Problem 17 - Essential Calculus

Question

Use the shell method to set up and evaluate the integral that gives the volume of the solid generated by revolving the plane region about the \(x\) -axis. $$ y=x^{3}, \quad x=0, \quad y=8 $$

Step-by-Step Solution

Answer

The volume of the solid is \(\frac{64 \pi}{5}\) cubic units.

Step 1: Identify the Regions

The plane region is bound by the curve \(y = x^{3}\), the line \(x = 0\) (the y-axis), and horizontally by \(y = 8\), which is the point where \(x = 2\) on the curve \(y = x^{3}\). So, the solid is rotated about the \(x\)-axis from \(x = 0\) to \(x = 2\).

Step 2: Set Up the Integral for Volume

For the shell method, the height of the shell \(h(x)\) is given by 'y' from the curve \(y = x^{3}\), and the radius 'r(x)' from the \(x\)-axis is \(x\). Thus the volume can be given by \[V = 2 \pi \int_{0}^{2} x \cdot x^{3} dx\]

Step 3: Evaluate the Integral

To evaluate the integral, first simplify inside the integral to \[V = 2 \pi \int_{0}^{2} x^{4} dx.\] Using the power rule for integration, the integral of \(x^{4}\) is \(\frac{1}{5}x^{5}\). Evaluate this from \(0\) to \(2\), which gives us: \[V = 2 \pi \left[ \frac{1}{5} \cdot (2^{5}) - \frac{1}{5} \cdot (0^{5}) \right].\]

Step 4: Simplify

On simplifying, we get: \[V = 2 \pi \left[ \frac{32}{5} - 0 \right] = 2 \pi \cdot \frac{32}{5} = \frac{64 \pi}{5}.\]

Key Concepts

Integral Setup

When you start using the shell method to find the volume of a rotating solid, setting up the integral correctly is crucial. It's like assembling the right blueprint before constructing a building. The integral's structure can make or break the solution.
To set up the integral using the shell method, you first need to understand the relevant region. In the given problem, the plane region is bounded by the curve \(y = x^3\), and it rotates around the \(x\)-axis.

• The boundary line \(x=0\) represents one edge of this region.
• When we reach \(y=8\), this equates to \(x=2\) on the curve, providing us with the other bound of integration.

For the shell method, visualize the shells:

• The height \(h(x)\) of each shell is equal to \(y\) from the curve.
• The radius \(r(x)\) is the distance from the axis of rotation, which is simply \(x\).

Putting it all together, the integral you set up is \(2\pi \int_{0}^{2} x \cdot x^3 \, dx\). This setup allows you to calculate the desired volume accurately.

Rotating Solids

Rotating a two-dimensional region around an axis to form a three-dimensional object is a fascinating concept in calculus. This process results in what we call "rotating solids," or more precisely, solids of revolution.
Creating these solids involves the rotation of a planar region about a specific axis. For the given problem, the plane region enclosed by the curve \(y = x^3\) is rotated about the \(x\)-axis. Imagine wrapping paper around an object to understand this better; as you wrap, the paper forms a three-dimensional shape.

• The method of shells is just one approach to determine the volume of such a solid.
• It's particularly useful when the region being rotated is easier to express in terms of \(x\) rather than \(y\).

Embrace the mental image of cylindrical shells being stacked to create the full volume. By summing up the volumes of these shells over the interval \([0, 2]\), we gain a comprehensive understanding of the entire solid's volume.

Power Rule for Integration

Once the integral is set up, evaluating it properly requires using certain calculus rules, one of which is the Power Rule for Integration.
This rule is immensely helpful for simplifying integrals where the integrand is an algebraic expression. For a function \(x^n\), the power rule tells us that its integral is given by \(\frac{1}{n+1} x^{n+1}\), where \(neq -1\).
In the context of this exercise, the simplified form inside the integral is \(x^4\). The power rule allows us to integrate this easily:

• The integral of \(x^4\) becomes \(\frac{1}{5} x^5\).
• We then evaluate this from \(0\) to \(2\).

Great care should be taken when applying these limits for accurate results: Substitute the limits into the antiderivative to get \(\left[\frac{1}{5}(2^5) - \frac{1}{5}(0^5)\right]\). After simplifying, the final volume \(\frac{64\pi}{5}\) emerges, which beautifully demonstrates how integration rules convert a theoretical problem into a tangible answer.