Problem 35 - Calculus Single Variable

Question

In each of Exercises 31-36, 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 between the \(y\) -axis and the curve \(x=y\) \(\exp \left(y^{3}\right), 0 \leq y \leq 1\)

Step-by-Step Solution

Answer

The volume is \( \frac{2 \pi}{3} (e - 1) \).

Step 1: Understand the Problem

We need to find the volume of the solid obtained by rotating the region \( \mathcal{R} \) between the \( y \)-axis and the curve \( x = y e^{y^3} \) from \( y = 0 \) to \( y = 1 \) about the \( x \)-axis.

Step 2: Set Up the Cylindrical Shells Formula

The method of cylindrical shells calculates volume using the formula: \[ V = 2 \pi \int_{a}^{b} y \cdot x(y) \, dy \] where \( x(y) \) is the expression for \( x \) in terms of \( y \), given as \( x = y e^{y^3} \), and the limits of \( y \) are from \( y = 0 \) to 1.

Step 3: Substitute into the Volume Formula

Substituting into the formula, we have: \[ V = 2 \pi \int_{0}^{1} y \cdot (y e^{y^3}) \, dy = 2 \pi \int_{0}^{1} y^2 e^{y^3} \, dy \] We now need to evaluate this integral.

Step 4: Evaluate the Integral

To evaluate \( \int_{0}^{1} y^2 e^{y^3} \, dy \), use the substitution \( u = y^3 \), then \( du = 3y^2 \, dy \) or \( y^2 \, dy = \frac{1}{3} du \). Changing limits: when \( y = 0, u = 0 \) and when \( y = 1, u = 1 \). Thus, we have: \[ V = 2 \pi \int_{0}^{1} \frac{1}{3} e^u \, du \]

Step 5: Integrate and Find Volume

Now, compute the integral: \[ V = \frac{2 \pi}{3} \left[ e^u \right]_{0}^{1} \] This becomes: \[ V = \frac{2 \pi}{3} \left( e^1 - e^0 \right) = \frac{2 \pi}{3} \left( e - 1 \right) \] Thus, the volume of the solid is \( \frac{2 \pi}{3} (e - 1) \).

Key Concepts

Volume of Solids of Revolution

When we talk about the volume of solids of revolution, we're referring to a method of finding volume by revolving a two-dimensional area around an axis. Imagine you have a piece of paper showing a graph. If you rotate this paper around a line, like the x-axis, you will create a 3D shape. This process is called rotating a region to create a solid of revolution.

One way to calculate the volume of these solids is through the method of cylindrical shells. This method is particularly useful when the axis of rotation is horizontal or vertical, and the solid extends in the direction perpendicular to the axis. Essentially, you imagine 'slicing' the solid into thin cylindrical shells. For each shell, you calculate its volume, and then you sum all these volumes together.

• Each cylindrical shell has a small thickness, making it nearly a cylinder.
• The height of each shell corresponds to the value of the function at that point.
• The radius of each shell is the distance from the axis of rotation.

By adding up all these individual shell volumes from one endpoint to the other, you get the total volume of the solid.

Integral Calculus

Integral calculus is all about adding things up. In our case, it's about summing up those cylindrical shells we mentioned earlier. The definite integral is our mathematical tool for doing this. Pretty neat, right?

With integral calculus, we can calculate areas, volumes, central points, and many other useful things. It is defined in terms of limits, and it helps us compute a multitude of real-world applications by breaking down massive quantities into infinitesimally small amounts which are then summed together.

• In the context of cylindrical shells, we're interested primarily in volume.
• The definite integral gives us a precise representation of accumulated quantities, such as volume.

The integral captures the accumulation of small pieces' volume (which are our cylindrical shells) over a defined range, providing us with the total volume when rotating a region around an axis.

Definite Integrals

A definite integral is a way of summing up stuff over a certain interval, from start to finish. It's like tracking the quantity continuously accumulated over a range. 

In this exercise, we used the definite integral to find out how much space a rotated region takes up. By setting up the integral from the lower bound (y = 0) to the upper bound (y = 1), you collect all the 'shell volumes' between these bounds.

The notation\[ \int_{a}^{b} f(x) \, dx \]represents the definite integral of function \(f(x)\) from a to b:

• \(a\) and \(b\) are your limits of integration, defining the start and end of accumulation.
• \(f(x)\) is your function, describing how each shell fits into the total volume.
• The result gives a numerical value describing the total quantity accumulated.

The usage of the definite integral in the context of cylindrical shells allows us to solve complex geometry problems in a straightforward manner, providing precise volume calculations.