Problem 55 - Calculus of a Single Variable

Question

Dividing a Solid In Exercises 55 and 56 , consider the solid formed by revolving the region bounded by \(y=\sqrt{x}, y=0,\) and \(x=4\) about the \(x\) -axis. Find the value of \(x\) in the interval \([0,4]\) that divides the solid into two parts of equal volume. Find the value of \(x\) in the interval \([0,4]\) that divides the solid into two parts of equal volume.

Step-by-Step Solution

Answer

The value of \(x\) that splits the solid into two parts of equal volume is \(x = 2\sqrt{2}\).

Step 1: Setup of integral for volume

We express the volume of the solid, V, as the integral of the area of each infinitesimally small disk, with radius \(\sqrt{x}\) and thickness \(dx\), from \(x=0\) to \(x=4\). This can be given by the following expression (given by the Disc Method formula): \(V = \int_0^4 \pi (\sqrt{x})^2 \, dx\).

Step 2: Calculation of total volume

Evaluating the integral yields the total volume of the solid, V. Substituting \((\sqrt{x})^2\) for x we obtain: \(V = \int_0^4 \pi x \, dx = \pi \left[\frac{1}{2} x^2\right]_0^4 = 8\pi\) units^3.

Step 3: Calculation of half the volume

Since the solid needs to be divided into two equal volumes, this volume should simply be half of the total volume, i.e. \(\frac{8\pi}{2} = 4\pi\) units^3.

Step 4: Setup of integral for x

Setting up another volume integral similar to step 1, this time from \(x=0\) to an arbitrary point \(x{'}\) yields the following expression: \(4\pi = \int_0^{x'} \pi x \, dx\). The integrand is still the area of an infinitesimal disk.

Step 5: Find x-value

Solving for \(x{'}\) would yield the value of x which splits the solid into two equal volumes. Cancelling \(\pi\) from both sides, we have \(4 = [\frac{1}{2} x^{2}]_0^{x'}\). Solving for \(x{'}\), we get \(x'= \sqrt{8} = 2\sqrt{2}\).

Key Concepts

Solid Revolution

Imagine spinning a two-dimensional shape around an axis. The three-dimensional object you get is called a solid of revolution. These can be visualized as objects like vases, bowls, or even car tires that result from revolving a flat profile. When a curve, such as the one defined by the equation \(y=\sqrt{x}\), is revolved around the x-axis, it creates a three-dimensional solid. The shape formed from revolving the specified region in the exercise between \(y=\sqrt{x}, y=0,\) and \(x=4\) around the x-axis is called a solid of revolution. In integral calculus, we can calculate the volume of this solid using techniques that sum up countless infinitesimally thin discs—each representing a cross-section of the solid. This approach transforms a seemingly complex three-dimensional problem into a solvable one-dimensional integral.
Visual aids, like sketches or computer software, can enhance understanding of these shapes. Remember, constructing a solid of revolution is like layering an infinite number of discs along the axis of rotation.

Integral Calculus

Integral calculus is a branch of mathematics focused on the accumulation of quantities and the spaces under and between curves. In the context of finding volumes, integral calculus allows the summation of infinite slices — imagined as paper-thin discs — of a three-dimensional object. The method for finding volumes by setting up an integral is powerful because it breaks down complex shapes into manageable pieces with known areas.
In our exercise, the integral \(V = \int_0^4 \pi (\sqrt{x})^2 \, dx\) represents the sum of the areas of countless tiny circles (disks) along the x-axis. By computing the integral, we calculate the total area of these disks, hence finding the volume of the entire solid. Without integral calculus, determining the volume of irregularly shaped objects would be much more challenging.

Volume of Solid

The volume of a solid is the measure of the space it occupies. In simpler terms, if you filled the hollow inside of a vase with water, the amount of water it can hold is its volume. For regular shapes like cubes or spheres, specific formulas exist for volume calculation. However, for irregular solids, especially those generated by the rotation of a function, we rely on integral calculus to find how much space they occupy.
The exercise involves a solid formed by revolving a region bound by a curve around an axis. By determining the volume of an infinitesimally small disc at a particular location and integrating that expression over the desired range, we find the solid's total volume. The disc method is especially useful here, as it simplifies the volume calculation to an integral one can solve analytically or numerically.

Disc Method Calculus

The disc method in calculus is a technique used to calculate the volume of a solid of revolution. This method involves slicing the solid parallel to the axis of revolution — creating discs whose volume is easy to determine. The formula for the volume of each thin disc is the area of the disc's face (a circle) multiplied by its infinitesimal thickness. This translates to the mathematical expression \(\pi r^2 \, dx\), where \(r\) is the radius of the circle and \(dx\) is the disc's thickness.

Breaking Down the Disc Method: Step by Step

• Identify the function representing the boundary of the shape you want to revolve.
• Determine the axis of revolution, which tells you how the area will sweep out the solid.
• Express the radius of the typical disc in terms of the variable of integration (\(x\) or \(y\)).
• Set up the integral with the correct bounds that correspond to the range over which the solid extends.
• Evaluate the integral to find the volume of the solid.

In our specific problem, once we've calculated the volume using the disc method, we then use the concept of equal volumes to determine the point of division along the x-axis. Integral calculus combined with the disc method turns a complex spatial problem into a manageable calculation.