Question

Sketch the region \(R\) bounded by the graphs of the given equations, and show a typical vertical slice. Then find the volume of the solid generated by revolving \(R\) about the \(x\) -axis. \(y=\frac{x^{2}}{\pi}, x=4, y=0\)

Short answer

The volume is \( \frac{1024}{5\pi^2} \).

Step-by-step solution

Identify the Boundaries of the Region R

The region is bounded by the graph of the equation \( y = \frac{x^2}{\pi} \), the vertical line \( x = 4 \), and the line \( y = 0 \). The graph \( y = \frac{x^2}{\pi} \) is a parabola that opens upwards. The line \( x = 4 \) is a vertical line. The line \( y = 0 \) is the x-axis, which acts as a lower bound for the region.

Visualize and Sketch the Region R

Sketch the parabola \( y = \frac{x^2}{\pi} \) in the first quadrant, followed by the vertical line \( x = 4 \), and the x-axis \( y = 0 \). Shade the region bounded by these three curves, which forms a closed shape. This is the region \( R \).

Show a Typical Vertical Slice

Draw a representative vertical slice (rectangle) from the curve \( y = \frac{x^2}{\pi} \) to the x-axis. This slice has a thickness \( \Delta x \) and height \( \frac{x^2}{\pi} \). This slice will be used for the integration to find the volume.

Set Up the Integral for Volume

The volume of the solid generated by revolving the region around the x-axis is found using the washer method. The volume \( V \) is obtained by: \[ V = \pi \int_{0}^{4} \left( \frac{x^2}{\pi} \right)^2 \, dx \]This expression comes from revolving the function \( y = \frac{x^2}{\pi} \) around the x-axis.

Simplify the Integral Expression

Simplify the integral expression:\[ V = \int_{0}^{4} \frac{x^4}{\pi^2} \, dx \]Factor out \( \frac{1}{\pi^2} \) from the integral for simplification:\[ V = \frac{1}{\pi^2} \int_{0}^{4} x^4 \, dx \]

Evaluate the Integral

Evaluate \( \int_{0}^{4} x^4 \, dx \):\[ \int_{0}^{4} x^4 \, dx = \left[ \frac{x^5}{5} \right]_{0}^{4} = \frac{4^5}{5} - \frac{0^5}{5} = \frac{1024}{5} \]

Find the Volume

Substitute the evaluated integral back into the expression for \( V \):\[ V = \frac{1}{\pi^2} \times \frac{1024}{5} = \frac{1024}{5\pi^2} \]This is the volume of the solid formed by revolving the region around the x-axis.

Key concepts

Volume of Solids of Revolution

The volume of solids of revolution involves finding the space or content occupied by a 3D object created by rotating a 2D region around an axis. When a shape in the plane is rotated, it traces out a solid three-dimensional object. To visualize this, imagine spinning a simple 2D shape like a rectangle around a line such as the x-axis. The resulting figure could be similar to a cylinder or a barrel, depending on the shape and rotation. To compute the volume, we use calculus, calculating the sum of all tiny slices of the solid. Each slice resembles a thin disk or washer, and since all slices are very thin, they come together to form the entire volume. This integration of infinitesimal slices allows for precise and detailed volume measurements even for complex shapes. Understanding this concept is crucial as it combines geometry with integral calculus for practical applications.

Washer Method

The washer method is an integral technique used to find the volume of a solid of revolution. It is particularly used when the region is bounded on one side by a curve and possibly by another different graph or the y-axis. When rotating around an axis, each slice becomes a washer, which is like a flat ring. The washer method is used when there is a hollow center, as opposed to a complete disk. In practice, the volume of a washer is determined by subtracting the volume of the inner disk (hole) from the outer disk. To set up the integral:

• Identify the functions that create the inner and outer radius of the washers.
• The outer radius is often given by the upper function, while the inner radius is given by any possible lower function or axis.
• Repeat the integration over the interval of interest along the axis of rotation.

Using this method allows us to accurately compute a volume that accounts for any hollow areas in the solid, resulting in an exact volume of the "washed out" region.

Integral Calculus

Integral calculus is a branch of calculus concerned with the concept of integration. Integration helps us find quantities like areas, volumes, central points, and many other useful things. The main idea of integral calculus is to add up many infinitely small things to find the total. The definite integral of a function gives the accumulated sum over an interval, which can represent physical quantities like area under a curve or the total volume of a solid. In the context of volume calculation, integration evaluates the sum of tiny volume elements (slices) to find the total volume. Here's a step-by-step approach:

• Establish the function or functions that describe the bounds of your region.
• Determine your interval, which is the section of the x-axis over which you will integrate.
• Set up the integral formula, aligning each x-value within your limits with its corresponding y-value from the function.
• Calculate the integral, resulting in the desired total measurement, such as area or volume.

Understanding integral calculus empowers you to tackle real-world problems that involve accumulating quantities.

Parabolas

A parabola is a symmetric curve on a plane that can be observed in various natural and engineered systems. It represents the graph of a quadratic function of the form \( y = ax^2 + bx + c \). Parabolas are U-shaped: they either open upwards or downwards depending on the coefficient of \( x^2 \). In the exercise, we specifically deal with the equation \( y = \frac{x^2}{\pi} \), which shows a parabola opening upwards.Key features of parabolas include:

• The vertex, the point where the parabola turns. For \( y = \frac{x^2}{\pi} \), the vertex is at the origin (0,0).
• An axis of symmetry, a vertical line passing through the vertex. For this exercise, it's the y-axis or \( x = 0 \).
• The focus and directrix which can be used in certain geometric constructions, though not critical for volume calculations.

This specific parabola describes part of the region that will form a solid upon rotation, demonstrating the wide application of parabolas in calculus and geometry. Understanding parabolas is fundamental when tackling problems involving regions bounded by these curves.