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Chapter 6: Problem 13

Find the areas of the surfaces generated by revolving the curves about the indicated axes. If you have a grapher, you may want to graph these curves to see what they look like. $$y=x^{3} / 9, \quad 0 \leq x \leq 2 ; \quad x \text { -axis }$$

Short Answer

Expert verified
Use the formula \( A = 2\pi \int_{0}^{2} \frac{x^3}{9} \sqrt{1 + \left( \frac{x^2}{3} \right)^2} \, dx \) and evaluate numerically.

Step by step solution

01

Understand the Formula for Surface Area of Revolution

To find the surface area of a curve revolved around the x-axis, we use the formula: \( A = 2\pi \int_{a}^{b} y \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \, dx \). Here, \( y = f(x) \) is the curve, and \( \frac{dy}{dx} \) is its derivative.
02

Identify the given function and its derivative

The given function is \( y = \frac{x^3}{9} \). To find the derivative, apply the power rule:\[ \frac{dy}{dx} = \frac{d}{dx} \left( \frac{x^3}{9} \right) = \frac{3x^2}{9} = \frac{x^2}{3} \].
03

Substitute into the Surface Area Formula

Substitute \( y = \frac{x^3}{9} \) and \( \frac{dy}{dx} = \frac{x^2}{3} \) into the formula:\[ A = 2\pi \int_{0}^{2} \left( \frac{x^3}{9} \right) \sqrt{1 + \left( \frac{x^2}{3} \right)^2} \, dx \].
04

Simplify the Expression inside the Integral

Inside the integral, we need to simplify \( \sqrt{1 + \left( \frac{x^2}{3} \right)^2} \):\[ \left( \frac{x^2}{3} \right)^2 = \frac{x^4}{9} \].Now, \[ 1 + \frac{x^4}{9} = \frac{9}{9} + \frac{x^4}{9} = \frac{9 + x^4}{9} \], so the integral becomes:\[ A = \frac{2\pi}{9} \int_{0}^{2} x^3 \sqrt{\frac{9 + x^4}{9}} \, dx \].
05

Evaluate the Integral

Rewrite \( \sqrt{\frac{9 + x^4}{9}} \) as \( \frac{1}{3} \sqrt{9 + x^4} \), and the integral becomes:\[ A = \frac{2\pi}{27} \int_{0}^{2} x^3 \sqrt{9 + x^4} \, dx \].To evaluate this integral, you could use numerical methods or a calculator as it doesn't have a simple antiderivative.
06

Approximate the Integral Numerically

If using numerical methods or graphing calculators, compute \( \int_{0}^{2} x^3 \sqrt{9 + x^4} \, dx \) to approximate the surface area value. Use a tool to calculate and find the area numerically.

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Key Concepts

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

Surface Area of Revolution
When you revolve a curve around an axis, it forms a 3D shape. The surface area of this shape is what we're looking to find. For this exercise, we're focusing on revolving around the x-axis.
To get the surface area, we use a special formula:
  • First, identify your function, in this case, it's \( y = x^3/9 \).
  • Use the surface area of revolution formula: \( A = 2\pi \int_{a}^{b} y \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \, dx \).
This formula helps us find the area on the curved surface by multiplying around the circular loops formed by revolving each point on the curve.
Integration
Integration is a core concept in calculus that allows us to sum small pieces to find a whole. It's used here to find the total surface area of the revolution.
  • To find the surface area, integrate over the function from the start to the end of the interval, here from \( x = 0 \) to \( x = 2 \).
  • The function inside the integral tells us how to add up those small slices.
  • For our formula \( A = 2\pi \int_{a}^{b} y \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \, dx \), \( y \) and its derivative are both part of the expression we integrate.
Integration helps combine these small surface slices into the total area of the 3D surface.
Derivative
The derivative tells us how a function changes at any point and is essential for surface area calculations.
For this exercise, we worked with the curve given by the equation \( y = \frac{x^3}{9} \). To calculate its derivative:
  • Apply the power rule of differentiation to find \( \frac{dy}{dx} \).
  • Here, \( \frac{dy}{dx} = \frac{x^2}{3} \).
The derivative is crucial in our formula, helping us account for the curve's slope, which affects how much area is covered when the curve is revolved. Without it, we can't accurately compute the integral.
Numerical Approximation
Sometimes, calculating an integral exactly isn't feasible with basic calculus techniques.
That's where numerical approximation comes in handy. For complex integrals like \( \int_{0}^{2} x^3 \sqrt{9 + x^4} \, dx \), we often don't find a simple antiderivative.
  • Numerical methods, such as using a calculator or software like graphing tools, allow us to estimate the value.
  • Tools like the trapezoidal rule or Simpson's rule give us approximate results which are often sufficient for practical purposes.
This approximation ensures that even when exact answers are tough to find, we can still get a trustworthy estimate of the surface area.

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Most popular questions from this chapter

Use the shell method to find the volumes of the solids generated by revolving the regions bounded by the given curves about the given lines. \(y=x+2, \quad y=x^{2}\) a. The line \(x=2\) b. The line \(x=-1\) c. The \(x\) -axis d. The line \(y=4\)

Find the length of the curve $$ y=\int_{0}^{x} \sqrt{\cos 2 t} d t $$ from \(x=0\) to \(x=\pi / 4\)

Find the volumes of the solids generated by revolving the regions bounded by the lines and curves about the \(y\) -axis. $$\begin{aligned}&\text { The region enclosed by } x=\sqrt{\cos (\pi y / 4)}, \quad-2 \leq y \leq 0\\\&x=0 \end{aligned}$$

Find the volume of the solid generated by revolving each region about the given axis. The region in the first quadrant bounded above by the curve \(y=x^{2},\) below by the \(x\) -axis, and on the right by the line \(x=1\) about the line \(x=-1\)

As found in Exercise 37 the centroid of the semicircle \(y=\sqrt{a^{2}-x^{2}}\) lies at the point \((0,2 a / \pi) .\) Find the area of the surface swept out by revolving the semicircle about the line \(y=a\).
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