Question

Set up the iterated integral that computes the surface area of the given surface over the region \(R .\) \(f(x, y)=\frac{1}{x^{2}+y^{2}+1} ; \quad R\) is bounded by the circle \(x^{2}+\) \(y^{2}=9\).

Short answer

The iterated integral is: \[ A = \int_0^{2\pi} \int_0^3 \sqrt{1 + \frac{4r^2}{(r^2 + 1)^4}} \, r \, dr \, d\theta \]

Step-by-step solution

Understand the given problem

To find the surface area of the surface defined by the function \( f(x, y) = \frac{1}{x^2 + y^2 + 1} \) over the region \( R \), we need to set up an iterated integral. The region \( R \) is a circular area defined by \( x^2 + y^2 = 9 \).

Use the surface area integral formula

The surface area \( A \) of a surface defined by \( z = f(x, y) \) above a region \( R \) in the \( xy \)-plane is given by the integral formula: \[ A = \iint_R \sqrt{1 + \left( \frac{\partial f}{\partial x} \right)^2 + \left( \frac{\partial f}{\partial y} \right)^2} \, dA \]

Compute partial derivatives of \( f(x, y) \)

Calculate \( \frac{\partial f}{\partial x} \) and \( \frac{\partial f}{\partial y} \). For \( f(x, y) = \frac{1}{x^2 + y^2 + 1} \), \( \frac{\partial f}{\partial x} = \frac{-2x}{(x^2 + y^2 + 1)^2} \) and \( \frac{\partial f}{\partial y} = \frac{-2y}{(x^2 + y^2 + 1)^2} \).

Substitute partial derivatives into the surface area formula

Substitute the partial derivatives into the formula: \[ A = \iint_R \sqrt{1 + \left(\frac{-2x}{(x^2 + y^2 + 1)^2}\right)^2 + \left(\frac{-2y}{(x^2 + y^2 + 1)^2}\right)^2} \, dA \] Simplify the expression under the square root.

Simplify the expression under the square root

The expression under the square root becomes: \[ 1 + \frac{4x^2}{(x^2 + y^2 + 1)^4} + \frac{4y^2}{(x^2 + y^2 + 1)^4} \] Combine terms: \[ 1 + \frac{4(x^2 + y^2)}{(x^2 + y^2 + 1)^4} \] and further simplify this.

Switch to polar coordinates

The region \( R \) is a circle centered at the origin with radius 3. Switch to polar coordinates where \( x = r \cos \theta \) and \( y = r \sin \theta \). The limits for \( r \) are from 0 to 3, and for \( \theta \) from 0 to \( 2\pi \).

Write the iterated integral in polar coordinates

Substitute \( x = r\cos\theta \) and \( y = r\sin\theta \) into the integral. The differential \( dA \) becomes \( r \, dr \, d\theta \). The integrand is: \[ \sqrt{1 + \frac{4r^2}{(r^2 + 1)^4}} \] The iterated integral becomes: \[ A = \int_0^{2\pi} \int_0^3 \sqrt{1 + \frac{4r^2}{(r^2 + 1)^4}} \, r \, dr \, d\theta \]

Finalize the setup of the iterated integral

The iterated integral is fully set up, but evaluating it requires further computation or numerical methods: \[ A = \int_0^{2\pi} \int_0^3 \sqrt{1 + \frac{4r^2}{(r^2 + 1)^4}} \, r \, dr \, d\theta \] This integral will provide the surface area when solved.

Key concepts

Iterated Integrals

An iterated integral is a method used in calculus to evaluate a multiple integral by breaking it down into a sequence of single integrals. In our exercise, it is used to calculate the surface area of the region defined by the function \( f(x, y) = \frac{1}{x^2 + y^2 + 1} \) over a circular region \( R \). This involves performing integration in a step-by-step manner, first integrating with respect to one variable and then the other.
To set up an iterated integral, you define limits for each variable. In Cartesian coordinates, the limits are aligned with the boundaries of region \( R \).
Once we switch to polar coordinates, which are more suitable for circular regions, the iterated integral becomes:

• \( \int_0^{2\pi} \int_0^3 \sqrt{1 + \frac{4r^2}{(r^2 + 1)^4}} \, r \, dr \, d\theta \)

With \( r \) being the radial distance from the origin and \( \theta \) the angle, we first integrate with respect to \( r \) and then with respect to \( \theta \). This setup simplifies evaluating the area of curves and surfaces, especially when the boundaries are symmetric, as they are for a circle.

Partial Derivatives

Partial derivatives are used in multivariable calculus to measure how a function changes as individual variables change, while keeping other variables constant. In our exercise, the partial derivatives \( \frac{\partial f}{\partial x} \) and \( \frac{\partial f}{\partial y} \) are crucial in determining the slope of the surface described by \( f(x, y) = \frac{1}{x^2 + y^2 + 1} \).
To compute these, we treat all other variables but one as constants. For instance:

• \( \frac{\partial f}{\partial x} = \frac{-2x}{(x^2 + y^2 + 1)^2} \)
• \( \frac{\partial f}{\partial y} = \frac{-2y}{(x^2 + y^2 + 1)^2} \)

These derivatives help us find the slope in the \( x \)- and \( y \)-directions, respectively.
In setting up the integral for surface area, these partial derivatives contribute to the expression under the square root, giving us insight into how steeply the surface rises over the circular region defined by \( R \).

Polar Coordinates

Polar coordinates provide a powerful way to handle circular regions, like the one specified in this problem. Unlike Cartesian coordinates, which use \((x, y)\) pairs, polar coordinates use \((r, \theta)\) where:

• \( r \) is the distance from the origin.
• \( \theta \) is the angle measured from the positive x-axis.

For our region \( R \), which is the circle \( x^2 + y^2 = 9 \), using polar coordinates simplifies integration because:

• The radius \( r \) varies from 0 to 3 (since \( 9 = 3^2 \)).
• The angle \( \theta \) completes a full circle from 0 to \( 2\pi \).

In polar form, the transformation \( x = r \cos{\theta} \) and \( y = r \sin{\theta} \) means that the differential area element \( dA \) is equal to \( r \, dr \, d\theta \). This setup accommodates natural symmetry and simplifies many problems in circular regions.

Multivariable Calculus

Multivariable calculus is the branch of calculus dealing with functions of several variables. It extends the concepts of one-dimensional calculus, like derivatives and integrals, to higher dimensions. The exercise focuses on a two-variable function \( f(x, y) \) and demonstrates how to compute features like surface area.
With multivariable calculus, we can:

• Understand how functions change as we vary input variables together.
• Calculate surface areas, volumes, and other properties of multidimensional shapes.

When dealing with problems such as computing the surface area over a region, tools of multivariable calculus such as iterated integrals and partial derivatives simplify the process substantially.
These tools enable us to frame real-world phenomena in mathematical terms by integrating over regions that may not be readily manageable in simpler calculus. They open doors to solving complex geometry problems, optimizing functions with constraints, and much more in scientific and engineering contexts.