Question

For the following regions \(R\), determine which is greater - the volume of the solid generated when \(R\) is revolved about the \(x\) -axis or about the y-axis. \(R\) is bounded by \(y=4-2 x,\) the \(x\) -axis, and the \(y\) -axis.

Short answer

Answer: The volume generated by revolving the region R around the y-axis is greater.

Step-by-step solution

Find the intersection points between the curves and axes

To find the limits of integration, we need to find the points where the curve \(y=4-2x\) intersects the \(x\)-axis and the \(y\)-axis. For the \(x\)-axis, set \(y=0\) and solve for \(x\): $0 = 4 - 2x \\ x = 2$ For the \(y\)-axis, set \(x=0\) and solve for \(y\): $y = 4-2(0)\\ y = 4$ So, the region R is bounded by the points \((0, 0), (2, 0)\), and \((0, 4)\).

Set up the integral for volume when revolved around the x-axis

To find the volume when revolved around the x-axis, we will use the disk method. The formula for the volume is: \(V_x = \pi \int_a^b [f(x)]^2 dx\) Since we are given the function \(f(x) = 4-2x\), we can then set up the integral: \(V_x = \pi \int_0^2 (4-2x)^2 dx\)

Calculate the volume for x-axis revolution

Now, we can evaluate the integral to find the volume: $V_x = \pi \int_0^2 (4-2x)^2 dx \\ V_x = \pi \int_0^2 (16 - 16x + 4x^2) dx \\ V_x = \pi[ \frac{4}{3}x^3 - 8x^2 + 16x]_0^2 \\ V_x = \pi(\frac{4}{3}(8) - 8(4) + 16(2)) \\ V_x = \frac{32}{3}\pi$

Set up the integral for the volume when revolved around the y-axis

To find the volume when revolved around the \(y\)-axis, we need to first write \(x\) in terms of \(y\). From the given equation, we have: \(x = \frac{4-y}{2}\) Now use the disk method to set up the integral for the volume when revolved around the y-axis: \(V_y = \pi \int_c^d [g(y)]^2 dy\) Using the new function \(g(y) = \frac{4-y}{2}\) and the limits of integration found in Step 1: \(V_y = \pi \int_0^4 (\frac{4-y}{2})^2 dy\)

Calculate the volume for y-axis revolution

Now, we can evaluate the integral to find the volume: $V_y = \pi \int_0^4 (\frac{4-y}{2})^2 dy \\ V_y = \pi \int_0^4 (\frac{16 - 8y + y^2}{4}) dy \\ V_y = \pi [\frac{y^3}{12} -\frac {4y^2}{3} + 4y]_0^4 \\ V_y = \pi (\frac{64}{12} - \frac {4(16)}{3} + 4(4)) \\ V_y = \frac{40}{3}\pi$

Compare the volumes

Now that we have calculated the volumes for both x-axis and y-axis revolutions, we can compare them: $V_x = \frac{32}{3}\pi \\ V_y = \frac{40}{3}\pi$ Since \(\frac{40}{3}\pi > \frac{32}{3}\pi\), the volume of the solid generated when the region R is revolved around the y-axis is greater than the volume when revolved around the x-axis.

Key concepts

Disk Method

The disk method is a powerful technique in calculus used to find the volume of a solid of revolution. Imagine a two-dimensional area being rotated around an axis, creating a three-dimensional shape. The disk method helps determine the volume of such a shape by approximating it as a series of thin disks stacked along the axis of rotation.

To compute the volume using the disk method, follow these steps:

• Identify the curve that defines the boundary of the region you're interested in revolving.
• Determine the axis of rotation (usually the x-axis or y-axis).
• Divide the solid into an infinite number of infinitesimally thin disks or cylinders.
• Calculate the volume of each disk and then take the sum as an integral over the interval of interest.

In mathematical terms, if you're revolving around the x-axis and the function is given as \(f(x)\), the formula for the volume \(V\) is:\[ V = \pi \int_{a}^{b} [f(x)]^2 \, dx \]This method is ideally used when the solid is rotated about its horizontal axis, leading to a simple interpretation where radius and height of the disks meet the function directly.

Integral Calculus

Integral calculus is a branch of mathematics that deals with integrals, which are used to find areas, volumes, and central points. Integral calculus plays a crucial role when calculating volumes of solids of revolution, such as in the disk method.

In essence, integrals help us "sum up" infinitely many small quantities. For example, when calculating the area under a curve, integrals add up all the narrow slices stacked from left to right.

Why is this important in finding volumes? Consider each small slice as a disk; an integral allows you to calculate the total volume of all such disks as one smooth operation:

• The definite integral \(\int_{a}^{b} f(x) \, dx\) sums up the function's values over a set interval \([a, b]\).
• Each disk's volume is a small slice \(\Delta x\) thick with an area \(\pi (f(x))^2\), contributing to the total volume.

Integral calculus is not only significant for calculating areas and volumes but also in a broad range of applications like physics, statistics, and engineering. It provides the mathematical tools needed to quantify continuous change and accumulation scenarios.

Coordinate Geometry

Coordinate geometry, also known as analytic geometry, involves using algebra and calculus to solve geometric problems related to curves and surfaces. It bridges the gap between algebraic expressions and geometric figures. This powerful tool allows us to easily transition between equations and graphs for analysis and problem-solving.

In the case of the original exercise, coordinate geometry helps us:

• Locate the points where the boundary line equation \(y = 4 - 2x\) intersects the axes, helping identify the region of interest.
• Determine the shape and area of the region to be revolved, ensuring accuracy when using methods like the disk method.
• Transform coordinates to suit different rotation axes (from \(x\) to \(y\) and vice versa), which is essential when setting up integrals for revolving solids.

Using coordinate geometry means equations and graphs work together, allowing for precise area and volume calculations when regions are defined by these curves. It’s crucial when needing to switch between Cartesian coordinates \((x, y)\) and other forms that require different perspectives on the same geometric problem.