Problem 114 - Calculus Volume 1

Question

For the following exercise, find the volume generated when the region between the two curves is rotated around the given axis. Use both the shell method and the washer method. Use technology to graph the functions and draw a typical slice by hand. [T] Over the curve of y = 3x, x = 0, and y = 3 rotated around the y-axis.

Step-by-Step Solution

Answer

The volume is \( \pi \) cubic units.

Step 1: Identify the region

The region is bounded by the curves \( y = 3x \), \( x = 0 \), and \( y = 3 \). We can translate these into the x-interval: \( x = 0 \) to \( x = 1 \) since when \( y = 3 \), \( x = 1 \). This forms a triangular region in the first quadrant.

Step 2: Apply the Washer Method

For the washer method, we rotate around the y-axis. The inner radius is zero because the region starts from the y-axis, there's no "hole," and the outer radius is the x-coordinate \( x = \frac{y}{3} \) from \( y = 0 \) to \( y = 3 \). Volume is given by the integral \[ V = \pi \int_{0}^{3} \left( \left(\frac{y}{3}\right)^2 \right) \, dy = \pi \int_{0}^{3} \frac{y^2}{9} \, dy. \]

Step 3: Integrate for Washer Method

Calculate the definite integral obtained:\[ V = \pi \left[ \frac{y^3}{27} \right]_0^3 = \pi \left( \frac{27}{27} - 0 \right) = \pi. \]

Step 4: Apply the Shell Method

For the shell method, consider a line parallel to the y-axis. The height of the shell is \( 3 - 3x \), the radius \( x \), and thickness \( \Delta x \). Express volume with height, radius, and calculate \[ V = 2\pi \int_{0}^{1} x(3 - 3x) \, dx = 2\pi \int_{0}^{1} (3x - 3x^2) \, dx. \]

Step 5: Integrate for Shell Method

Compute the integral calculated in the shell method expression:\[ V = 2\pi \left[ \frac{3x^2}{2} - x^3 \right]_0^1 = 2\pi \left( \frac{3}{2} - 1 \right) = \pi. \]

Step 6: Compare Both Methods

Both the shell and washer methods give the same volume answer, ensuring consistency in calculations: volume is \( \pi \).

Key Concepts

Shell Method

The shell method is a powerful tool for finding the volume of a solid generated by rotating a region around an axis. This method is especially useful when the axis of rotation is parallel to the slicing direction. Let's break it down to make it understandable.

Imagine cutting a region into thin vertical slices. Each slice is just like a cylindrical shell, and when rotated, it forms a hollow cylinder. Our goal is to calculate the volume of each shell and sum them up.

• The height of each shell is the difference in y-values over the region, in this case, given by the function.
• The radius is the distance from the y-axis to the shell, typically represented as the x-coordinate.
• The thickness is a small change in x, expressed as \( \Delta x \).

The formula for the volume of each shell is:
\[ V = 2\pi \int_a^b x \cdot (f(x)) \, dx \]In our solution, the function is \( 3x \), which means every shell has a height of \( 3 - 3x \). Therefore, our integral ranges from 0 to 1. Calculating shows the volume is \( \pi \).

Washer Method

The washer method is used similarly to the shell method but applies better for rotations perpendicular to the slicing direction. The concept is based on subtracting the volume of a hole from a disk to form a washer, making it suitable for regions between two curves.

Picture each horizontal slice as a flat washer. By rotating around the y-axis, we create these washers whose volume is easily computed by considering the difference between the outer and inner radii.

• Outer Radius: Represents the far edge of the region from the axis of rotation, defined by \( x = \frac{y}{3} \).
• Inner Radius: Since there's no inner boundary curve in this specific problem, it is zero.
• The thickness is a tiny change in y or \( \Delta y \).

The formula for the volume of a washer is:
\[ V = \pi \int_a^b \left( R^2 - r^2 \right) \, dy \]In this scenario, the volume is evaluated from y = 0 to y = 3. The resulting calculations also yield a volume of \( \pi \).

Definite Integrals

Definite integrals are essential in calculating volumes, as they allow us to sum up infinitely many small contributions. In both the shell and washer methods, integrals calculate the cumulative effect of all slices or shells.

A definite integral takes a function, examines it over a specified interval, and finds the "total change".

• In the washer method, it calculates the net volume of washers from one end to the other of the interval.
• For the shell method, it sums up the volume of all shells across the range.

The interesting part is that despite differing methods and integrands, definite integrals often deliver consistent and reliable results.
Understanding how they function within each method gives clarity on why both methods arrive at the same volume—\( \pi \). In this exercise, comparing both shows us the integrals both integrate over similar ranges, maintaining consistency.