Problem 45 - Calculus Single Variable

Question

In each of Exercises 43-48, use the method of cylindrical shells to calculate the volume \(V\) of the solid that is obtained by rotating the given planar region \(\mathcal{R}\) about the \(x\) -axis. \(\mathcal{R}\) is the region that is bounded on the left by the curve \(y=5-x^{2}\) for \(-2 \leq x \leq-1,\) on the right by the curve \(y=5-x^{2}\) for \(1 \leq x \leq 2,\) above by the line segment \(y=4\) for \(-1 \leq x \leq 1\), and below by the line segment \(y=1\) for \(-2 \leq x \leq 2\)

Step-by-Step Solution

Answer

The volume is \(\frac{656\pi}{15}\).

Step 1: Understand the Region of Rotation

To find the volume, first understand the region of interest. The region R is bounded by a parabola \(y=5-x^2\) for \(-2 \leq x \leq 2\), the line \(y=4\) for \(-1 \leq x \leq 1\), and the line \(y=1\) for \(-2 \leq x \leq 2\). This region is symmetric about the \(x\)-axis, and we will use the method of cylindrical shells to find the volume of the solid formed when this region is rotated about the \(x\)-axis.

Step 2: Define the Height of the Shells

For the cylindrical shell method, the height of a shell at a certain \(y\)-value is the horizontal distance between the curve \(y=5-x^2\) and the line perpendicularly intersecting the \(y\)-axis (i.e., the line \(x=\pm \sqrt{5-y}\)). Since the region is bounded by \(y=1\) and \(y=4\), the height of the cylindrical shell is \(2\sqrt{5-y}\).

Step 3: Express the Radius and Thickness of the Shells

The radius of each cylindrical shell is \(y\) and the thickness of the shell is \(dy\). These are determined since the shell method is compatible with horizontal strips (across the \(y\)-axis), and this formulation is necessary to calculate volume by integration.

Step 4: Setup the Integral

The volume \(V\) of the solid using cylindrical shells is given by the integral formula: \[ V = \int_{y_{min}}^{y_{max}} 2\pi (radius)(height) \cdot (thickness) \]Here, \(y_{min} = 1\) and \(y_{max} = 4\). So, the integral becomes: \[ V = 2\pi \int_{1}^{4} y \cdot 2\sqrt{5-y} \, dy \]

Step 5: Compute the Integral

Simplify and calculate the integral:\[ V = 4\pi \int_{1}^{4} y\sqrt{5-y} \, dy \]To solve this, use substitution. Let \(u = 5-y\), thus \(du = -dy\) and the limits for \(u\) are from \(u=4\) when \(y=1\) to \(u=1\) when \(y=4\). Therefore, \[ V = 4\pi \int_{4}^{1} (5-u)\sqrt{u} \, (-du) \]Switch the bounds of integration:\[ V = 4\pi \int_{1}^{4} (5-u)\sqrt{u} \, du \]Distribute within the integral:\[ V = 4\pi \left( \int_{1}^{4} 5\sqrt{u} \, du - \int_{1}^{4} u\sqrt{u} \, du \right) \]

Step 6: Evaluate the Integrals

Calculate each integral:\[ \int 5\sqrt{u} \, du = 5 \cdot \frac{2}{3} u^{3/2} + C \]Evaluate from \(1\) to \(4\):\[ 5 \times \frac{2}{3} [ (4)^{3/2} - (1)^{3/2} ] = \frac{10 \times (8 - 1)}{3} = \frac{70}{3} \]Next integral:\[ \int u^{3/2} \, du = \frac{2}{5} u^{5/2} + C \]Evaluate from \(1\) to \(4\):\[ \frac{2}{5} [ (4)^{5/2} - (1)^{5/2} ] = \frac{2 \times (32 - 1)}{5} = \frac{62}{5} \]

Step 7: Combine Results and Final Calculation

Now subtract the results of the two integrals:\[ \text{Result} = \frac{70}{3} - \frac{62}{5} \]Combine over a common denominator:\[ \frac{350}{15} - \frac{186}{15} = \frac{164}{15} \]Calculate the volume:\[ V = 4\pi \cdot \frac{164}{15} = \frac{656\pi}{15} \]

Step 8: Conclusion

The volume of the solid obtained by rotating the region \(\mathcal{R}\) about the \(x\)-axis is \(\frac{656\pi}{15}\).

Key Concepts

Volume of Solids of Revolution

Finding the volume of solids of revolution is a captivating application of calculus that allows us to visualize and compute the volume of a 3D object created by rotating a 2D region around an axis. One of the most effective methods for calculating these volumes is the **cylindrical shell method**. In this exercise, the method is applied to the region bounded by the curve, lines, and the axis itself.

The process essentially involves breaking down the solid into coaxial cylindrical shells. Imagine slicing the solid into thin rings, each with a different radius corresponding to its distance from the axis of rotation. The summed volumes of these rings, as they become infinitesimally thin and the number of rings approaches infinity, total to the volume of the solid.

The formula used for the cylindrical shell method is:

• Volume, \( V = 2\pi \int_{a}^{b} x \, y(x) \, dx \) for solids revolving around the x-axis, where \( x \) is the radius of the shell and \( y(x) \) is its height.

In practical terms, as was demonstrated in the original exercise, setting up the integral correctly represents the biggest challenge. But once the correct shells are outlined and the appropriate limits identified, the process becomes a straightforward integration task.

Definite Integrals

Definite integrals are a fundamental concept in calculus, used here to calculate the exact volume of a solid of revolution. This involves calculating an integral that has set limits, capturing the net area —or actual volume, in this context— within these bounds.

In this exercise, the integral was defined by constraints set by the region between curves and lines. These constraints determine the limits of the integral, which are as crucial as the function itself for accurately computing the volume.

The volume computation here uses a definite integral between 1 and 4:

• \( V = 2\pi \int_{1}^{4} y(2\sqrt{5-y}) \, dy \)

This integral provides the volume of the cylindrical shells between these y-values, effectively corresponding to the height of each shell and where they are positioned along the y-axis.

Via techniques like substitution, the calculation simplifies, converting the real world representation into a mathematical solution. Substitution is often used to simplify complex integrands into more manageable forms, as seen when substituting \( u = 5-y \).

Integral Calculus

Integral calculus is the branch of calculus involved in finding the integral of functions, represented as areas under curves or the volumes of solids, such as our example solid of revolution. Within integral calculus, there are techniques and formulae that simplify the integration process.

In this exercise, integral calculus came into play during the setup and solving of our definite integral to compute the volume. Techniques such as integration by substitution made complex integrals manageable, allowing them to be broken down into simpler parts.

For learners, the key to mastering integral calculus lies in understanding:

• **Basic integration rules and techniques** like substitution and integration by parts.
• The ability to **interpret real-world problems** like solid revolutions into mathematical terms.
• **Connect the dots** between the calculus concepts and their practical applications.

Appreciating these concepts helps in tackling exercises with confidence, enabling learners to progress from basic to advanced calculus topics. Integral calculus doesn't only compute areas or volumes; it offers insights into the dynamic systems and changes in mathematics and beyond.