Problem 129 - Calculus Volume 1

Question

For the following exercises, use shells to find the volumes of the given solids. Note that the rotated regions lie between the curve and the x-axis and are rotated around the y-axis. $$y=5 x^{3}-2 x^{4}, x=0, \text { and } x=2$$

Step-by-Step Solution

Answer

The volume is approximately \( 21.3334\pi \).

Step 1: Understand the Problem

We need to find the volume of the solid formed by rotating the region bounded by the curve \( y = 5x^3 - 2x^4 \), the x-axis, and the vertical lines \( x=0 \) and \( x=2 \) around the y-axis. This requires using the method of cylindrical shells.

Step 2: Set Up the Integral for Volume Using Shells

For the cylindrical shell method, the volume \( V \) is given by:\[ V = 2\pi \int_{a}^{b} (radius)(height) \, dx \]The radius of a shell at a point \( x \) is \( x \), and the height is the function value \( y = 5x^3 - 2x^4 \). Thus, the integral becomes:\[ V = 2\pi \int_{0}^{2} x(5x^3 - 2x^4) \, dx \]

Step 3: Simplify the Integral Expression

Simplify the integrand:\[ 2\pi \int_{0}^{2} (5x^4 - 2x^5) \, dx \]This gives us the expression \( 2\pi \int_{0}^{2} (5x^4 - 2x^5) \, dx \) to evaluate.

Step 4: Evaluate the Integral

Evaluate the integral:\[ \int (5x^4 - 2x^5) \, dx = \frac{5}{5}x^5 - \frac{2}{6}x^6 = x^5 - \frac{1}{3}x^6 \]Now, apply the limits from 0 to 2:\[ V = 2\pi \left[ \left(x^5 - \frac{1}{3}x^6 \right) \bigg|_0^2 \right] \]

Step 5: Calculate the Definite Integral

Substitute the limits into the evaluated integral:\[ V = 2\pi \left[ \left(2^5 - \frac{1}{3}(2)^6 \right) - \left(0^5 - \frac{1}{3}(0)^6 \right) \right] = 2\pi \left[ 32 - \frac{1}{3}(64) \right] \]\[ V = 2\pi \left[ 32 - 21.3333 \right] = 2\pi (10.6667) \]

Step 6: Final Calculation

Calculate the final volume:\[ V = 2\pi \times 10.6667 \approx 21.3334\pi \]Thus, the volume of the solid is approximately \( 21.3334\pi \).

Key Concepts

Volume of Solids

When studying calculus, the concept of volume of solids is crucial, especially when dealing with shapes obtained by revolving a region around an axis. This process forms what we call a solid of revolution. Understanding how these volumes are computed helps reveal the connection between geometry and calculus. In many problems, such as the one we are examining, the area between a curve and an axis is rotated to form a three-dimensional shape.
One popular method to calculate these volumes is the method of cylindrical shells. This approach simplifies the problem by breaking down the solid into numerous cylindrical shells. Calculating the volume of these individual shells and summing them up gives us the total volume of the solid. This is particularly helpful when the solid does not have a straightforward geometric shape. In our problem, the use of cylindrical shells allows us to handle the complexity of the volume obtained by revolving a non-linear curve around an axis.

Rotating Regions

Rotating regions around an axis is an essential concept in calculus when dealing with the formation of solids. In our exercise, the given region under a specific curve is rotated around the y-axis to form a new three-dimensional shape. This practice of rotational symmetry is fundamental in visualizing how two-dimensional areas transform into three-dimensional solids.
The axis around which a region rotates can significantly affect the volume and shape of the resulting solid. Consequently, understanding the impact of different axes is critical. For instance, if the region between the curve and the x-axis had been rotated around the x-axis instead, the shape of the formed solid and its calculated volume would have differed significantly. Analyzing these rotations helps us understand how changes in dimensions occur through calculus.

Calculus Integrals

In the world of calculus, integrals play a vital role. They enable us to compute areas, volumes, and many other quantities. In this exercise, integrals help calculate the volume of a complex shape formed by revolving a region. The integral setup reflects the product of radius and height for cylindrical shells, which is key for deriving the volume of the solid.
The integral formula used here is expressed as:

• \[ V = 2\pi \int_{a}^{b} (radius) \cdot (height) \, dx \]

In our example, the radius is simply the x-coordinate of the shell, while the height corresponds to the value of the function denoting the region's boundary. Crucially, calculus integrals allow us to tackle the complexity involved in summing the contributions of infinitely small shells to find the total volume. This process emphasizes the power of calculus in solving problems involving continuous change.

Definite Integrals

Definite integrals are a specific type of integral where the limits of integration are clearly defined. They allow us to find the total accumulation of quantities over a specified interval. In problems like ours, using definite integrals is vital for determining the exact volume of a solid formed by rotating a region.
For the exercise, the definite integral is evaluated from the lower boundary x = 0 to the upper boundary x = 2. These limits define the region under consideration, ensuring that only the portion of the curve between these x-values contributes to the volume.

• By evaluating the definite integral:\[\int_{0}^{2} (5x^4 - 2x^5) \, dx\]we can precisely account for the entire volume encompassed within these boundaries, resulting in an accurate computation of the solid's volume.

Thus, the definite integral not only aids in computing exact values but also facilitates understanding how accumulation works over a range.