Problem 13 - Single Variable Calculus

Question

Use the method of cylindrical shells to find the volume of the solid generated by revolving the region bounded by the graphs of the equations and/or inequalities about the indicated axis. Sketch the region and a representative rectangle. \(y=\frac{1}{x^{2}+1}, \quad x=0, \quad x=2 ;\) the \(y\) -axis

Step-by-Step Solution

Answer

The volume of the solid generated by revolving the region bounded by \(y=\frac{1}{x^{2}+1}, \quad x=0, \quad x=2\), and the \(y\)-axis can be found using the method of cylindrical shells. The integral representing the volume of the solid is given by: \[V = 2\pi \int_{1/5}^{1} y \sqrt{\frac{1}{y}-1}\, dy\] After making appropriate substitutions and simplifying, we obtain: \[V = 2\pi \int_{0}^{4} \frac{\sqrt{u}}{(u+1)^3} du\] Evaluating the integral, we get: \[V = \frac{-\pi}{180}\]

Step 1: Sketch the region and representative rectangle

We are given the region between \(y=\frac{1}{x^2+1}, \quad x=0, \quad x=2\), and the \(y\)-axis. To sketch the region, plot the function \(y=\frac{1}{x^{2}+1}\) for \(0\leq x\leq2\). This will be a smooth curve that starts at \((0,1)\) and approaches the \(x\)-axis as \(x\) goes to 2. We also include vertical lines at \(x = 0\) and \(x=2\) to form a closed region. When drawing the representative rectangle (which will become a cylindrical shell), make sure that it is parallel to the axis of rotation, which in this case is the \(y\)-axis.

Step 2: Set up the integral using the method of cylindrical shells

In the method of cylindrical shells, we must first determine the thickness of the shell, which is \(\Delta x\). Considering the equation \(x = h(y)\), we notice that since we're revolving around the \(y\)-axis and the domain is given in terms of \(x\): \[x = h(y)\] To obtain this inverse function, first solve for \(x\): \[x=\sqrt{\frac{1}{y}-1}\] So, \(h(y)=x\), and the equation becomes: \[h(y) = \sqrt{\frac{1}{y}-1}\] The height of the rectangular strip is given by the length in the horizontal direction, which in this case is \(h(y)\), and the circumference of the shell produced is \(2\pi y\). With that, we can write the volume of each shell in terms of the limits of integration and the integrand: \[V = \int_{a}^{b} 2\pi y h(y) dy\] Since \(a\) would be the graph’s lowest \(y\) value and \(b\) the highest one, we now must find the range of \(y\). As we’re looking for the intersection points of \(y=\frac{1}{x^{2}+1}\) with lines \(x=0\) and \(x=2\), we can replace \(x\) in the function for these values. Doing so will result in: \[y=\frac{1}{0^{2}+1}=1 \Rightarrow a=1\] \[y=\frac{1}{2^{2}+1}=\frac{1}{5} \Rightarrow b=\frac{1}{5}\] Now, we have everything needed to set up the integral: \[V = \int_{1/5}^{1} 2\pi y \sqrt{\frac{1}{y}-1}\, dy\]

Step 3: Evaluate the integral

Given that our integral can be separated and written in the form: \[V = 2\pi \int_{1/5}^{1} y \sqrt{\frac{1}{y}-1}\, dy\] We’ll now proceed to make a substitution, let: \[u = 1/y -1 \Rightarrow y = \frac{1}{u+1} \Rightarrow dy = - \frac{1}{(u+1)^2} du\] Plugging the substitution into the integral, we get: \[V = 2\pi \int_{4}^{0} \frac{1}{u+1} \sqrt{u}\, \left(-\frac{1}{(u+1)^2}\right) du\] Simplifying the integrand, we have: \[V = -2\pi \int_{4}^{0} \frac{\sqrt{u}}{(u+1)^3} du\] Change the limits of integration to make them positive: \[V = 2\pi \int_{0}^{4} \frac{\sqrt{u}}{(u+1)^3} du\] Now we can use integration by parts or substitute again to solve this integral. Let's choose the latter method: \[v = u+1 \Rightarrow u = v-1 \Rightarrow du = dv\] The integral now becomes: \[V = 2\pi \int_{1}^{5} \frac{\sqrt{v-1}}{v^3} dv\] This integral is easier to evaluate, so we proceed with integration by parts, letting: \[p(v) = \sqrt{v-1} \Rightarrow p'(v) = \frac{1}{2 \sqrt{v-1}}\] \[q'(v) = \frac{1}{v^3} \Rightarrow q(v) = -\frac{1}{2v^2}\] Now we apply integration by parts: \[V = 2\pi \left[p(v) q(v) \bigg|_{1}^{5} - \int_{1}^{5} p'(v) q(v) dv\right]\] Solving the integral (we will omit the calculus steps, but they are quite standard): \[V = 2\pi (-\frac{1}{20} - \frac{1}{18})\] Finally, the volume is: \[V = \frac{-\pi}{180}\]

Key Concepts

Volume of Solids

When you revolve a region in a plane around an axis, you create a 3D solid. Imagine this as a series of stacked thin disks or shells. This becomes a convenient way to measure volume.
The volume of the solid generated by revolving a region can be calculated using various methods, one of which is the method of cylindrical shells.
This method is particularly useful when the axis of rotation is parallel to the region's length. In our example, revolving around the y-axis forms cylindrical shells.
The core idea is simple: calculate the volume of these infinitesimally small cylinders and sum them up using integration. The resulting integral gives us the total volume of the solid.
In essence, it's about breaking down a complex 3D shape into manageable cylindrical sections and then adding them together for the total volume.

Definite Integrals

In calculus, integrals are used to find areas, volumes, and more. A definite integral specifically gives us the accumulated total of a function's values over a specific interval.
This accumulated total is crucial in determining the volume of a solid, as it effectively adds up the volumes of infinitely many infinitesimal pieces.

• Lower and upper limits are fundamental parts of a definite integral. They define the start and end of the interval over which the integration is performed.
• The value of a definite integral can be thought of as the total sum of function values, weighted by their small, but infinite, delta widths.

In this problem, the definite integral helps us sum up the volumes of cylindrical shells, over the interval of y-values from 1/5 to 1.

Integration Techniques

Advanced integration techniques are often required to solve more complex integrals. In the exercise, solving the integral involves several steps and substitutions.
One such technique is substitution, where we make a change of variables to simplify the integral.
Another key technique used is integration by parts, which is ideal for dealing with products of functions.

• Substitution allows us to simplify integrals involving composites of functions.
• Integration by parts breaks down an integral of a product of functions into simpler parts.

In this solution, these techniques help us handle the complex expression \[ \frac{\sqrt{u}}{(u+1)^3} \] and eventually solve it to find the volume.

Revolving Regions

Revolving regions around an axis transforms a flat shape into a 3D object. Understanding this concept is fundamental when finding volumes of solids.
In our exercise, the region defined by the function \( y = \frac{1}{x^2 + 1} \) is revolved around the y-axis.
This transforms the original flat area into a layered cake-like structure made of cylindrical shells. These shells are created by revolving vertical strips of the region around the axis.

• Every point on these strips traces out a circle, forming the surface of the cylindrical shell.
• The method of cylindrical shells leverages this cylindrical form to simplify the volume calculation.

By setting up an integral across the range of y-values, the volume of each imaginary shell is calculated, summing up to the total volume of the solid formed.