha\pi$, and we need to find the value of $\alpha$.

Calculating Surface Area

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

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

Partial Derivatives

First, calculate the partial derivatives of $z$ with respect to $x$ and $y$:

• $ \frac{\partial z}{\partial x} = \frac{\partial}{\partial x}(2x + 2y - 100) = 2 $
• $ \frac{\partial z}{\partial y} = \frac{\partial}{\partial y}(2x + 2y - 100) = 2 $

Surface Area Element

Now, substitute these derivatives into the surface area formula's integrand:

$ \sqrt{1 + \left(\frac{\partial z}{\partial x}\right)^2 + \left(\frac{\partial z}{\partial y}\right)^2} = \sqrt{1 + (2)^2 + (2)^2} = \sqrt{1 + 4 + 4} = \sqrt{9} = 3 $

The integrand is a constant value, 3.

Defining the Region D

The region $D$ is the projection onto the xy-plane of the surface $S$. Since the surface lies inside the cylinder $x^2 + y^2 = 1$, the region $D$ is the disk defined by $x^2 + y^2 \leq 1$. This is a circle with radius $r=1$.

Evaluating the Integral

The surface area integral becomes:

$ \text{Surface Area} = \iint_D 3 \,dA $

Since 3 is a constant, we can pull it out of the integral:

$ \text{Surface Area} = 3 \iint_D \,dA $

The integral $\iint_D \,dA$ represents the area of the region $D$. The area of the disk $D$ (with radius 1) is $ \pi r^2 = \pi (1)^2 = \pi $.

Therefore, the surface area is:

$ \text{Surface Area} = 3 \times \pi = 3\pi $

Determining Alpha

We are given that the surface area is $ \alpha\pi $. By comparing this with our calculated surface area:

$ \alpha\pi = 3\pi $

Dividing both sides by $\pi$, we find:

$ \alpha = 3 $

The value of $\alpha$ is 3.