Problem 16 - 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. \(y=3^{x^{2}}, \quad y=0, \quad x=0, \quad x=1 ; \quad\) the \(y\) -axis

Step-by-Step Solution

Answer

The volume of the solid generated by revolving the region bounded by \(y = 3^{x^2}\), \(y = 0\), \(x = 0\), and \(x = 1\) around the \(y\)-axis is \(\frac{2π}{\ln{3}}\).

Step 1: Visualize the region

Draw the graph of \(y = 3^{x^2}\), the x-axis (y=0), the vertical line \(x=0\), and the vertical line \(x=1\). The resulting area bounded by these curves will be revolved around the y-axis to form the solid.

Step 2: Identify a representative rectangle

For the method of cylindrical shells, you need a representative rectangle parallel to the axis of rotation. In this case, it means a vertical rectangle. Consider the vertical rectangle with one side on the curve \(y = 3^{x^2}\), and another side on the line \(x=1\). Now let's set up the cylindrical shells volume formula.

Step 3: Set up the volume formula

The formula for the volume of cylindrical shells is given by: \[V = 2π \int_a^b x \cdot f(x) \, dx\] where \(f(x)\) is the function we are revolving, and \(a\) and \(b\) are the integration limits (in this case, \(x=0\) to \(x=1\)). The height of each shell (cylinder) is given by \(y=f(x)=3^{x^2}\). Now we can plug in the values: \[V = 2π \int_0^1 x \cdot 3^{x^2} \, dx\]

Step 4: Evaluate the integral

Now we need to evaluate the integral to find the volume: \[V = 2π \int_0^1 x \cdot 3^{x^2} \, dx\] Make a substitution by letting \(u = x^2\); therefore, \(\frac{du}{dx}=2x\), and \(dx = \frac{du}{2x}\): \[V = 2π \int_0^1 x \cdot 3^{u} \frac{du}{2x}\] The x in the numerator and denominator cancel each other out, and we integrate with respect to u from 0 to 1: \[V = π \int_0^1 3^{u} \, du\] Now, we can evaluate the integral: \[V = π \left[\frac{1}{\ln{3}} (3^u)\right]_0^1\] \[V = π \left(\frac{1}{\ln{3}} (3^1) - \frac{1}{\ln{3}} (3^0)\right)\] \[V = π \left(\frac{3 - 1}{\ln{3}}\right) = π \cdot \frac{2}{\ln{3}}\] So, the volume of the solid is \(\frac{2π}{\ln{3}}\).

Key Concepts

Volume of Solids

Calculus provides powerful methods for calculating the volume of solids created by rotating a region around an axis. This is particularly useful in fields like engineering and physics, where these calculations can represent real-world situations.
To determine the volume of a solid formed by rotation, we use integration. One of the approaches is the method of cylindrical shells, a technique that breaks the solid into component cylindrical shapes.

• The region to be rotated is described by specific boundaries—often involving graphs of mathematical functions.
• The axis of rotation determines how the volume is calculated geometrically.

Understanding these volumes assists in predictions about structures and materials when faced with complex geometrical designs.

Integral Calculus

Integral calculus is a branch of mathematics focused on the concept of summation, especially finding the total or whole from its parts. In the context of volume calculation, integral calculus helps us sum the infinitesimally small elements creating the solid.
This discipline of calculus is essential because:

• It allows us to calculate areas under curves, which is essentially what we're doing when finding volumes of solids.
• Integral calculus lets us handle continuous data, meaning we're able to model smoothly varying natural phenomena.

Moreover, we use definite integrals, as one of the solutions involves evaluating an integral over a set boundary, making integral calculus central to volume calculations.

Calculus Problems

Calculus problems, such as finding the volume of a solid of revolution, often challenge students with multistep and abstract operations. These problems involve:

• Interpreting the function and its graphical representation.
• Determining limits, substitutions, and transformations to simplify equations.

Understanding calculus problems revolves around recognizing patterns and applying standard solutions, like the cylindrical shell method, keeping track of the boundaries and transformations.
Solving these challenges not only improves comprehension of mathematical concepts but also enhances problem-solving skills, beneficial across various scientific disciplines.

Cylindrical Integration

Cylindrical integration is a direct application of the method of cylindrical shells. This method transforms a two-dimensional area into a three-dimensional solid by integration:

• A "shell" is effectively a hollow cylinder, formed by revolving a vertical or horizontal strip around an axis.
• Instead of calculating the volume of every thin disk, which can be cumbersome, cylindrical integration uses the surface area of shells to find the volume immediately.

For example, given a function like \(y = 3^{x^2}\), revolving it around the y-axis requires setting up a formula like \[V = 2\pi \int_a^b x \cdot f(x) \, dx\]. This advantageously reduces the amount of work necessary to find the volume, as demonstrated by the simplification techniques using substitutions in calculus.