Problem 31 - Calculus Early Transcendentals

Question

Use the shell method to find the volume of the following solids. The ellipsoid formed when that part of the ellipse \(x^{2}+2 y^{2}=4\) with \(x \geq 0\) is revolved about the \(y\) -axis

Step-by-Step Solution

Answer

Answer: The volume of the ellipsoid is \(\frac{4}{3}\pi\) cubic units.

Step 1: Parametrize the ellipse function

We are given the ellipse function \(x^{2}+2 y^{2}=4\) with \(x\geq0\). To parametrize the ellipse function, we can rewrite it in the following form: $$ x^{2}+2 y^{2}=4 \Rightarrow x=\sqrt{4-2y^{2}} $$ Now, we can write \(x\) as a function of \(y\), in which \(x=f(y)\): $$ f(y)=\sqrt{4-2y^{2}} $$ The range of \(y\) will be \(-\frac{\sqrt{2}}{2} \leq y \leq \frac{\sqrt{2}}{2}\) since \(0 \leq x \leq 2\).

Step 2: Calculate the volume of the cylinder

We will use the shell method to calculate the volume of the ellipsoid. In the shell method, we find the volume of the cylinder created by revolving a small section around the y-axis. The volume of a single cylinder with radius \(x(y)\) and height \(dy\) is given by: $$ dV=2\pi \cdot x(y) \cdot dy $$ In our case, the function \(x(y) = f(y) = \sqrt{4-2y^{2}}\). So, the volume of the cylinder is: $$ dV=2\pi \cdot \sqrt{4-2y^{2}} \cdot dy $$

Step 3: Use integral to sum the volumes of the cylinders

We now integrate the volume of the cylinders over the range of \(y\) to find the volume of the ellipsoid. The integral is as follows: $$ V=\int_{-\frac{\sqrt{2}}{2}}^{\frac{\sqrt{2}}{2}} 2\pi \cdot \sqrt{4-2y^{2}} \cdot dy $$ To solve the integral, we use a substitution method. Let \(u = 4-2y^{2}\). Then, \(du = -4y \, dy\), or \(dy = \frac{1}{-4y}du\). The new integral would be: $$ V=\int_{2}^{0} 2\pi \cdot \sqrt{u} \cdot \frac{1}{-4y} \, du $$ Use integration by parts, we get: $$ V=-\frac{1}{2}\pi \int_{2}^{0} u^{\frac{1}{2}} \, du $$ Now, integrate with respect to \(u\): $$ V=-\frac{1}{2}\pi \left[\frac{2}{3} u^{\frac{3}{2}}\right]_{2}^{0} $$ Finally, evaluate the integral and obtain the volume of the ellipsoid: $$ V=-\frac{1}{2}\pi \left(\frac{2}{3}(0)^{\frac{3}{2}}-\frac{2}{3}(2)^{\frac{3}{2}}\right) = \frac{4}{3}\pi $$ The volume of the ellipsoid is \(\frac{4}{3}\pi\) cubic units.

Key Concepts

Shell Method

The shell method is a clever technique used to calculate the volume of solids of revolution. It is especially useful when the solid is formed by revolving a region around an axis. The key idea involves slicing the solid into cylindrical shells rather than disks or washers. Each shell has a height and a varying radius that depend on the axis of rotation.

Here's how it works:

• Consider a region that you want to revolve around an axis.
• Cylinder shells: Imagine slicing the region into thin strips parallel to the axis of rotation. When these strips are revolved around the axis, they form a shell.
• Calculate the volume of each shell: The formula for a cylindrical shell of height \(f(y)\), radius \(y\), and thickness \(dy\) is \(dV=2\pi \cdot y \cdot f(y) \cdot dy\).
• Integrate: Sum the volumes of all shells using an integral over the range of the variable, such as \( y \in [a, b] \).

The shell method is often preferred when it's more convenient to integrate with respect to a different variable than the one determining the axis of rotation. When dealing with revolutions about the \ y \-axis, it's practical because you parameterize your function in terms of \ y \.

Parametrization of Ellipse

Parametrization is a way to describe geometric shapes using a single variable, usually an angle or a parameter. It is particularly useful for ellipses, which are more complex than simple circles and straight lines. In our exercise, we're working with the ellipse equation \(x^{2}+2y^{2}=4\), which is not perfectly circular, as seen by the different coefficients on \(x\) and \(y\).
Parametrizing serves multiple purposes:

• Express in terms of one variable: Rewriting the equation allows for evaluation along a single axis, simplifying computations.
• Substitute variables: For the given ellipse, solving for \(x\) in terms of \(y\) gives us \(x = \sqrt{4-2y^{2}}\). This allows us to study changes in \(x\) directly as \(y\) changes.
• Boundaries of ellipses: Determine the range of \(y\) by imposing conditions on \(x\), such as non-negativity, to get meaningful intervals.

Using parametrization, you can easily adapt the problem for integration and revolutions, mapping out which sections of the ellipse contribute to the solid's shape.

Integration by Parts

Integration by parts is a mathematical technique used to solve integrals. It is particularly handy when the product of two functions is involved. The overarching principle comes from the product rule for differentiation.
To use integration by parts, remember the formula:\[\int u \, dv = uv - \int v \, du\]Here's a step-by-step:

• Choose \(u\) and \(dv\) wisely: Fascinating integrals often stem from selecting \(u\) as the component that simplifies upon differentiation, while \(dv\) is the easily integratable part.
• Differentiate and integrate: Find \(du\) and \(v\) from your choices. For instance, differentiating \(u\) yields \(du\), and integrating \(dv\) gives \(v\).
• Substitute into formula: Use the integration-by-parts formula to rewrite the integral.
• Solve the resulting integral: Continue the process, or use other techniques until the integral is resolved.

It's particularly useful when paired with other techniques, such as substitution, to solve more complex integrals, as seen with the ellipsoid volume calculation.

Integral Calculus

Integral calculus is a fundamental branch of calculus that concerns itself with the accumulation of quantities, such as areas under curves, volumes, and more. Through integration, we can sum infinite infinitesimal parts to find these accumulations.
Integral calculus has many applications:

• Finding areas: Integrals compute the area enclosed by curves in a plane.
• Determining volumes: Techniques like the shell method use integrals to find volumes of solids of revolution.
• Accumulating change: Integrals represent a total quantity from a rate of change, crucial for mechanics, economics, and more.
• In mathematics: Integrals help solve differential equations, find averages, and analyze graphs.

Mastering the concepts and techniques of integration allows you to handle various problems involving accumulation and change. Whether using substitution, integration by parts, or understanding definite and indefinite integrals' properties, integral calculus is a powerful set of tools in the mathematician's toolkit.