Question

Find the volume of the solid generated by revolving the region in the first quadrant bounded by \(y=x^{3}\) and \(y=4 x\) about (a) the \(x\) -axis\( (b) the line \)y=8 .

Short answer

The volume of the solid formed by revolving the region bounded by \(y=x^{3}\) and \(y=4x\) around the x-axis is \(3\pi\) cubic units, and the volume when revolving around the line \(y=8\) is \(32\pi/5\) cubic units.

Step-by-step solution

Identify intersection points

First, we need to solve for the intersection points of \(y=x^{3}\) and \(y=4x\). This can be done by setting these two equations equal to each other and solving for x: \(x^{3} = 4x\), or \(x^{3} - 4x = 0\). Factoring out an x, we get \(x(x - 2)(x + 2) = 0\). The solutions are x = 0 and x = 2.

Apply the disk method

The disk method formula (for revolvement around the x-axis) is given by \[V = \pi\int_{a}^{b}[f(x)]^{2}dx\]. Here, \(f(x)=4x-x^{3}\) (the upper function minus the lower function), a = 0, and b = 2. Substituting these into the formula, you compute the integral to get the volume for part (a).

Solve for volume for part (b)

For part (b), we also use the disk method, but this time revolving around y=8. Now, your radius is \(8-x^{3}\) (line y minus lower function) for the disk formed by \(y=x^{3}\) and \(8-4x\) (line y minus upper function) for the disk formed by \(y=4x\). These give your new \(f(x)\) for each integral. You integrate these two volumes separately (since they are separate areas bounded by line y=8) and sum the two volumes together to get the total volume revolving around y=8.

Key concepts

Disk Method

The disk method is a powerful tool used to find the volume of a solid of revolution. When a region in the plane is revolved around a line, the resulting solid can be split into a series of thin disks, resembling poker chips stacked on top of one another. Each disk's volume is given by the formula \( V = \pi\int_{a}^{b}[f(x)]^{2}\,dx \), where \( a \) and \( b \) are the bounds of integration, and \( f(x) \) is the function being rotated.

To use the disk method, first imagine a thin slice perpendicular to the axis of revolution which forms a disk when revolved around the axis. The radius of this disk is determined by the distance from the axis of revolution to the function value. By integrating the area of these disks over the interval from \( a \) to \( b \)—effectively stacking them all together—you get the volume of the solid. This method is particularly useful because it simplifies the intricate three-dimensional problem into a one-dimensional integral.

Integration of Volume

Integration in calculus is a significant tool for calculating volumes, especially when dealing with complex shapes like solids of revolution. With solids generated by revolved regions, the cross-sections perpendicular to the axis of rotation are typically circular. The volume of each circular disk can be expressed as \( \pi r^{2}h \) where \( r \) is the radius of the disk, and \( h \)—the height—is an infinitesimally small value that represents the disk's thickness.

The process of integrating involves adding up these infinitely thin volumes of disks along the axis of rotation from one endpoint to the other of the object's boundary. This requires setting up an integral with limits corresponding to those endpoints. Finding the volume of a solid of revolution is then a matter of performing this integration, which you calculate by taking the definite integral of the area formula of the disk over the specified interval.

Solving Intersection Points

Solving for the intersection points of two curves is crucial in setting up the boundaries for integration when finding the volume of a solid of revolution. Intersection points determine the limits of integration (\( a \) and \( b \) in our integrals) necessary for computing the volume.

To find these points, set the two functions equal to each other and solve for the variable, typically \( x \) or \( y \) depending on the orientation. In the given problem, setting \( y=x^{3} \) and \( y=4x \) gives us the equation \( x^{3}-4x=0 \) which, after factoring, leads us to the x-values where the curves intersect. These values create the integral's bounds for calculating the solid's volume. It's a fundamental step because accurate limits ensure that the appropriate region is being revolved and thus that the resultant volume is correct.

Calculus Applications

The applications of calculus are vast and highly practical in various science and engineering fields. Calculus helps to explain and predict natural phenomena in physics and biology, for example, and is essential in economics to find optimum points for cost and profit. In engineering, it plays a critical role in calculating stresses and strains in materials.

In the context of finding volumes through integration, calculus moves beyond theoretical mathematics to provide solutions to real-world problems. It enables us to find volumes of not just basic geometrical shapes, but also irregular objects that cannot be otherwise calculated using simple geometry. The methodology of setting up an integral to compute the volume of a solid of revolution exemplifies how calculus is used to transform a geometrically complex problem into a solvable mathematical equation.