Question

Sketch the region \(R\) bounded by the graphs of the given equations, and show a typical vertical slice. Then find the volume of the solid generated by revolving \(R\) about the \(x\) -axis. \(y=x^{3}, x=3, y=0\)

Short answer

The volume is approximately 982.45 cubic units.

Step-by-step solution

Identify the Given Equations and Bounds

The problem provides three equations which define the region: \(y = x^3\), which is a cubic curve, \(x = 3\), a vertical line, and \(y = 0\), the x-axis. These boundaries enclose the region \(R\) that we are interested in.

Visualize and Sketch the Region

First, sketch the graph of \(y = x^3\) starting from \(x = 0\) to \(x = 3\). The line \(x = 3\) serves as the right boundary, and the x-axis \(y = 0\) is the lower boundary. The region \(R\) is enclosed between \(x = 0\), \(x = 3\), above the x-axis and below the cubic curve.

Set Up the Integral for Volume

To find the volume of the solid generated by revolving region \(R\) about the x-axis, we use the disk method. The volume \(V\) is given by the integral: \[ V = \pi \int_{a}^{b} [f(x)]^2 \, dx \]where \(f(x) = x^3\), \(a = 0\), and \(b = 3\).

Compute the Integral

Substitute \(f(x) = x^3\) into the formula:\[ V = \pi \int_{0}^{3} (x^3)^2 \, dx = \pi \int_{0}^{3} x^6 \, dx \].Integrate \(x^6\):\[ \int x^6 \, dx = \frac{x^7}{7} + C \].Now evaluate from \(0\) to \(3\):\[ V = \pi \left[ \frac{3^7}{7} - \frac{0^7}{7} \right] = \pi \frac{2187}{7} \].

Calculate the Numerical Value

Finally, compute the numerical value:\( \pi \frac{2187}{7} \approx 982.45 \). Thus, the volume of the solid is approximately 982.45 cubic units.

Key concepts

Disk Method

When solving problems involving the volume of solids of revolution, the disk method is a fundamental technique. It allows us to calculate volumes by decomposing the solid into a series of thin disk-like slices.

Each slice is perpendicular to the axis of rotation. In this exercise, our region defined by cubic function revolves about the x-axis. Each disk then has a radius equal to the function value at that point, or in this case, the function value of the cubic equation, namely, \( y = x^3 \).

To find the volume using the disk method:

• Each disk's volume is approximated as the product of its area and thickness.
• The area of a disk is \( \pi [f(x)]^2 \) and the thickness is \( dx \).

Hence, the integral \( \pi \int_a^b [f(x)]^2 \, dx \) sums up the contributions of all disks from the lower limit \( a \) to the upper limit \( b \).

In this example, our function is \( x^3 \), making the integral setup \( \pi \int_0^3 (x^3)^2 \, dx \).

Cubic Functions

Cubic functions are a type of polynomial function where the highest degree is three. Mathematically, they have the form \( y = ax^3 + bx^2 + cx + d \). In our exercise, the function provided is the simplified version \( y = x^3 \).

Cubic functions, like \( x^3 \), often exhibit certain symmetrical patterns in their graphs. They start from one end of the graph and curve, possibly with two inflection points, before moving to the opposing end.

Understanding the behavior and characteristics of cubic functions is crucial, especially when sketching them. They generally:

• Increase continuously in one direction without bouncing back.
• Have a single real root if other terms are absent, otherwise up to three different roots in certain configurations.

Seeing how the function \( y = x^3 \) increases helps to visualize the region we are revolving around the axis.

Integral Calculus

Integral calculus is an essential tool for solving volumes of solids, particularly when using rotation methods like the disk method. It allows us to sum up an infinite number of infinitesimally small quantities.

In our problem, the function \( x^3 \) revolved around the x-axis forms a region with a flat bottom and a curved top. We need to compute the integral to determine the solid's volume.

The integral formed \( \pi \int_0^3 x^6 \, dx \) results from squaring the function \( x^3 \) and represents the summation of areas of all disks.

To perform this integration:

• Start by finding the antiderivative of \( x^6 \).
• The antiderivative is \( \frac{x^7}{7} \). Don't forget the constant of integration \( + C \) which is irrelevant for definite integrals.
• Evaluate this from \( 0 \) to \( 3 \) to find the total volume.

Such an understanding of integral calculus is vital when calculating areas under curves and, by extension, volumes.

Graph Sketching

Graph sketching provides a visual interpretation of algebraic functions. This helps in problem-solving by illustrating how different regions interact or are bounded.

In this exercise, the first step is sketching the graph of \( y = x^3 \) from \( x = 0 \) to \( x = 3 \). Here are key points to focus on:

• The point where \( x = 0 \) starts the curve at the origin.
• The curve is continuous and rises steeply as \( x \) increases, passing through points like \( (1,1) \) and \( (3,27) \).
• The vertical line \( x = 3 \) is the boundary, constraining our region to revolve.
• The intersection of the curve with the x-axis (\( y = 0 \)) defines the entire lower boundary of the solid.

Sketching isn't just about drawing; it involves knowing how the function moves and how it interacts with other shapes, like the bounds \( x = 3 \) or \( y = 0 \). Such graphical understanding simplifies solving equations and identifying regions for rotation.