Question

Let \(z=f(x, y)\) and \(z=g(x, y)=2 f(x, y) .\) Why is the surface area of \(g\) over a region \(R\) not twice the surface area of \(f\) over \(R ?\)

Short answer

The surface area of \( g \) isn't twice \( f \) because the derivatives contribute quadratically, not linearly, to the area.

Step-by-step solution

Understanding Surface Area Formula

The surface area of a function \( z = f(x, y) \) over a region \( R \) in the xy-plane is given by the integral \[ \int \int_{R} \sqrt{1 + \left( \frac{\partial f}{\partial x} \right)^2 + \left( \frac{\partial f}{\partial y} \right)^2 } \ dA, \] where \( dA \) represents an infinitesimal area element in the region \( R \).

Applying the Surface Area Formula to g(x, y)

When \( z = g(x, y) = 2 f(x, y) \), the new partial derivatives become \( \frac{\partial g}{\partial x} = 2 \frac{\partial f}{\partial x} \) and \( \frac{\partial g}{\partial y} = 2 \frac{\partial f}{\partial y} \). Therefore, the surface area integral for \( g(x, y) \) is \[ \int \int_{R} \sqrt{1 + (2 \frac{\partial f}{\partial x})^2 + (2 \frac{\partial f}{\partial y})^2 } \ dA. \]

Simplifying the Surface Area Formula for g(x, y)

The expression inside the square root becomes \( 1 + 4 \left( \frac{\partial f}{\partial x} \right)^2 + 4 \left( \frac{\partial f}{\partial y} \right)^2 \). Hence, the integral for \( g \) is \[ \int \int_{R} \sqrt{1 + 4 \left( \frac{\partial f}{\partial x} \right)^2 + 4 \left( \frac{\partial f}{\partial y} \right)^2 } \ dA. \]

Comparing the Terms and Surface Areas

The terms inside the square root for \( g(x, y) \) are not simply double those for \( f(x, y) \), due to the quadratic increase in the partial derivative terms. This makes the expression inside the square root in \( g \) fundamentally different from simply twice the expression for \( f \). As a result, the surface area of \( g \) is not simply twice the surface area of \( f \).

Key concepts

Partial Derivatives

In multi-variable calculus, partial derivatives are used to understand how a function changes when one of the variables is slightly adjusted, while all other variables are held constant. This is especially useful in analyzing functions of more than one variable, like the surface described by \( z = f(x, y) \). Imagine you're hiking on a mountain. The slope of the path in the direction of one axis can be considered a partial derivative. It tells you how steep the mountain is in that very direction.
For a function \( f(x, y) \), the partial derivative with respect to \( x \) is denoted as \( \frac{\partial f}{\partial x} \), and the partial derivative with respect to \( y \) is \( \frac{\partial f}{\partial y} \).

• \( \frac{\partial f}{\partial x} \): Measures the rate of change of the function as the \( x \)-coordinate changes, maintaining the \( y \)-coordinate constant.
• \( \frac{\partial f}{\partial y} \): Indicates the rate of change with changes in \( y \) while \( x \) stays fixed.

These derivatives become part of the formula to calculate the surface area of a function, contributing to understanding how the function "bends" across the surface.

Gradient Vector

The gradient vector is a powerful tool that combines the partial derivatives of a function. This vector points in the direction of the steepest ascent on a surface, and its magnitude tells us how steep the slope is in that direction. Think of it as a compass that guides you uphill.
For a function \( f(x, y) \), the gradient is represented as \( abla f \) and is given by the vector:\[ abla f = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right) \]

• Direction: The gradient indicates the direction in which the function increases most rapidly. If you want to climb the mountain as quickly as possible, follow the direction of the \( abla f \).
• Magnitude: The length of the gradient vector denotes how steep the hill is. Larger magnitudes mean steeper ascents.

For functions describing surfaces, the gradient vector plays a crucial role in determining how changes in \( x \) and \( y \) interplay to affect the shape of the surface.

Integrals

Integrals are fundamental in calculating areas under curves or, in this case, the surface area of parametric surfaces. In the context of multi-variable functions like \( f(x, y) \), the double integral computes the total surface area over a defined region \( R \). Imagine covering the entire area under a surface with tiny rectangles and summing them up; that’s what the integral does.
The specific formula for a surface area integral for a function \( z = f(x, y) \) is:\[\int \int_{R} \sqrt{1 + \left( \frac{\partial f}{\partial x} \right)^2 + \left( \frac{\partial f}{\partial y} \right)^2 } \ dA\]

• Element \( dA \): Represents a tiny piece of area over which the function is integrated.
• Expression Inside the Integral: \( \sqrt{1 + \left( \frac{\partial f}{\partial x} \right)^2 + \left( \frac{\partial f}{\partial y} \right)^2 } \) gives the adjustment needed to account for the slope of \( f(x, y) \).

Understanding integrals helps you grasp why scaling \( z \) to \( g(x, y) = 2f(x, y) \) doesn’t just double the surface area — the adjustment terms inside the square root grow quadratically with \( x \) and \( y \), altering the computed area substantially.