Question

The region bounded by the graph of \(f(x)=\frac{1}{1+x^{2}}\) and the \(x\) -axis between \(x=0\) and \(x=1\) is revolved about the \(x\) axis. Find the volume of the solid that is generated.

Short answer

The volume is \( \frac{\pi}{4} \).

Step-by-step solution

Identify the Formula for Volume of Revolution

To find the volume of the solid formed by revolving a region around the x-axis, use the disk method. The formula is \( V = \pi \int_{a}^{b} [f(x)]^2 \, dx \), where \( f(x) \) is the function representing the curve and \([a, b]\) is the interval of integration.

Define the Function and Interval

The function given is \( f(x) = \frac{1}{1+x^2} \). The volume is calculated over the interval \([0, 1]\) because we are revolving the region between \( x = 0 \) and \( x = 1 \).

Apply the Formula

Plug the function and interval into the volume formula: \[V = \pi \int_{0}^{1} \left( \frac{1}{1+x^2} \right)^2 \, dx \]

Simplify the Integrand

Simplify \( \left( \frac{1}{1+x^2} \right)^2 \) to \( \frac{1}{(1 + x^2)^2} \). The integral to solve is now:\[V = \pi \int_{0}^{1} \frac{1}{(1+x^2)^2} \, dx \]

Solve the Integral

To solve the integral, use the substitution method. Let \( u = 1 + x^2 \), then \( du = 2x \, dx \). However, note \( x = 0 \) results in \( u = 1 \) and \( x = 1 \) results in \( u = 2 \), so we adjust accordingly:\[= \frac{1}{2} \int_{1}^{2} u^{-2} \, du\]Computing the integral, we have:\[V = \frac{1}{2} \left[ -u^{-1} \right]_{1}^{2} = \frac{1}{2} \left[ -\frac{1}{2} - (-1) \right] = \frac{1}{2} \cdot \frac{1}{2} = \frac{1}{4} \]

Multiply by \( \pi \) to Find the Volume

Finally, multiply the result by \( \pi \) to find the volume:\[V = \pi \cdot \frac{1}{4} = \frac{\pi}{4} \]

Key concepts

disk method

The disk method is a technique used in calculus to find the volume of a solid of revolution. It involves slicing the solid into thin, circular disks and summing up their volumes. This is especially helpful when you have a function rotated around an axis, like the x-axis in our problem.

• Think of each disk as a flat cylinder.
• The formula for the volume of each disk is based on the area of the circle offormula is considered part of the base of this cylinder.
• The radius of the disk is given by the value of the function at a specific point, which changes across the interval.

This method relies heavily on integration as a means to aggregate the total volume from each infinitesimal disk over the interval given.

integration by substitution

Integration by substitution is a powerful method used to simplify the process of integration by making a change of variables. It's similar to the chain rule in differentiation.The idea is to simplify the integrand by replacing a function of a variable with a single variable.

• You start by identifying a part of the integrand that can be replaced with a simpler expression.
• For our problem, we let \( u = 1 + x^2 \), hence \( du = 2x \, dx \).
• Adjust the bounds of integration according to this substitution, changing from \( x \) bounds to \( u \) bounds. When \( x = 0 \, \rightarrow \, u = 1 \) and \( x = 1 \, \rightarrow \, u = 2 \).

This substitution technique not only simplifies solving the integral but also makes complex problems more manageable.

volume of a solid of revolution

The volume of a solid of revolution refers to the three-dimensional space occupied by a solid that is formed by revolving a 2D region around an axis. This concept adds a layer of sophistication to understanding the geometry of curves in calculus.

• The entire curve is seen as sweeping out a volume rather than just tracing a path.
• The disk method helps us analyze this concept through slicing the volume into thin disks.
• In simple terms, each point of the curve generates a circle as it revolves around the axis, contributing to the total volume.

Understanding this concept helps improve spatial reasoning and the ability to visualize how objects exist in a 3D space originating from 2D areas.

calculus problem-solving

Calculus problem-solving involves various techniques and skills to tackle a wide range of problems that involve rates of change and areas under curves. Approaching a problem in calculus generally involves:

• Identifying the type of problem and the correct methods to apply. For volumes, it often involves methods like the disk or shell method as seen here.
• Setting up the appropriate integral by understanding the underlying geometry or rate of change being represented.
• Applying algebraic and calculus techniques such as substitution to simplify and solve the problem.
• Double-checking parts of the solution such as substituting correctly and ensuring all numerical calculations are accurate.

Developing these skills is crucial for mastering calculus, enabling students to solve complex problems with confidence and a systematic approach.