Finding Volume By Using Integration

Welcome! Integral calculus is a powerful tool that allows us to move from two-dimensional concepts like area to three-dimensional concepts like volume. This guide provides a step-by-step approach to understanding and applying different integration methods to calculate the volume of complex 3D shapes.

The Core Idea: Slicing

Imagine a loaf of bread. You can find its total volume by slicing it into thin pieces, finding the volume of each slice, and then adding them all up. Integration does the same thing, but with infinitely thin slices.

We will explore three primary methods for finding the volume of solids, particularly "solids of revolution," which are formed by rotating a 2D area around an axis.

Method 1: The Disk Method

The Disk Method is used when the solid of revolution is solid all the way through (no holes). We slice the solid perpendicular to the axis of rotation, and each slice is a thin circular disk.

When to Use It:

• The solid is generated by rotating a region around an axis.
• The cross-sections perpendicular to the axis of rotation are solid circles (disks).
• The region to be rotated is flush against the axis of rotation.

The Formula:

The volume of a single disk is \( \pi r^2 h \). In our case, the radius \( r \) is the function value \( f(x) \) and the height \( h \) is an infinitesimally small change in x, which we call \( dx \).

For rotation about the x-axis: $$ V = \pi \int_{a}^{b} [R(x)]^2 \,dx $$ Where \( R(x) \) is the radius of the disk at a given x-value.

For rotation about the y-axis: $$ V = \pi \int_{c}^{d} [R(y)]^2 \,dy $$ Where \( R(y) \) is the radius of the disk at a given y-value.

Example: The Disk Method

Question: Find the volume of the solid generated by revolving the region bounded by \( y = \sqrt{x} \), \( x = 4 \), and the x-axis (\( y = 0 \)) about the x-axis.

Solution:

1. Identify the Method: The region is flush against the axis of rotation (the x-axis), so the solid will be solid. We use the Disk Method.
2. Determine the Radius \( R(x) \): The radius of each disk is the distance from the x-axis to the curve, which is simply the function value. So, \( R(x) = \sqrt{x} \).
3. Set up the Integral: The region extends from \( x = 0 \) to \( x = 4 \). We use the formula for rotation about the x-axis. $$ V = \pi \int_{0}^{4} (\sqrt{x})^2 \,dx $$
4. Evaluate the Integral: $$ V = \pi \int_{0}^{4} x \,dx $$ $$ V = \pi \left[ \frac{x^2}{2} \right]_{0}^{4} $$ $$ V = \pi \left( \frac{4^2}{2} - \frac{0^2}{2} \right) $$ $$ V = \pi \left( \frac{16}{2} - 0 \right) = 8\pi $$

Answer: The volume of the solid is \( 8\pi \) cubic units.

Method 2: The Washer Method

The Washer Method is an extension of the Disk Method. It is used when the solid of revolution has a hole in the middle, creating a shape like a washer or a donut. This happens when the region being rotated is not flush against the axis of rotation.

The Formula:

The volume of a washer is the volume of the outer disk minus the volume of the inner disk. The radius of the outer disk is \( R(x) \) and the radius of the inner disk is \( r(x) \).

For rotation about the x-axis: $$ V = \pi \int_{a}^{b} ([R(x)]^2 - [r(x)]^2) \,dx $$ Where \( R(x) \) is the outer radius and \( r(x) \) is the inner radius.

Example: The Washer Method

Question: Find the volume of the solid generated by revolving the region bounded by the curves \( y = x \) and \( y = x^2 \) about the x-axis.

Solution:

1. Identify the Method: There is a gap between the region and the axis of rotation for some parts, so the solid will have a hole. We use the Washer Method.
2. Determine the Radii \( R(x) \) and \( r(x) \):
   • The outer radius \( R(x) \) is the distance from the x-axis to the outer curve, which is \( y = x \). So, \( R(x) = x \).
   • The inner radius \( r(x) \) is the distance from the x-axis to the inner curve, which is \( y = x^2 \). So, \( r(x) = x^2 \).
3. Find the Limits of Integration: Find where the curves intersect by setting them equal: \( x = x^2 \Rightarrow x^2 - x = 0 \Rightarrow x(x-1) = 0 \). They intersect at \( x = 0 \) and \( x = 1 \).
4. Set up the Integral: $$ V = \pi \int_{0}^{1} (x^2 - (x^2)^2) \,dx $$ $$ V = \pi \int_{0}^{1} (x^2 - x^4) \,dx $$
5. Evaluate the Integral: $$ V = \pi \left[ \frac{x^3}{3} - \frac{x^5}{5} \right]_{0}^{1} $$ $$ V = \pi \left( (\frac{1^3}{3} - \frac{1^5}{5}) - (0) \right) $$ $$ V = \pi \left( \frac{1}{3} - \frac{1}{5} \right) = \pi \left( \frac{5-3}{15} \right) = \frac{2\pi}{15} $$

Answer: The volume of the solid is \( \frac{2\pi}{15} \) cubic units.

Method 3: The Cylindrical Shell Method

The Cylindrical Shell Method involves slicing the solid into nested cylindrical shells, like the layers of an onion. This method is often easier to use when rotating a region about the y-axis, but the functions are defined in terms of x.

The Formula:

The volume of a single cylindrical shell is \( 2 \pi \times \text{radius} \times \text{height} \times \text{thickness} \).

For rotation about the y-axis: $$ V = 2\pi \int_{a}^{b} (\text{radius}) \cdot (\text{height}) \,dx $$ $$ V = 2\pi \int_{a}^{b} x \cdot h(x) \,dx $$ Where the radius is \( x \) and the height is \( h(x) \).

Example: The Shell Method

Question: Find the volume of the solid generated by revolving the region bounded by \( y = 2x^2 - x^3 \) and the x-axis about the y-axis.

Solution:

1. Identify the Method: We are rotating about the y-axis, and our function is in terms of x. Solving for x would be difficult. The Shell Method is ideal here.
2. Determine Radius and Height:
   • The radius of a shell at a given x is simply \( x \).
   • The height of the shell is the function value, \( h(x) = 2x^2 - x^3 \).
3. Find the Limits of Integration: Find where the curve hits the x-axis: \( 2x^2 - x^3 = 0 \Rightarrow x^2(2-x) = 0 \). The region is bounded by \( x = 0 \) and \( x = 2 \).
4. Set up the Integral: $$ V = 2\pi \int_{0}^{2} x (2x^2 - x^3) \,dx $$ $$ V = 2\pi \int_{0}^{2} (2x^3 - x^4) \,dx $$
5. Evaluate the Integral: $$ V = 2\pi \left[ \frac{2x^4}{4} - \frac{x^5}{5} \right]_{0}^{2} $$ $$ V = 2\pi \left[ \frac{x^4}{2} - \frac{x^5}{5} \right]_{0}^{2} $$ $$ V = 2\pi \left( (\frac{2^4}{2} - \frac{2^5}{5}) - (0) \right) $$ $$ V = 2\pi \left( \frac{16}{2} - \frac{32}{5} \right) = 2\pi \left( 8 - \frac{32}{5} \right) $$ $$ V = 2\pi \left( \frac{40-32}{5} \right) = 2\pi \left( \frac{8}{5} \right) = \frac{16\pi}{5} $$

Answer: The volume of the solid is \( \frac{16\pi}{5} \) cubic units.

Summary: Which Method to Use?

Choosing the right method is key. Here's a quick guide:

Method	Best For	Key Idea
Disk Method	Solid shapes (no holes), region flush with axis of rotation.	Sum of volumes of thin disks: \( \pi r^2 \).
Washer Method	Shapes with a hole in the center.	Sum of volumes of washers: \( \pi (R^2 - r^2) \).
Shell Method	Rotating around a vertical axis when functions are in terms of x (or vice-versa).	Sum of volumes of cylindrical shells: \( 2\pi r h \).