AP Calculus BC 8.9 Volume with Disc Method: Revolving Around the x- or y-Axis Study Notes - New Syllabus
AP Calculus BC 8.9 Volume with Disc Method: Revolving Around the x- or y-Axis Study Notes- New syllabus
AP Calculus BC 8.9 Volume with Disc Method: Revolving Around the x- or y-Axis Study Notes – AP Calculus BC- per latest AP Calculus BC Syllabus.
LEARNING OBJECTIVE
- Definite integrals allow us to solve problems involving the accumulation of change over an interval.
Key Concepts:
- Volume with Disc Method: Revolving Around the x- or y-Axis
Volume with Disc Method: Revolving Around the x- or y-Axis
Volume with Disc Method: Revolving Around the \( x \)-Axis
The disc method is used to find the volume of a solid of revolution when a region in the plane is revolved around the \( x \)-axis (or \( y \)-axis), and the cross sections perpendicular to the axis of revolution are solid discs. The volume is obtained by integrating the area of each disc along the axis of revolution.
If the cross section is a disc of radius \( R(x) \), then the area of the cross section is:
\( A(x) = \pi \left[ R(x) \right]^2 \)
For revolution around the \( x \)-axis:
\( V = \displaystyle \int_{a}^{b} \pi \left[ R(x) \right]^2 \, dx \)
Here, \( R(x) \) is the distance from the \( x \)-axis to the curve \( y = f(x) \).
Example:
Find the volume of the solid generated when the region bounded by \( y = \sqrt{x} \), \( x = 4 \), and the \( x \)-axis is revolved around the \( x \)-axis.
▶️ Answer/Explanation
Here, \( R(x) = \sqrt{x} \).
\( V = \displaystyle \int_{0}^{4} \pi \left( \sqrt{x} \right)^2 dx \)
\( V = \pi \displaystyle \int_{0}^{4} x \, dx \)
\( V = \pi \left[ \dfrac{x^2}{2} \right]_{0}^{4} \)
\( V = \pi \left[ \dfrac{16}{2} – 0 \right] = 8\pi \)
Final Answer: \( 8\pi \) cubic units.
Example:
Find the volume of the solid generated by revolving the region bounded by \( y = 2x \), \( x = 0 \), and \( x = 3 \) about the \( x \)-axis.
▶️ Answer/Explanation
Here, \( R(x) = 2x \).
\( V = \displaystyle \int_{0}^{3} \pi \left( 2x \right)^2 dx \)
\( V = \pi \displaystyle \int_{0}^{3} 4x^2 \, dx \)
\( V = 4\pi \left[ \dfrac{x^3}{3} \right]_{0}^{3} \)
\( V = 4\pi \left( \dfrac{27}{3} – 0 \right) \)
\( V = 36\pi \)
Final Answer: \( 36\pi \) cubic units.
Example:
The region bounded by \( y = x \), \( x = 0 \), and \( x = 2 \) is revolved around the \( x \)-axis. What is the volume of the solid formed?
A) \( \dfrac{8\pi}{3} \) cubic units
B) \( \dfrac{16\pi}{3} \) cubic units
C) \( 8\pi \) cubic units
D) \( 4\pi \) cubic units
▶️ Correct Answer & Explanation
Here, \( R(x) = x \).
\( V = \displaystyle \int_{0}^{2} \pi \left( x \right)^2 dx \)
\( V = \pi \left[ \dfrac{x^3}{3} \right]_{0}^{2} \)
\( V = \pi \left( \dfrac{8}{3} – 0 \right) \)
\( V = \dfrac{8\pi}{3} \) cubic units
Correct Option: A
Volume with Disc Method: Revolving Around the \( y \)-Axis
When a region in the \( xy \)-plane is revolved around the \( y \)-axis and the cross sections perpendicular to the \( y \)-axis are solid discs, the volume can be computed using:
\( V = \displaystyle \int_{c}^{d} \pi \left[ R(y) \right]^2 dy \)
Here:
- \( R(y) \) is the radius of the disc, measured horizontally from the axis of rotation to the curve.
- \( [c, d] \) is the interval for \( y \).
- If the curve is given as \( x = g(y) \), then \( R(y) = g(y) \).
Example:
The region bounded by \( x = y^2 \) and \( x = 0 \) for \( 0 \leq y \leq 3 \) is revolved around the \( y \)-axis. Find the volume of the solid.
▶️ Answer/Explanation
Here, \( R(y) = y^2 \).
\( V = \displaystyle \int_{0}^{3} \pi \left( y^2 \right)^2 dy \)
\( V = \pi \displaystyle \int_{0}^{3} y^4 dy \)
\( V = \pi \left[ \dfrac{y^5}{5} \right]_{0}^{3} \)
\( V = \pi \cdot \dfrac{243}{5} \)
\( V = \dfrac{243\pi}{5} \ \text{cubic units} \)
Example:
The region bounded by \( x = \sqrt{y} \), \( y = 0 \), and \( y = 4 \) is revolved around the \( y \)-axis. Find the volume of the solid.
▶️ Answer/Explanation
Here, \( R(y) = \sqrt{y} \).
\( V = \displaystyle \int_{0}^{4} \pi \left( \sqrt{y} \right)^2 dy \)
\( V = \pi \displaystyle \int_{0}^{4} y \, dy \)
\( V = \pi \left[ \dfrac{y^2}{2} \right]_{0}^{4} \)
\( V = \pi \cdot \dfrac{16}{2} \)
\( V = 8\pi \ \text{cubic units} \)
Example:
The region bounded by \( x = 2 \), \( x = y \), and \( y = 0 \) to \( y = 2 \) is revolved around the \( y \)-axis. Find the volume of the solid
▶️ Answer/Explanation
Outer radius \( R(y) = 2 \), inner radius \( r(y) = y \).
\( V = \displaystyle \int_{0}^{2} \pi \left[ R(y)^2 – r(y)^2 \right] dy \)
\( V = \pi \displaystyle \int_{0}^{2} \left( 4 – y^2 \right) dy \)
\( V = \pi \left[ 4y – \dfrac{y^3}{3} \right]_{0}^{2} \)
\( V = \pi \left( 8 – \dfrac{8}{3} \right) \)
\( V = \dfrac{16\pi}{3} \ \text{cubic units} \)