Problem 19 - Calculus and its applications

Question

Find the volume generated by rotating the area bounded by the graphs of each set of equations around the \(y\) -axis. $$ y=2 x, x=0, x=3 $$

Step-by-Step Solution

Answer

The volume generated by rotating the region is \(36\pi\) cubic units.

Step 1: Understand the Region to be Rotated

The region is bounded by the line \(y = 2x\), the line \(x = 0\), and the line \(x = 3\). This region is a triangular area in the first quadrant.

Step 2: Set Up the Integral Expression for Volume

We use the method of cylindrical shells to find the volume of the solid of revolution. The formula for the volume using cylindrical shells when rotating about the \(y\)-axis is \[ V = \int_{a}^{b} 2\pi x (f(x)) \, dx \]where \(f(x)\) is the function defining the height of the shell and \([a, b]\) is the interval of \(x\). Here, \(f(x) = 2x\), \(a = 0\), and \(b = 3\).

Step 3: Substitute the Function into the Equation

Using the function \(f(x) = 2x\), substitute into the integral expression:\[ V = \int_{0}^{3} 2\pi x (2x) \, dx = \int_{0}^{3} 4\pi x^2 \, dx \]

Step 4: Evaluate the Integral

Evaluate the integral \(V = \int_{0}^{3} 4\pi x^2 \, dx\):First, find the antiderivative:\[ \int 4\pi x^2 \, dx = 4\pi \frac{x^3}{3} = \frac{4\pi}{3} x^3 \]Now, evaluate from 0 to 3:\[ V = \left[ \frac{4\pi}{3} x^3 \right]_{0}^{3} = \frac{4\pi}{3} (3^3) - \frac{4\pi}{3} (0^3) = \frac{4\pi}{3} (27) \]

Step 5: Simplify and Compute the Volume

Simplify and compute:\[ V = \frac{4\pi}{3} \times 27 = 36\pi \] cubic units.

Key Concepts

Cylindrical Shell Method

The Cylindrical Shell Method is a fascinating technique used to find the volume of a solid of revolution. When you rotate a region around a given axis, this method helps you calculate the volume of the resulting solid. It is particularly useful for rotating around the y-axis, especially when the shapes or regions involve complex curves or vertical strips.

• This method involves dissecting the solid into tiny cylindrical shells.
• Each shell's volume is calculated and then summed across the interval using integration.
• By multiplying each cylindrical shell's circumference, height, and thickness, you get its volume.

The formula to compute the volume in this case is:\[ V = \int_{a}^{b} 2\pi x (f(x)) \, dx \]where:

• \(x\) is the radius of a cylindrical shell,
• \(f(x)\) is the height of the shell,
• The bounds \([a, b]\) cover the region's x-values.

The method efficiently handles solids originating from rotating a region around axes parallel to the specified boundary.

Definite Integral

A Definite Integral is a crucial concept in calculus, especially when calculating areas, volumes, and other quantities over a specific interval. It represents the accumulation of a quantity, such as area or volume, and is used to sum up infinitesimally small changes over a continuous range.
In the problem at hand, the definite integral helps sum up the volumes of each of those tiny cylindrical shells between the limits. By evaluating the integral expression:\[V = \int_{0}^{3} 4\pi x^2 \, dx\]you get the total volume for the solid after rotation.
To evaluate a definite integral:

• Find the antiderivative of the function inside the integral.
• Apply the limits of integration to the antiderivative, calculating the result as \(F(b) - F(a)\).

The result gives you the accumulated value over the specified interval, in our case, the volume of the solid of revolution.

Solid Geometry

Solid Geometry touches on the study of three-dimensional shapes and solids, including how they are calculated and visualized. When you transform two-dimensional regions into three-dimensional solids through rotations, as with revolutions, solid geometry principles help clarify these shapes' properties and structures.
In this exercise, a triangle defined by lines \(y=2x\), \(x=0\), and \(x=3\) is rotated around the y-axis to give a three-dimensional solid whose volume you seek. Understanding solid geometry allows you to:

• Visualize how changes in dimensions create different geometric entities.
• Connect planar and solid shapes to understand depth and volume.
• Relate properties like symmetry and boundaries to form more complex shapes via transformations.

Edge-to-edge understanding of solid geometry ensures precise computations of volumes, surface areas, and the characteristics of the resulting shapes when dealing with revolutions and rotations.

Calculus Integration Techniques

Calculus Integration Techniques are the tools that allow you to compute areas, volumes, and other quantities involving continuous change. These techniques are fundamental in problems involving the cylindrical shell method and definite integrals.
In the given problem, integration involves finding the antiderivative of the integrand \(4\pi x^2\) and using it to evaluate the definite integral. Key integration steps include:

• Identifying the form of the function to determine the appropriate integration method.
• Applying power rule when the function is a polynomial, such as \(x^2\), by increasing the exponent by one and dividing by the new exponent.
• Substituting back into the integral boundaries to find the final volume.

By mastering these techniques, including using substitution, integration by parts, and partial fractions, you can solve a wide array of calculus problems, making sense of complex 2D and 3D relationships.