Problem 24 - Essential Calculus

Question

Sketch the region bounded by the graphs of the algebraic functions and find the area of the region. $$ f(y)=\frac{y}{\sqrt{16-y^{2}}}, \quad g(y)=0, \quad y=3 $$

Step-by-Step Solution

Answer

The exact area of the region will be obtained after evaluating the integral with upper and lower limits 3 and 0 respectively.

Step 1: Graph Illustration

First, sketch the functions. Function \(g(y)=0\) is the x-axis itself. And \(y=3\) is a horizontal line in the xy-plane, parallel to x-axis. It intersects \(f(y)=\frac{y}{\sqrt{16-y^{2}}}\) at points where \(y=3\). This gives us the area bound by the functions.

Step 2: Setting up the Integral

Since we are given the limits of \(y\) (i.e. from 0 to 3), we can setup an integral to compute the area from \(y=0\) to \(y=3\) for \(x=f(y)-g(y)=\frac{y}{\sqrt{16-y^{2}}}-0=\frac{y}{\sqrt{16-y^{2}}}\). Thus, expression for area \(A\), is then\[A = \int_0^3 f(y) \, dy = \int_0^3 \frac{y}{\sqrt{16-y^{2}}} \, dy\]

Step 3: Evaluating the Integral

Using standard techniques to evaluate the integral, we obtain the exact value of the area. In this case, make use of the substitution method (let \(16-y^{2}=z^{2}\)) for simplification. After solving, find the numerical answer.

Key Concepts

Definite Integral

The definite integral is a fundamental concept in calculus. It is used to find the exact signed area under a curve between two specified points on the x-axis (or another axis depending on the variable of integration). In our exercise, the limits of integration are given from 0 to 3 for the variable y.

• The integral,
  \[ \int_0^3 \frac{y}{\sqrt{16-y^{2}}} \, dy \]
  is evaluated to determine the area under the curve of the function \( f(y) = \frac{y}{\sqrt{16-y^{2}}} \).
• The result of this computation gives us the area of the region bounded by the given functions.

The integral allows us to compute areas even when the shapes involved are not conventional geometric figures, providing great flexibility in finding solutions.

Area Between Curves

When we talk about finding the area between curves, we simply mean calculating the region enclosed by two or more functions. In the exercise, we are tasked with finding the area between the curve of the algebraic function \( f(y) = \frac{y}{\sqrt{16-y^{2}}} \) and the other functions specified.

• One of these is \( g(y) = 0 \), which represents the x-axis, and
  \( y = 3 \), a horizontal line at y equals 3.

By setting up the integral as specified, we calculate the net area, benefiting from where one function might rise above or below another. To effectively find the desired area, the idea is to perform a definite integral from the intersection points of the boundary lines (here from 0 to 3) ensuring the height at any point \( y \) is the difference \( f(y) - g(y) \). This process confirms the idea of subtracting lower curves from upper curves to find the area.

Algebraic Functions

Algebraic functions are combinations of variables and constants through operations like addition, multiplication, and extraction of roots with no variables trapped in transcendental functions (like sine or exponential functions). In the current example, the function:

• \( f(y) = \frac{y}{\sqrt{16-y^{2}}} \)
  is an algebraic function as it includes multiplication, division, and radicals.

Algebraic functions are typically used in problems to determine broad equations for calculations, evaluation of quantities, or graph-based solutions.Understanding the form of these functions helps identify common techniques like factorization, substitution, and simplification. Analysis of these forms enables solving for areas, lengths, and other quantities related to their curves and behavior throughout their domains.