Problem 66 - Essential Calculus

Question

Find the area of the surface generated by revolving the curve about each given axis. $$ x=\frac{1}{3} t^{3}, y=t+1, \quad 1 \leq t \leq 2, \quad y \text { -axis } $$

Step-by-Step Solution

Answer

The surface area of the revolution is defined by the definite integral \(2 \pi\int_{1}^{2} \frac{1}{3}(y-1)^3 \sqrt{1+ [(y-1)^2]^2} dy\), which could be solved by using various techniques of integration.

Step 1: Transform the coordinate equation

Given the equations \(x = \frac{1}{3}t^3\) and \(y = t+1\), we need to write x as a function of y. First, solve the equation \(y = t+1\) for \(t\), obtaining \(t = y-1\). Next, substitute \(t = y-1\) into the first equation and simplify to obtain \(x = \frac{1}{3}(y-1)^3\).

Step 2: Calculate the derivative of x

To calculate the surface area of revolution, we need the derivative of \(x\) with respect to \(y\). Differentiating \(x = \frac{1}{3}(y-1)^3\) using the chain rule yields \(x' = (y-1)^2\).

Step 3: Apply the formula for surface area of revolution

Substitute into the surface area of revolution formula, we get \(2 \pi\int_{1}^{2} \frac{1}{3}(y-1)^3 \sqrt{1+ [(y-1)^2]^2} dy\). Use integral calculus to evaluate this expression.

Step 4: Calculate the definite integral

The computation of definite integral require techniques of integration. This is a non-trivial task and might require numerical approximation.

Key Concepts

Definite Integrals

Definite integrals are a powerful tool in calculus that allow us to calculate the net area under a curve between two points. When dealing with revolved surfaces, definite integrals help us find the surface area generated by a curve revolving around an axis.
In our given exercise, we substitute derived parametric equations into the surface area formula to compute a well-defined integral. This involves evaluating a curve from one point to another, typically across a specified interval like from 1 to 2, as seen in the exercise.

• The limits of integration are taken from the range assigned to the parameter, in this case, 1 to 2 for the variable t.
• The definite integral takes into account the entire curve by covering all the infinitesimal strips between these points.

These strips, revolving around the axis, build up the surface area. While solving, numerical integration methods can be used if the integral isn't easily solvable analytically, due to complexity or non-standard integrals.

Parametric Equations

In mathematical modeling, parametric equations define a set of quantities as explicit functions of one or more independent parameters. This allows for flexibility in describing curves and surfaces in mathematics.
For the problem at hand, parametric equations were initially stated as:

• \(x = \frac{1}{3}t^3\)
• \(y = t+1\)

These provide a method to describe curves independently of the traditional Cartesian coordinates, splitting complex motions into simpler, one-dimensional movements controlled by the parameter, t. By transforming these equations appropriately, such as solving \(y = t + 1\) for \(t\) and substituting back, we derive another parametric form involving \(x\) and \(y\) alone.
It's particularly valuable when dealing with curves and computing associated values such as lengths, areas, and volumes when revolution occurs.

Chain Rule

The chain rule is an essential differentiation technique used when dealing with composite functions. This is especially useful in problems involving parametric equations and derivatives.
Consider our transformed equation, \(x = \frac{1}{3}(y-1)^3\). It is necessary to differentiate this before moving to integration for surface area. The chain rule simplifies this process by allowing us to break down derivatives into manageable parts, which in turn helps computing complex expressions:

• Since \(x\) depends on \(y\), the derivative \(x'\) is found using the chain rule, resulting in \(x' = (y-1)^2\).
• This differentiated term is then plugged into the integral for calculating the surface area of revolution.

Without the chain rule, managing large expressions that arise from parametric equations could become cumbersome. It becomes a cornerstone of calculus, enabling us to easily maneuver through advanced problems involving integrals.