Question

Set up the iterated integral that computes the surface area of the given surface over the region \(R .\) \(f(x, y)=x^{2}-y^{2} ; \quad R\) is the rectangle with opposite corners (-1,-1) and (1,1).

Short answer

The iterated integral for the surface area is \( \int_{-1}^{1} \int_{-1}^{1} \sqrt{1 + 4x^2 + 4y^2} \, dx \, dy \).

Step-by-step solution

Understand the Formula

To find the surface area of a surface given by a function \( f(x, y) \) over a region \( R \), we use the formula: \[ \iint_R \sqrt{1 + \left( \frac{\partial f}{\partial x} \right)^2 + \left( \frac{\partial f}{\partial y} \right)^2} \, dx \, dy \].

Key concepts

Iterated Integrals

Iterated integrals are an integral of an integral, often used to calculate volumes or areas in multivariable calculus. Here, they help us compute surface areas by integrating over both variables associated with a surface function. In our case, the function is \( f(x, y) = x^2 - y^2 \). When computing the surface area using iterated integrals, each variable gets its own integral operation, one inside the other.

• The integral over the \( x \)-direction: This represents calculating the area slice by slice in the horizontal direction.
• The integral over the \( y \)-direction: Here, the area is computed vertically across the region.

By setting limits from -1 to 1 for both \( x \) and \( y \), we calculate the surface area over the rectangle \( R \) with opposite corners (-1, -1) and (1,1). Iterating both integrals gives a total surface area as an accumulation of these tiny area segments. This concept helps simplify calculus problems into more manageable parts by focusing on one dimension at a time.

Partial Derivatives

Partial derivatives represent the change of a function concerning one of its variables while keeping others constant, which is essential when analyzing functions of multiple variables like \( f(x, y) = x^2 - y^2 \). In our problem, we calculate two partial derivatives: \( \frac{\partial f}{\partial x} \) and \( \frac{\partial f}{\partial y} \).

• \( \frac{\partial f}{\partial x} = 2x \): This shows how the function changes as \( x \) changes, keeping \( y \) fixed.
• \( \frac{\partial f}{\partial y} = -2y \): This captures the function's variation with \( y \), keeping \( x \) static.

By utilizing partial derivatives, we obtain a gradient of the surface, necessary for calculating more complex aspects like the rate of change. These are plugged back into the surface area formula, giving us the needed components to compute the surface area over region \( R \). Partial derivatives therefore form the foundational tools in multivariable calculus, giving insights into surface behavior and characteristics.

Surface Area Calculation

Calculating the surface area of a function over a specified region involves several key steps, particularly when using multivariable calculus techniques. For the function \( f(x, y) = x^2 - y^2 \) over region \( R \), we use a formula integrating the function's complexity into concise calculations: \[\iint_R \sqrt{1 + \left( \frac{\partial f}{\partial x} \right)^2 + \left( \frac{\partial f}{\partial y} \right)^2} \, dx \, dy\]

• Plug the partial derivatives from earlier into the formula: \( \frac{\partial f}{\partial x} = 2x \) and \( \frac{\partial f}{\partial y} = -2y \).
• Calculate the expression under the square root: \( 1 + (2x)^2 + (-2y)^2 \).
• Simplify this to form: \( 1 + 4x^2 + 4y^2 \).
• Perform iterated integration over \( x \) and \( y \) from -1 to 1 for the entire area, effectively capturing the surface's entirety in its domain.

This approach highlights the power of calculus to transform complex spatial problems into tractable calculations. It allows accurate modeling and computation of surfaces, vital for scientific and engineering applications.