AP Calculus BC 8.12 Volume with Washer Method: Revolving Around Other Axes Study Notes - New Syllabus
AP Calculus BC 8.12 Volume with Washer Method: Revolving Around Other Axes Study Notes- New syllabus
AP Calculus BC 8.12 Volume with Washer Method: Revolving Around Other Axes 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 Other Axes
Volume with Washer Method: Revolving Around Other Axes
Volume with Washer Method: Revolving Around Other Axes
When a region is revolved around a line other than the coordinate axes (for example, \( y = k \) or \( x = h \)), the washer method can still be applied by adjusting the radii to account for the shift of the axis of revolution.
The general formula for washers is:
\( V = \pi \displaystyle \int_{a}^{b} \left[ R(x)^2 – r(x)^2 \right] \, dx \) (if integrating with respect to \( x \))
or
\( V = \pi \displaystyle \int_{c}^{d} \left[ R(y)^2 – r(y)^2 \right] \, dy \) (if integrating with respect to \( y \))
Key Adjustments for Other Axes:
If revolving around \( y = k \) (horizontal axis):
- Outer radius: \( R(x) = |k – \text{lower curve}| \) or \( |k – \text{upper curve}| \) depending on orientation.
- Inner radius: \( r(x) = |k – \text{opposite curve}| \).
If revolving around \( x = h \) (vertical axis):
- Outer radius: \( R(y) = |h – \text{left curve}| \).
- Inner radius: \( r(y) = |h – \text{right curve}| \).
Example:
The region bounded by \( y = \sqrt{x} \), \( y = 0 \), and \( x = 4 \) is revolved around the line \( y = 3 \). Find the volume of the resulting solid.
▶️ Answer/Explanation
Outer radius: \( R(x) = 3 – 0 = 3 \)
Inner radius: \( r(x) = 3 – \sqrt{x} \)
Volume:
\( V = \pi \displaystyle \int_{0}^{4} \left[ (3)^2 – (3 – \sqrt{x})^2 \right] dx \)
\( = \pi \displaystyle \int_{0}^{4} \left[ 9 – (9 – 6\sqrt{x} + x) \right] dx \)
\( = \pi \displaystyle \int_{0}^{4} \left[ 6\sqrt{x} – x \right] dx \)
\( = \pi \left[ 6 \cdot \dfrac{2}{3}x^{3/2} – \dfrac{x^2}{2} \right]_{0}^{4} \)
\( = \pi \left[ 4 \cdot 8 – \dfrac{16}{2} \right] \)
\( = \pi \left[ 32 – 8 \right] = 24\pi \)
Example:
The region bounded by \( y = x^2 \), \( y = 0 \), and \( x = 1 \) is revolved around \( x = -2 \). Find the volume.
▶️ Answer/Explanation
Step 1: We integrate with respect to \( y \) since the axis is vertical.
From \( y = 0 \) to \( y = 1 \), left curve: \( x = -2 \) (axis), right curve: \( x = \sqrt{y} \)
Outer radius: \( R(y) = \sqrt{y} + 2 \)
Inner radius: \( r(y) = 2 \)
Volume:
\( V = \pi \displaystyle \int_{0}^{1} \left[ (\sqrt{y} + 2)^2 – (2)^2 \right] dy \)
\( = \pi \displaystyle \int_{0}^{1} \left[ y + 4\sqrt{y} + 4 – 4 \right] dy \)
\( = \pi \displaystyle \int_{0}^{1} \left[ y + 4\sqrt{y} \right] dy \)
\( = \pi \left[ \dfrac{y^2}{2} + 4 \cdot \dfrac{2}{3}y^{3/2} \right]_{0}^{1} \)
\( = \pi \left[ \dfrac{1}{2} + \dfrac{8}{3} \right] = \pi \cdot \dfrac{19}{6} \)
Example:
The region bounded by \( y = x^2 \), \( y = 4 \) is revolved around \( y = -1 \). Find the volume.
▶️ Answer/Explanation
Outer radius: \( R(x) = (4) – (-1) = 5 \)
Inner radius: \( r(x) = x^2 – (-1) = x^2 + 1 \)
Interval: Find \( x \) where \( x^2 = 4 \) → \( x = -2 \) to \( x = 2 \).
\( V = \pi \displaystyle \int_{-2}^{2} \left[ 25 – (x^2 + 1)^2 \right] dx \)
\( = \pi \displaystyle \int_{-2}^{2} \left[ 25 – (x^4 + 2x^2 + 1) \right] dx \)
\( = \pi \displaystyle \int_{-2}^{2} \left[ 24 – 2x^2 – x^4 \right] dx \)
Since the function is even, double the integral from \( 0 \) to \( 2 \):
\( V = 2\pi \displaystyle \int_{0}^{2} \left[ 24 – 2x^2 – x^4 \right] dx \)
\( = 2\pi \left[ 24x – \dfrac{2x^3}{3} – \dfrac{x^5}{5} \right]_{0}^{2} \)
\( = 2\pi \left[ 48 – \dfrac{16}{3} – \dfrac{32}{5} \right] \)
\( = 2\pi \left[ \dfrac{720 – 80 – 96}{15} \right] = 2\pi \cdot \dfrac{544}{15} \)
\( = \dfrac{1088\pi}{15} \)