Question

Find the area of the given surface over the region \(R\). \(f(x, y)=x^{2}+y^{2}+10 ; R\) is bounded by the circle \(x^{2}+y^{2}=\) \(16 .\)

Short answer

The area of the surface is \( 81\pi \) square units.

Step-by-step solution

Understand the Problem

We need to find the area of the surface defined by the function \( f(x, y) = x^2 + y^2 + 10 \) over the circular region \( R \) where \( x^2 + y^2 \leq 16 \). This is essentially finding a surface integral over the given region.

Parametrize the Region

The circular region \( R \) is centered at the origin with radius 4, as \( x^2 + y^2 \leq 16 \). Since this is a circle, it is easier to use polar coordinates where \( x = r \cos \theta \) and \( y = r \sin \theta \), with \( 0 \leq r \leq 4 \) and \( 0 \leq \theta \leq 2\pi \).

Surface Integral Formula

The surface area of a function \( z = f(x, y) \) over a region \( R \) is given by the integral \[ \iint_R \sqrt{1 + (f_x)^2 + (f_y)^2} \, dA \] where \( f_x \) and \( f_y \) are partial derivatives of \( f \).

Calculate Partial Derivatives

Compute the partial derivatives of \( f(x, y) \) with respect to \( x \) and \( y \):- \( f_x = \frac{\partial}{\partial x}(x^2 + y^2 + 10) = 2x \).- \( f_y = \frac{\partial}{\partial y}(x^2 + y^2 + 10) = 2y \).

Set Up the Integral

Substitute the partial derivatives into the surface integral formula:\[ \iint_R \sqrt{1 + (2x)^2 + (2y)^2} \, dA = \iint_R \sqrt{1 + 4x^2 + 4y^2} \, dA \].Convert to polar coordinates:\[ \int_0^{2\pi} \int_0^4 \sqrt{1 + 4r^2} \, r \, dr \, d\theta \].

Evaluate the Integral

Evaluate the inner integral with respect to \( r \):\[ \int_0^4 \sqrt{1 + 4r^2} \, r \, dr \].Let \( u = 1 + 4r^2 \), then \( du = 8r \, dr \), hence \( r \, dr = \frac{1}{8} \, du \). The limits for \( u \) are 1 to 65.This integral becomes \( \frac{1}{8} \int_1^{65} \sqrt{u} \, du \), solve to get the antiderivative:\( \frac{1}{8} \left[ \frac{2}{3} u^{3/2} \right]_1^{65} \). Calculate this: \( \frac{1}{8} \left( \frac{2}{3} (65^{3/2} - 1^{3/2}) \right) \).

Evaluate Outer Integral

The result from the inner integral is the value \( V \). Now evaluate:\[ \int_0^{2\pi} V \, d\theta = V \times (2\pi) \].Multiply the result of the inner evaluated integral by \( 2\pi \) to obtain the final surface area.

Key concepts

Polar Coordinates

To tackle problems involving circular regions, polar coordinates are an invaluable tool. Instead of using the traditional Cartesian coordinates, where we describe points using \(x\) and \(y\), polar coordinates allow us to express a point in terms of its distance \(r\) from the origin and the angle \(\theta\) it makes with the positive x-axis. This is particularly useful for circular areas.

For the given problem, the circular region defined by \(x^2 + y^2 \leq 16\) is a circle centered at the origin with radius 4. Using polar coordinates, we switch to \(x = r \cos \theta\) and \(y = r \sin \theta\). The variable \(r\) varies from 0 to 4, while \(\theta\) covers the full circle, from 0 to \(2\pi\).

This transformation simplifies the process of integration over circular domains, as it naturally fits the geometry and makes calculus operations more straightforward.

Partial Derivatives

Partial derivatives are a fundamental concept in multivariable calculus, involving functions of more than one variable. They measure how the function changes as you vary one variable, holding the others constant.

For our function \(f(x, y) = x^2 + y^2 + 10\), the partial derivative with respect to \(x\), denoted \(f_x\), is found by differentiating \(f\) while treating \(y\) as a constant. This gives:

\(f_x = \frac{\partial}{\partial x}(x^2 + y^2 + 10) = 2x\)

In a similar fashion, the partial derivative with respect to \(y\), \(f_y\), results in:

\(f_y = \frac{\partial}{\partial y}(x^2 + y^2 + 10) = 2y\)

These partial derivatives are crucial when calculating the surface integral, as they help determine how the surface rises and falls in different directions.

Integral Calculus

Integral calculus helps us calculate quantities like areas, volumes, and other accumulations. For surface area calculation over a region, we utilize a multiple integration approach.

In this exercise, we need to compute the surface area of the function \(f(x, y)\) over the circular region \(R\). The surface integral we use is:

\[ \iint_R \sqrt{1 + (f_x)^2 + (f_y)^2} \, dA \]

This expression takes into account the slopes in the x and y directions, providing a way to add up "pieces" of area on our surface. Converting to polar coordinates makes integration over circular regions seamless, by using \(r\, dr\, d\theta\) as the new area element \(dA\).

This integral simplification enables us to tackle complex area problems, ensuring accurate measurements over geometrically challenging regions.

Surface Area Calculation

Surface area calculation in calculus involves evaluating integrals that sum infinitesimal elements of a surface. For our problem, replacing the double integral with polar coordinates leads to the following integral over the circular region:

\[ \int_0^{2\pi} \int_0^4 \sqrt{1 + 4r^2} \, r \, dr \, d\theta \]

This setup now requires evaluating the inner integral with respect to \(r\). We use substitution to simplify it, by letting \(u = 1 + 4r^2\), which reformulates the integral into:

\[ \frac{1}{8} \int_1^{65} \sqrt{u} \, du \]

The calculated result then gets multiplied by \(2\pi\) from the outer integral, culminating in the total surface area. This layered approach demonstrates how integral calculus enables us to accurately determine the surface area spanning a curved surface, through systematic substitution and careful integration.