8.9 Volume with Disc Method: Revolving Around the x- or y-Axis

5 min read•january 29, 2023

Anusha Tekumulla

ethan_bilderbeek

Anusha Tekumulla

ethan_bilderbeek

AP Calculus AB/BC ♾️

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The Disc Method

The Disc Method is a technique used to find the volume of a solid formed by revolving a region around the x- or y-axis. This method involves slicing the solid into many thin discs, and then finding the volume of each disc. The total volume of the solid is found by adding up the volumes of all of these discs.

Here are the steps to use the Disc Method:

Identify the region that is being revolved to form the solid. This region should be defined by a function, f(x) or g(y), and should be bounded by two lines, x = a and x = b or y = c and y = d.
Decide on the axis of revolution. If the region is being revolved around the x-axis, the width of each disc will be dx and the radius of each disc will be f(x). If the region is being revolved around the y-axis, the width of each disc will be dy and the radius of each disc will be g(y).
Find the volume of each disc by multiplying the area of the disc by the width of the disc. The area of the disc is found by multiplying π by the square of the radius.
Use the definite integral to find the total volume of the solid by integrating the function for the volume of each disc with respect to x or y.

Here are a few examples:

Example 1: Consider a region defined by the function f(x) = x^2, revolved around the x-axis from x = 0 to x = 1. The width of each disc is dx and the radius of each disc is f(x) = x^2. The volume of each disc is found by multiplying π by the square of the radius and the width of the disc. This gives us πx^2 * dx. To find the total volume of the solid, we integrate this function from x = 0 to x = 1. This gives us the definite integral from 0 to 1 of πx^2 dx = (π/3)

Example 2: Consider a region defined by the function g(y) = √(4-y^2), revolved around the y-axis from y = 0 to y = 2. The width of each disc is dy and the radius of each disc is g(y) = √(4-y^2). The volume of each disc is found by multiplying π by the square of the radius and the width of the disc. This gives us π(4-y^2) dy. To find the total volume of the solid, we integrate this function from y = 0 to y = 2. This gives us the definite integral from 0 to 2 of π(4-y^2) dy = 4π

Example 3: Consider a region defined by the function f(x) = x, revolved around the x-axis from x = 1 to x = 2. The width of each disc is dx and the radius of each disc is f(x) = x. The volume of each disc is found by multiplying π by the square of the radius and the width of the disc. This gives us πx^2 * dx. To find the total volume of the solid, we integrate this function from x = 1 to x = 2. This gives us the definite integral from 1 to 2 of πx^2 dx = (π/3)

It's important to note that in the above examples, the limits of integration were chosen based on the region given.

Example 4: Consider a region defined by the function f(x) = x^3, revolved around the x-axis from x = 0 to x = 2. The width of each disc is dx and the radius of each disc is f(x) = x^3. The volume of each disc is found by multiplying π by the square of the radius and the width of the disc. This gives us πx^6 * dx. To find the total volume of the solid, we integrate this function from x = 0 to x = 2. This gives us the definite integral from 0 to 2 of πx^6 dx = (4π/7)

Example 5: Consider a region defined by the function g(y) = 2 - y^2, revolved around the y-axis from y = -1 to y = 1. The width of each disc is dy and the radius of each disc is g(y) = 2 - y^2. The volume of each disc is found by multiplying π by the square of the radius and the width of the disc. This gives us π(2 - y^2)^2 * dy. To find the total volume of the solid, we integrate this function from y = -1 to y = 1. This gives us the definite integral from -1 to 1 of π(2 - y^2)^2 dy = 4π/3.

Example 6: Consider a region defined by the function f(x) = x^(1/2), revolved around the x-axis from x = 0 to x = 4. The width of each disc is dx and the radius of each disc is f(x) = x^(1/2). The volume of each disc is found by multiplying π by the square of the radius and the width of the disc. This gives us πx^(3/2) * dx. To find the total volume of the solid, we integrate this function from x = 0 to x = 4. This gives us the definite integral from 0 to 4 of πx^(3/2) dx = (4π/3).

Example 7: Consider a region defined by the function g(y) = 2 + y^2, revolved around the y-axis from y = -2 to y = 2. The width of each disc is dy and the radius of each disc is g(y) = 2 + y^2. The volume of each disc is found by multiplying π by the square of the radius and the width of the disc. This gives us π(2 + y^2)^2 * dy. To find the total volume of the solid, we integrate this function from y = -2 to y = 2. This gives us the definite integral from -2 to 2 of π(2 + y^2)^2 dy = 32π/3.

In conclusion, the Disc Method is a useful technique for finding the volume of a solid formed by revolution around either the x- or y-axis. To use this method, it's important to identify the region being revolved, choose the axis of revolution, find the volume of each disc, and use the definite integral to find the total volume of the solid. Through the use of examples, it's possible to see how the Disc Method can be applied in various scenarios to find the volume of different types of solids.

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