Question

Let \( S \) be the surface area of the portion of the plane \( z = x + y + 3 \), which lies inside the cylinder \( x^2 + y^2 = 1 \). Then, the value of \( \left( \frac{S}{\pi} \right)^2 \) is equal to ............. (rounded off to two decimal places).

Hint:

For surface area integrals, always use polar coordinates when dealing with circular regions. The integrand for a plane surface is \( \sqrt{1 + \left( \frac{\partial z}{\partial x} \right)^2 + \left( \frac{\partial z}{\partial y} \right)^2} \).

Answer 1

Step 1: Formula for surface area.
The surface area of a surface \( z = f(x, y) \) over a region \( R \) is given by: \[ S = \iint_R \sqrt{1 + \left( \frac{\partial z}{\partial x} \right)^2 + \left( \frac{\partial z}{\partial y} \right)^2} \, dx \, dy. \] For the given plane \( z = x + y + 3 \), we compute the partial derivatives: \[ \frac{\partial z}{\partial x} = 1, \quad \frac{\partial z}{\partial y} = 1. \] Thus, the integrand becomes: \[ \sqrt{1 + 1^2 + 1^2} = \sqrt{3}. \] Step 2: Set up the integral.
The region \( R \) is the disk \( x^2 + y^2 \leq 1 \), which is a circle with radius 1. Using polar coordinates, where \( x = r \cos \theta \), \( y = r \sin \theta \), and \( dx \, dy = r \, dr \, d\theta \), the surface area integral becomes: \[ S = \int_0^{2\pi} \int_0^1 \sqrt{3} \cdot r \, dr \, d\theta. \] Step 3: Evaluate the integral.
First, integrate with respect to \( r \): \[ \int_0^1 r \, dr = \frac{1}{2}. \] Now, integrate with respect to \( \theta \): \[ \int_0^{2\pi} d\theta = 2\pi. \] Thus, the surface area is: \[ S = \sqrt{3} \cdot \frac{1}{2} \cdot 2\pi = \pi \sqrt{3}. \] Step 4: Calculate \( \left( \frac{S}{\pi} \right)^2 \).
We have: \[ \frac{S}{\pi} = \sqrt{3}, \quad \left( \frac{S}{\pi} \right)^2 = 3. \] Final Answer: \[ \boxed{3}. \]