AP Calculus BC 8.11 Volume with Washer Method: Revolving Around the x- or y-Axis Study Notes - New Syllabus
AP Calculus BC 8.11 Volume with Washer Method: Revolving Around the x- or y-Axis Study Notes- New syllabus
AP Calculus BC 8.11 Volume with Washer 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 Washer Method: Revolving Around the x- or y-Axis
Volume with Washer Method: Revolving Around the x- or y-Axis
Volume with Washer Method : Revolving Around the \( x \)-axis
The washer method is used when a solid is generated by revolving a region around an axis and the region has a hole (inner radius) in the middle. The cross-sections perpendicular to the axis of revolution are washers discs with a circular hole in the center.
The area of a washer is given by the difference of the areas of the outer circle and the inner circle:
\( A(x) = \pi \left[ R(x)^2 – r(x)^2 \right] \)
- \( R(x) \) is the outer radius distance from the axis of revolution to the outer curve.
- \( r(x) \) is the inner radius distance from the axis of revolution to the inner curve.
The volume is then calculated by integrating the area along the given interval:
\( V = \pi \displaystyle \int_{a}^{b} \left[ R(x)^2 – r(x)^2 \right] \, dx \)
Steps for Solving:
- Sketch the region to visualize the outer and inner boundaries.
- Identify \( R(x) \) and \( r(x) \) relative to the \( x \)-axis.
- Square both radii and subtract to find the washer cross-sectional area.
- Integrate over the given bounds.
Example:
The region bounded by \( y = \sqrt{x} \) and \( y = \dfrac{x}{2} \) for \( 0 \leq x \leq 4 \) is revolved around the \( x \)-axis. Find the volume of the solid.
▶️ Answer/Explanation
Identify radii relative to the \( x \)-axis:
Outer radius: \( R(x) = \sqrt{x} \)
Inner radius: \( r(x) = \dfrac{x}{2} \)
\( V = \pi \displaystyle \int_{0}^{4} \left[ (\sqrt{x})^2 – \left( \dfrac{x}{2} \right)^2 \right] dx \)
\( V = \pi \displaystyle \int_{0}^{4} \left[ x – \dfrac{x^2}{4} \right] dx \)
\( V = \pi \left[ \dfrac{x^2}{2} – \dfrac{x^3}{12} \right]_{0}^{4} \)
\( V = \pi \left[ \dfrac{16}{2} – \dfrac{64}{12} \right] \)
\( V = \pi \left[ 8 – \dfrac{16}{3} \right] \)
\( V = \dfrac{8\pi}{3} \)
Example:
The region between \( y = 4 \) and \( y = x^2 \) for \( -2 \leq x \leq 2 \) is revolved around the \( x \)-axis. Find the volume.
▶️ Answer/Explanation
Outer radius: \( R(x) = 4 \)
Inner radius: \( r(x) = x^2 \)
\( V = \pi \displaystyle \int_{-2}^{2} \left[ (4)^2 – (x^2)^2 \right] dx \)
\( V = \pi \displaystyle \int_{-2}^{2} \left[ 16 – x^4 \right] dx \)
Since the function is even, double from 0 to 2:
\( V = 2\pi \displaystyle \int_{0}^{2} \left[ 16 – x^4 \right] dx \)
\( V = 2\pi \left[ 16x – \dfrac{x^5}{5} \right]_{0}^{2} \)
\( V = 2\pi \left[ 32 – \dfrac{32}{5} \right] \)
\( V = \dfrac{256\pi}{5} \)
Example:
The region between \( y = \cos x \) and \( y = \sin x \) for \( 0 \leq x \leq \dfrac{\pi}{4} \) is revolved around the \( x \)-axis. Find the volume.
▶️ Answer/Explanation
For \( 0 \leq x \leq \dfrac{\pi}{4} \), outer radius: \( R(x) = \cos x \), inner radius: \( r(x) = \sin x \).
\( V = \pi \displaystyle \int_{0}^{\pi/4} \left[ (\cos x)^2 – (\sin x)^2 \right] dx \)
Recall identity: \( \cos^2 x – \sin^2 x = \cos 2x \)
\( V = \pi \displaystyle \int_{0}^{\pi/4} \cos 2x \, dx \)
\( V = \pi \left[ \dfrac{\sin 2x}{2} \right]_{0}^{\pi/4} \)
\( V = \pi \left( \dfrac{\sin \dfrac{\pi}{2}}{2} – 0 \right) \)
\( V = \dfrac{\pi}{2} \)
Volume with Washer Method: Revolving Around the \( y \)-axis
When a region in the \( xy \)-plane is revolved around the \( y \)-axis, the washer method is used to find the volume if the solid has a hollow center. A washer is like a disc with a hole in it, and its cross-sectional area is the area of the outer disc minus the area of the inner disc.
For revolution around the \( y \)-axis:
\( V = \pi \displaystyle \int_{c}^{d} \left[ R(y)^{2} – r(y)^{2} \right] \, dy \)
- \( R(y) \) = Outer radius (distance from the \( y \)-axis to the outer curve)
- \( r(y) \) = Inner radius (distance from the \( y \)-axis to the inner curve)
- \( c \) and \( d \) are the \( y \)-bounds of the region
Steps to Solve:
- Sketch the region and identify the axis of revolution.
- Express the outer radius \( R(y) \) and inner radius \( r(y) \) as functions of \( y \).
- Square both radii, subtract the inner square from the outer square.
- Integrate with respect to \( y \) over the given bounds.
Example:
Find the volume when the region bounded by \( x = y^{2} \) and \( x = 4 \) is revolved around the \( y \)-axis.
▶️ Answer/Explanation
Identify bounds: \( y \) ranges from \( -2 \) to \( 2 \).
Outer radius \( R(y) = 4 \), inner radius \( r(y) = y^{2} \).
Volume formula:
\( V = \pi \displaystyle \int_{-2}^{2} \left[ (4)^{2} – (y^{2})^{2} \right] \, dy \)
\( V = \pi \displaystyle \int_{-2}^{2} \left[ 16 – y^{4} \right] \, dy \)
Since the function is even: \( V = 2\pi \displaystyle \int_{0}^{2} \left[ 16 – y^{4} \right] \, dy \)
\( V = 2\pi \left[ 16y – \dfrac{y^{5}}{5} \right]_{0}^{2} \)
\( V = 2\pi \left[ 32 – \dfrac{32}{5} \right] = 2\pi \cdot \dfrac{128}{5} = \dfrac{256\pi}{5} \)
Example:
The region bounded by \( x = \sqrt{y} \) and \( x = 2 \) for \( 0 \leq y \leq 4 \) is revolved about the \( y \)-axis. Find the volume.
▶️ Answer/Explanation
\( R(y) = 2 \), \( r(y) = \sqrt{y} \)
\( V = \pi \displaystyle \int_{0}^{4} \left[ 4 – y \right] \, dy \)
\( V = \pi \left[ 4y – \dfrac{y^{2}}{2} \right]_{0}^{4} \)
\( V = \pi \left[ 16 – 8 \right] = 8\pi \)
Example:
The region bounded by \( x = y \) and \( y = 3 \) is revolved around the \( y \)-axis. Find the volume.
▶️ Answer/Explanation
Outer radius \( R(y) = y \), inner radius \( r(y) = 0 \).
\( V = \pi \displaystyle \int_{0}^{3} \left[ y^{2} – 0 \right] \, dy \)
\( V = \pi \left[ \dfrac{y^{3}}{3} \right]_{0}^{3} = \pi \cdot 9 = 9\pi \)