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Chapter 1: Problem 123

Find the area of the surface generated by revolving \(x=t^{2}, y=2 t, 0 \leq t \leq 4\) about the \(x\) -axis.

Short Answer

Expert verified
The surface area is \( \frac{8\pi}{3} (17\sqrt{17} - 1) \).

Step by step solution

01

Set Up the Surface Area Formula

For a parametric curve given by \( x = f(t) \) and \( y = g(t) \), revolving around the \( x \)-axis, the formula for the surface area \( S \) is: \[ S = 2\pi \int_{a}^{b} |y| \sqrt{\left( \frac{dx}{dt} \right)^2 + \left( \frac{dy}{dt} \right)^2} \, dt \].
02

Identify the Functions

From the parametric equations, we have \( x = t^2 \) and \( y = 2t \). So, \( f(t) = t^2 \) and \( g(t) = 2t \). The limits of \( t \) are from 0 to 4.
03

Compute Derivatives

Find the derivatives \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \). We get: \[ \frac{dx}{dt} = \frac{d}{dt}(t^2) = 2t \] and \[ \frac{dy}{dt} = \frac{d}{dt}(2t) = 2. \]
04

Substitute in the Integral

Substitute the functions into the surface area formula: \[ S = 2\pi \int_{0}^{4} |2t| \sqrt{(2t)^2 + (2)^2} \, dt = 2\pi \int_{0}^{4} 2t \sqrt{4t^2 + 4} \, dt. \]
05

Simplify the Integral

Simplify the expression inside the integral: \[ \sqrt{4t^2 + 4} = \sqrt{4(t^2 + 1)} = 2\sqrt{t^2 + 1}. \] So, the integral becomes: \[ S = 8\pi \int_{0}^{4} t \sqrt{t^2 + 1} \, dt. \]
06

Use Substitution

Use a substitution method where \( u = t^2 + 1 \), then \( du = 2t \, dt \) or \( t \, dt = \frac{1}{2} \, du \) to transform the integral limits and the expression. \[ S = 4\pi \int_{1}^{17} \sqrt{u} \, du. \]
07

Integrate

Find the integral of \( \sqrt{u} \): \[ \int \sqrt{u} \, du = \frac{2}{3} u^{3/2} + C. \] Apply this from 1 to 17: \[ S = 4\pi \left[ \frac{2}{3} u^{3/2} \right]_{1}^{17}. \]
08

Evaluate the Definite Integral

Substitute the limits into the antiderivative: \[ S = 4\pi \left[ \frac{2}{3} (17)^{3/2} - \frac{2}{3} (1)^{3/2} \right]. \] Simplify: \[ S = 4\pi \left( \frac{2}{3} (17\sqrt{17}) - \frac{2}{3} \right). \]
09

Simplify the Expression

Factor out \( \frac{2}{3} \times 4 \pi \) and combine: \[ S = \frac{8\pi}{3} (17\sqrt{17} - 1). \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Parametric Equations
Parametric equations provide a convenient way to represent curves in a plane using a pair of functions that define the coordinates
  • Typically, these are denoted by functions of a parameter, often labelled as \( t \).
  • In the problem, the parametric equations are given as \( x = t^2 \) and \( y = 2t \).
These equations define a curve in the \( xy \)-plane as \( t \) varies over a specified interval. Understanding parametric equations allows you to describe complex curves by breaking them into simpler functions of one variable. Instead of solving for \( y \) as a function of \( x \), parametric equations give a direct way to handle more intricate path geometries.When dealing with parametric curves, recognizing the changes in \( t \) is crucial because they define how points move along the curve.
Substitution Method
In calculus, the substitution method is a valuable tool used to simplify the process of integrating functions, especially in definite integrals.
  • It involves replacing a complicated part of the integrand with a new variable, simplifying the integration process.
  • In this exercise, a substitution \( u = t^2 + 1 \) was used.
  • The derivative of \( u \) with respect to \( t \), \( du = 2t \, dt \), helps in transforming the integral.
When using substitution, it is important to:- Change the limits of integration to reflect the new variable.- Adjust the differential and properly substitute all \( t \, dt \) terms with \( du \).By doing so, this method transforms the integrand into a simpler form, which is easier to integrate.This method, thus, not only simplifies calculations but also helps in integrating functions that might otherwise be very complex.
Definite Integral
A definite integral gives the exact value of the area under a curve over a specific interval. Unlike an indefinite integral, it produces a numerical result rather than a generalized function.
  • In this exercise, we evaluate an area under the transformed curve from \( t = 0 \) to \( t = 4 \).
  • The integral becomes \( 4\pi \int_{1}^{17} \sqrt{u} \, du \) after substitution.
The definite integral represents the accumulated sum of infinitesimally small quantities over a given interval.Finding these integrals often involves:- Calculating anti-derivatives.- Using the Fundamental Theorem of Calculus to express the area as the difference between these anti-derivatives evaluated at the bounds.In practice, definite integrals require not only solving the integral but correctly substituting the limits as derived from the substitution method.
Revolving around the x-axis
When a curve is revolved around the x-axis, it forms a three-dimensional surface.Calculating the surface area of such a surface uses the formula:\[ S = 2\pi \int_{a}^{b} |y| \sqrt{\left( \frac{dx}{dt} \right)^2 + \left( \frac{dy}{dt} \right)^2} \, dt \]
  • The integral calculates the surface's area generated by the curve as it rotates around the \( x \)-axis from \( t = a \) to \( t = b \).
  • Applying this to our parametric equations involves substituting for \( x = t^2 \) and \( y = 2t \).
This approach effectively measures the amount of space occupied by the surface.For a smooth understanding, consider how the curve grows over the interval, expanding and tracing out a circular path around the \( x \)-axis.This method elegantly combines parametric descriptions, calculus tools, and geometric visualizations to solve complex problems in a methodical way.

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Most popular questions from this chapter

True or False? Justify your answer with a proof or a counterexample. Given \(x=f(t)\) and \(y=g(t),\) if \(\frac{d x}{d y}=\frac{d y}{d x},\) then \(f(t)=g(t)+\mathrm{C},\) where \(\mathrm{C}\) is a constant.

For the following exercises, find the slope of a tangent line to a polar curve \(r=f(\theta) .\) Let \(x=r \cos \theta=f(\theta) \cos \theta\) and \(y=r \sin \theta=f(\theta) \sin \theta, \quad\) so the polar equation \(r=f(\theta)\) is now written in parametric form. $$ r=1-\sin \theta ;\left(\frac{1}{2}, \frac{\pi}{6}\right) $$

For the following exercises, convert the rectangular equation to polar form and sketch its graph. $$x^{2}+y^{2}=16$$

Use a graphing utility to graph \(r=\frac{1}{1-\cos \theta}\)

For the following exercises, find the area of the surface obtained by rotating the given curve about the \(x\) -axis. $$ x=t^{3}, \quad y=t^{2}, \quad 0 \leq t \leq 1 $$
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