AP Calculus AB 8.9 Volume with Disk Method (y=sqrt(x), x=1, x=4 Rotate Around x-axis)

Please subscribe!    / nickperich   AP Calculus AB 8.9: *Volume with the Disc Method* The *Disc Method* is a technique used to calculate the volume of a solid of revolution when a region in the plane is revolved around an axis. The method involves slicing the solid into thin, circular discs, and then summing the volumes of these discs using integration. #### Key Concept: When a region is revolved around a horizontal or vertical axis, each infinitesimally thin slice forms a **circular disc**. The volume of a disc is given by the formula for the area of a circle, \( \pi r^2 \), where \( r \) is the radius of the disc. #### Steps to Use the Disc Method: 1. *Identify the function(s) that define the region* to be revolved and the axis of revolution. 2. **Determine the radius of the disc**: If the region is revolved around the x-axis (horizontal), the radius of a disc is the distance from the x-axis to the function (or curve). This distance is given by \( r(x) \), where \( r(x) = f(x) \) for the function \( y = f(x) \). If the region is revolved around the y-axis (vertical), the radius of the disc is the distance from the y-axis to the function \( x = g(y) \). 3. **Set up the integral**: The volume of the solid is the sum of the volumes of all the discs. The volume of a disc with a radius \( r(x) \) and thickness \( dx \) (for vertical slices) or \( dy \) (for horizontal slices) is: \[ \text{Volume of disc} = \pi [r(x)]^2 \, dx \quad \text{(for revolution around the x-axis)} \] or \[ \text{Volume of disc} = \pi [r(y)]^2 \, dy \quad \text{(for revolution around the y-axis)} \] 4. **Integrate to find the total volume**: The total volume is obtained by integrating the volume of the discs over the appropriate interval: \[ V = \int_{a}^{b} \pi [r(x)]^2 \, dx \quad \text{(for revolution around the x-axis)} \] or \[ V = \int_{c}^{d} \pi [r(y)]^2 \, dy \quad \text{(for revolution around the y-axis)} \] where \( a \) and \( b \) are the bounds for \( x \), and \( c \) and \( d \) are the bounds for \( y \). #### Example Problem: Find the volume of the solid formed by revolving the region bounded by \( y = x^2 \) and the x-axis from \( x = 0 \) to \( x = 2 \) around the x-axis. 1. **Identify the function and axis**: The function is \( y = x^2 \), and the region is revolved around the x-axis. 2. **Determine the radius of the disc**: The radius of each disc is \( r(x) = x^2 \), the function that defines the boundary of the region. 3. **Set up the integral**: The volume is: \[ V = \int_{0}^{2} \pi [x^2]^2 \, dx = \pi \int_{0}^{2} x^4 \, dx \] 4. **Evaluate the integral**: \[ V = \pi \left[ \frac{x^5}{5} \right]_{0}^{2} = \pi \left( \frac{32}{5} - 0 \right) = \frac{32\pi}{5} \] So, the volume of the solid is \( \frac{32\pi}{5} \) cubic units. Conclusion: The Disc Method is used to find the volume of solids of revolution by integrating the area of circular cross sections (discs) along the axis of revolution. It is a powerful tool in calculus for solving volume problems where the region is revolved around a line. I have many informative videos for Pre-Algebra, Algebra 1, Algebra 2, Geometry, Pre-Calculus, and Calculus. Please check it out: / nickperich Nick Perich Norristown Area High School Norristown Area School District Norristown, Pa #math #algebra #algebra2 #maths #math #shorts #funny #help #onlineclasses #onlinelearning #online #study