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Chapter 5: Problem 10

Find the volume generated by rotating the area bounded by the graphs of each set of equations around the \(x\) -axis. $$ y=\frac{1}{\sqrt{x}}, x=1, x=4 $$

Short Answer

Expert verified
The volume is \( \pi \ln(4) \).

Step by step solution

01

Setup the Integral

To find the volume generated by rotating the area around the x-axis, we use the disk method. The formula for the volume is: \[ V = \pi \int_{a}^{b} [f(x)]^2 \, dx \]In this case, \(f(x) = \frac{1}{\sqrt{x}}\), \(a = 1\), and \(b = 4\). Thus, the integral becomes:\[ V = \pi \int_{1}^{4} \left(\frac{1}{\sqrt{x}}\right)^2 \, dx = \pi \int_{1}^{4} \frac{1}{x} \, dx \]
02

Compute the Integral

Now, compute the integral \( \pi \int_{1}^{4} \frac{1}{x} \, dx \). The integral of \( \frac{1}{x} \) with respect to \(x\) is \( \ln|x| \). Therefore, the definite integral becomes:\[ \pi \left[ \ln|x| \right]_{1}^{4} = \pi \left( \ln(4) - \ln(1) \right) \]
03

Evaluate the Result

Since \( \ln(1) = 0 \), the expression simplifies to:\[ \pi ( \ln(4) - 0 ) = \pi \ln(4) \]This is the exact volume of the solid formed.

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Key Concepts

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

Volume of Revolution
When we talk about the volume of revolution, imagine a flat, 2D area being spun around a line to create a 3D shape. This is the volume of revolution. Think of it as turning a piece of paper into a cylinder by spinning it. Calculus helps us find these volumes efficiently.
For our problem, the region is spun around the x-axis. The x-axis acts like a skewer, and our flat region twirls around it to make a 3D "shell" or "disk." When the x-axis is used, we typically use the **Disk Method** to find the volume.
  • Disk Method: It considers cross-sectional slices of the solid. Each slice is a disk (like a short cylinder).
  • Cross-Sections: The area of each slice is a circle. The radius of each slice is determined by the distance from the x-axis to the curve, given by the function.
Visualizing how we create 3D shapes from 2D regions can make calculus more intuitive and fun! The method is powerful for calculating volumes in real-world scenarios too.
Definite Integral
In calculus, an integral can be used to add up an infinite number of tiny amounts to find a whole quantity. For volumes of revolution, we use something called a **definite integral**. This means we're interested in adding up all the pieces only between two specific points.
In our problem, the definite integral spans from when x equals 1 to x equals 4. This is because our region is bounded by these vertical lines.
  • Formula: The definite integral is written as \( \int_{a}^{b} f(x) \, dx \), where \(a\) and \(b\) mark the start and end of the interval on the x-axis.
  • Properties: The result gives the cumulative value (like area or volume) over the interval \([a, b]\).
So, in simpler terms, the definite integral sums up infinitely small disks from x=1 to x=4 to find the total volume. The integral transforms a complex problem into something more manageable.
Calculus Problem Solving
Solving calculus problems often feels like piecing together a puzzle. Each piece, or step, brings you closer to the solution. Understanding how to solve problems helps in mastering calculus.
Our specific problem involved using calculus to find a volume of revolution through these steps:
  • Identify the Method: Recognize this is a volume of revolution problem, where the disk method is needed.
  • Set Up the Integral: Translate the problem into a calculation. Use the formula \( V = \pi \int_{a}^{b} [f(x)]^2 \, dx \).
  • Compute the Integral: Evaluate the integral to find a numeric answer. Here, it involved integrating \( \frac{1}{x} \) to find \( \ln(x) \).
  • Evaluate the Bounds: Plug the limits into the integral result. This gives the final numerical volume.
Each step in problem-solving reflects a bit of logic and creativity, making calculus a tool for turning abstract ideas into concrete solutions.

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Most popular questions from this chapter

Ted and Edith Markey have a new grandchild, Kurt, and want to create a trust fund for him that will yield \(\$ 1,000,000\) on his 22 nd birthday. a) What lump sum should they deposit now at \(4.2 \%\), compounded continuously, to achieve \(\$ 1,000,000 ?\) b) The amount in part (a) is more than they can afford, so they decide to invest a constant amount, \(R(t)\) dollars per year. Find \(R(t)\) such that the accumulated future value of the continuous money stream is \(\$ 1,000,000,\) assuming an interest rate of \(4.2 \%\) compounded continuously.

Let \(x\) be a continuous random variable that is normally distributed with mean \(\mu=22\) and standard deviation \(\sigma=5 .\) Using Table A, find the following. $$ P(x \leq 22.5) $$

The capitalized cost, \(c,\) of an asset over its lifetime is the total of the initial cost and the present value of all maintenance expenses that will occur in the future. It is computed with the formula $$ c=c_{0}+\int_{0}^{L} m(t) e^{-k t} d t $$ where \(c_{0}\) is the initial cost of the asset, \(L\) is the lifetime (in years), \(k\) is the interest rate (compounded continuously), and \(m(t)\) is the annual cost of maintenance. Find the capitalized cost under each set of assumptions. $$ \begin{array}{l} c_{0}=\$ 600,000, k=4 \% \\ m(t)=\$ 40,000+\$ 1000 e^{0.01 t}, L=40 \end{array} $$

Let \(x\) be a continuous random variable that is normally distributed with mean \(\mu=22\) and standard deviation \(\sigma=5 .\) Using Table A, find the following. $$ P(x \geq 20) $$

Let \(x\) be a continuous random variable with a standard normal distribution. Using Table A, find each of the following. $$ P(0 \leq x \leq 2.13) $$
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