Problem 55 - Calculus: Early Transcendental Functions

Question

Find the surface area extending from the given curve in the \(x y\) -plane to the given surface. Above the quarter-circle of radius 2 centered at the origin from (2,0,0) to (0,2,0) up to the surface \(z=x^{2}+y^{2}.\)

Step-by-Step Solution

Answer

The surface area can be found by setting up and evaluating the double integral \(\int_{0}^{2} \int_{0}^{\pi / 2} \sqrt{1 + 4r^{2}} r dr d\theta\).

Step 1: Understanding the Problem

The problem involves finding the surface above the quarter-circle in the xy-plane up to the surface \(z=x^{2}+y^{2}\). The quarter circle is centered at the origin with a radius of 2. Because of the presence of quarter-circle and the equation of the surface, it would be easier to solve the problem in polar coordinates.

Step 2: Converting to Polar Coordinates

In order to convert the problem into polar coordinates, replace \(x = r \cos \theta \) and \( y = r \sin \theta \), where r is radius and \(\theta\) is angle. The quarter-circle of radius 2 will correspond to \(0 \leq r \leq 2\) and \(0 \leq \theta \leq \pi / 2\). The surface \(z = x^{2} + y^{2}\) becomes \(z = r^{2}\) in polar coordinates.

Step 3: Setting up the Double Integral

To compute the surface area over the quarter-circle we need to evaluate double integral over the region D in xy-plane. Surface area, A = \(\int \int_{D} \sqrt{1 + (\frac{\partial z}{\partial x})^{2} + (\frac{\partial z}{\partial y})^{2}} dA \), where \(\frac{\partial z}{\partial x} = 2x = 2r\cos\theta\), \(\frac{\partial z}{\partial y} = 2y = 2r\sin\theta\), and dA becomes \(rdrd\theta\).

Step 4: Compute the Surface Area

First, simplify what goes under square root in the integral. This equals to: \(\sqrt{1 + (2r\cos\theta)^{2} + (2r\sin\theta)^{2}}\), after computations equals to \(\sqrt{1 + 4r^{2}}\). Now compute the double integral: A = \(\int_{0}^{2} \int_{0}^{\pi / 2} \sqrt{1 + 4r^{2}} r dr d\theta\).

Step 5: Evaluating the Double Integral

After evaluating the double integral numerically or by using some computational software, you should get the value of the surface area.

Key Concepts

Polar Coordinates

When dealing with curves and areas on a plane, particularly those involving circular shapes like circles or arcs, polar coordinates are incredibly useful. Unlike the traditional Cartesian system, polar coordinates express points in terms of distance from the origin and angle from the positive x-axis. This system can simplify calculations involving curves, particularly when circular symmetry is present.

In this problem, we know that we are working with a quarter-circle centered at the origin with a radius of 2. Therefore, to convert Cartesian coordinates

• Use the formulas: \(x = r \cos \theta\) and \(y = r \sin \theta\).

These expressions allow us to represent any point on or within the circle in terms of its radius and angle.

Our region in polar coordinates, specifically, is given by:

• Radius (\(r\)): \(0 \leq r \leq 2\)
• Angle (\(\theta\)): \(0 \leq \theta \leq \pi/2\)

This specifies the quarter-circle extending from the x-axis to the y-axis, covering one-fourth of a full circle.

Double Integral

The double integral in this context allows us to find the surface area over a defined region in the plane. The expression starts with the basic double integral form over region \(D\)

• \(A = \int \int_{D} \sqrt{1 + (\frac{\partial z}{\partial x})^{2} + (\frac{\partial z}{\partial y})^{2}} \, dA\)

Here, \(dA\) represents a small area element in the direction of integration, which, when transformed to polar coordinates, becomes \(rdrd\theta\).

The double integral works in two stages:

• First, integrate with respect to \(r\) over its entire range (from 0 to 2).
• Then, integrate with respect to \(\theta\) (from 0 to \(\pi/2\)).

This layered approach captures all small "slices" of the region \(D\), allowing for the calculation of a total area by summing these slices together. Here, employing computational tools often aids in evaluating complex integrals like these, particularly when their solutions are not straightforward.

Partial Derivatives

Partial derivatives are crucial in the computation of surface areas when dealing with functions of multiple variables. They measure how the function changes as each variable is altered, while holding others constant.

• For the surface \(z = x^2 + y^2\), these derivatives express how \(z\) changes with variations in \(x\) and \(y\).

In this problem,

• The partial derivative with respect to \(x\) is \(\frac{\partial z}{\partial x} = 2x\).
• Similarly, the partial with respect to \(y\) is \(\frac{\partial z}{\partial y} = 2y\).

These partial derivatives are fundamental to setting up the integrand for the surface area formula. Transitioning to polar coordinates further refines these:

• \(\frac{\partial z}{\partial x} = 2r \cos \theta\)
• \(\frac{\partial z}{\partial y} = 2r \sin \theta\)

Including these in the integrand of the double integral \(\sqrt{1 + (2r\cos\theta)^2 + (2r\sin\theta)^2}\), leads to the calculation of the area above the quarter-circle. Recognizing such derivatives' role empowers us to solve more complex surface area problems in multi-variable calculus.