/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 1 The region bounded by \(y=e^{2 x... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 6: Problem 1

The region bounded by \(y=e^{2 x}, y=0, x=-2, x=0\) is rotated around the \(x\) -axis. Find the volume. volume \(=\) ___________.

Short Answer

Expert verified
The volume is \(V = \pi \left[ \frac{1}{4} - \frac{1}{4e^{8}} \right] \)

Step by step solution

01

Identify the General Formula

The volume of a solid of revolution around the x-axis is given by: \[ V = \int_{a}^{b} \pi [f(x)]^2 dx \]where the function is rotated around the x-axis.
02

Determine the Function and Interval

The function given is \(y = e^{2x}\) and the bounds are from \(x = -2\) to \(x = 0\).
03

Set Up the Integral

Substitute the function into the volume formula:\[ V = \pi \int_{-2}^{0} (e^{2x})^2 dx \]which simplifies to\[ V = \pi \int_{-2}^{0} e^{4x} dx \]
04

Integrate the Function

Integrate the function \(e^{4x}\):\[ \int e^{4x} dx = \frac{1}{4} e^{4x} \]Evaluate the integral from \(x = -2\) to \(x = 0\):\[ V = \pi \[ \frac{1}{4} e^{4x} \]_{-2}^{0} \]
05

Calculate the Definite Integral

Evaluate the expression:\[ V = \pi \left[ \frac{1}{4} e^{4(0)} - \frac{1}{4} e^{4(-2)} \right] = \pi \left[ \frac{1}{4} e^{0} - \frac{1}{4} e^{-8} \right] = \pi \left[ \frac{1}{4} - \frac{1}{4} e^{-8} \right] \]
06

Simplify the Result

Simplify the expression to find the volume:\[ V = \pi \left[ \frac{1}{4} - \frac{1}{4e^{8}} \right] \]

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

volume of solid of revolution
The volume of a solid of revolution is an important concept in calculus. It involves rotating a region or function around an axis to create a 3D shape. To find its volume, we use integration.

For the volume of a solid revolving around the x-axis, the formula used is:
\begin{align*}V = \int_{a}^{b} \pi [f(x)]^2 dx\end{align*}
Where
  • \[a \] and \[b \] are the bounds of the region along the x-axis
  • \[f(x)\] is the function being rotated

This setup allows us to slice the solid into thin disks and sum their volumes from \[a \] to \[b \]. Each disk has a volume of \[\pi [f(x)]^2 dx\.\]
definite integral
A definite integral computes the accumulation of quantities, often areas under curves. It鈥檚 symbolized as:
\begin{align*}\overint{a}{b} f(x) dx\end{align*}
Where
  • \[f(x)\] is the function being integrated
  • \[a\] and \[b\] represent the start and end points on the x-axis
  • \[dx\] indicates integration concerning x

For solids of revolution, integrating from \[a \] to \[b \] with \[dx \] finds the total volume.
In the example, integrating \[e^{4x} \] from \[-2 to 0\] helps calculate the shape鈥檚 volume formed by rotating \[y = e^{2x} \].
integration techniques
Effective integration techniques help solve definite integrals.
For the example problem, the integral is set up as:
\begin{align*}\overint{-2}{0} \pi e^{4x} dx\end{align*}
Key steps to solve it:
  • Identify appropriate formula: \[V = \pi \overint{a}{b} [f(x)]^2 dx\]
  • Transform function: Square the function and simplify, which changes \[e^{2x} \] to \[e^{4x} \]
  • Integrate: Use rules of exponential integration, knowing \[\overint e^{kx} dx = \frac{1}{k} e^{kx} \]
  • Evaluate at bounds: Substitute in the limits \[-2\] and \[0\] to get the result
The calculated volume integrates \[e^{4x} \] using these techniques.
exponential functions
Exponential functions, like \[y = e^{2x} \], are a common subject in calculus. They feature a constant base, \[e\] (approximately \[2.718\]), raised to a variable exponent.
Properties of exponential functions include:
  • Rapid Growth: They increase or decrease at a faster rate compared to linear or polynomial functions
  • Continuous Change: Useful in modeling real-world situations involving growth or decay
  • Integration and Differentiation Rules: Their integral and derivatives are straightforward \( \overint e^{kx} dx = \frac{1}{k} e^{kx} \)

In the given problem, the exponential function \[y = e^{2x} \] determines the curve's shape and the solid formed by rotating about the x-axis. Knowing how to integrate such functions is crucial for solving the volume.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Consider the curve \(f(x)=3 \cos \left(\frac{x^{3}}{4}\right)\) and the portion of its graph that lies in the first quadrant between the \(y\) -axis and the first positive value of \(x\) for which \(f(x)=0\). Let \(R\) denote the region bounded by this portion of \(f,\) the \(x\) -axis, and the \(y\) -axis. Assume that \(x\) and \(y\) are each measured in feet. a. Picture the coordinate axes rotated 90 degrees clockwise so that the positive \(x\) -axis points straight down, and the positive \(y\) -axis points to the right. Suppose that \(R\) is rotated about the \(x\) axis to form a solid of revolution, and we consider this solid as a storage tank. Suppose that the resulting tank is filled to a depth of 1.5 feet with water weighing 62.4 pounds per cubic foot. Find the amount of work required to lower the water in the tank until it is 0.5 feet deep, by pumping the water to the top of the tank. b. Again picture the coordinate axes rotated 90 degrees clockwise so that the positive \(x\) axis points straight down, and the positive \(y\) -axis points to the right. Suppose that \(R\), together with its reflection across the \(x\) -axis, forms one end of a storage tank that is 10 feet long. Suppose that the resulting tank is filled completely with water weighing 62.4 pounds per cubic foot. Find a formula for a function that tells the amount of work required to lower the water by \(h\) feet. c. Suppose that the tank described in (b) is completely filled with water. Find the total force due to hydrostatic pressure exerted by the water on one end of the tank.

A point mass of 1 grams located 6 centimeters to the left of the origin and a point mass of 3 grams located 7 centimeters to the right of the origin are connected by a thin, light rod. Find the center of mass of the system. Center of Mass = ___________.

A cylindrical tank, buried on its side, has radius 3 feet and length 10 feet. It is filled completely with water whose weight density is \(62.4 \mathrm{lbs} / \mathrm{ft}^{3},\) and the top of the tank is two feet underground. a. Set up, but do not evaluate, an integral expression that represents the amount of work required to empty the top half of the water in the tank to a truck whose tank lies 4.5 feet above ground. b. With the tank now only half-full, set up, but do not evaluate an integral expression that represents the total force due to hydrostatic pressure against one end of the tank.

Find the volume of the solid obtained by rotating the region bounded by the curves $$ y=x^{2}, \quad y=1 $$ about the line \(y=4\) Answer: ___________.

Find the volume of the solid obtained by rotating the region in the first quadrant bounded by \(y=x^{6}, y=1,\) and the \(y\) -axis around the \(x\) -axis. Volume = ___________.
See all solutions

Recommended explanations on Math Textbooks

Geometry

Read Explanation

Pure Maths

Read Explanation

Calculus

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Probability and Statistics

Read Explanation

Applied Mathematics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.