AP Calculus BC 8.10 Volume with Disc Method: Revolving Around Other Axes Study Notes - New Syllabus
AP Calculus BC 8.10 Volume with Disc Method: Revolving Around Other Axes Study Notes- New syllabus
AP Calculus BC 8.10 Volume with Disc 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 Disc Method: Revolving Around Other Axes
Volume with Disc Method: Revolving Around Other Axes
Volume with Disc Method: Revolving Around Other Axes
When the region is revolved around an axis that is parallel to the \( x \)-axis or \( y \)-axis but not on them, we adjust the radius by considering the distance from the curve to the axis of rotation.
General formula:
\( V = \displaystyle \int_{a}^{b} \pi \left[ R(\text{variable}) \right]^2 \ d(\text{variable}) \)
Where:
- \( R \) is the distance from the curve to the axis of rotation.
- The variable of integration depends on whether the slices are perpendicular to the \( x \)-axis or \( y \)-axis.
- If rotating about a line \( y = k \) (horizontal) or \( x = h \) (vertical), adjust \( R \) accordingly.
Example:
The region bounded by \( y = \sqrt{x} \), \( x = 0 \), and \( x = 4 \) is revolved around \( y = 3 \). Find the volume.
▶️ Answer/Explanation
Axis is \( y = 3 \) (horizontal). Radius \( R(x) = 3 – \sqrt{x} \).
\( V = \displaystyle \int_{0}^{4} \pi \left[ 3 – \sqrt{x} \right]^2 dx \)
\( V = \pi \displaystyle \int_{0}^{4} \left( 9 – 6\sqrt{x} + x \right) dx \)
\( V = \pi \left[ 9x – 4x^{3/2} + \dfrac{x^2}{2} \right]_{0}^{4} \)
\( V = \pi \left[ 36 – 32 + 8 \right] \)
\( V = 12\pi \ \text{cubic units} \)
Example:
The region bounded by \( y = x^2 \), \( y = 0 \), and \( x = 1 \) is revolved around \( x = 2 \). Find the volume.
▶️ Answer/Explanation
Axis is \( x = 2 \) (vertical). Radius \( R(y) = 2 – \sqrt{y} \).
We integrate with respect to \( y \) from \( 0 \) to \( 1 \):
\( V = \displaystyle \int_{0}^{1} \pi \left[ 2 – \sqrt{y} \right]^2 dy \)
\( V = \pi \displaystyle \int_{0}^{1} \left( 4 – 4\sqrt{y} + y \right) dy \)
\( V = \pi \left[ 4y – \dfrac{8}{3}y^{3/2} + \dfrac{y^2}{2} \right]_{0}^{1} \)
\( V = \pi \left( 4 – \dfrac{8}{3} + \dfrac{1}{2} \right) \)
\( V = \pi \cdot \dfrac{13}{6} \)
\( V = \dfrac{13\pi}{6} \ \text{cubic units} \)
Example:
The region bounded by \( y = x + 1 \), \( x = 0 \), and \( x = 2 \) is revolved around \( y = -1 \). Find the volume.
▶️ Answer/Explanation
Axis is \( y = -1 \) (horizontal). Radius \( R(x) = (x + 1) – (-1) = x + 2 \).
\( V = \displaystyle \int_{0}^{2} \pi \left[ x + 2 \right]^2 dx \)
\( V = \pi \displaystyle \int_{0}^{2} \left( x^2 + 4x + 4 \right) dx \)
\( V = \pi \left[ \dfrac{x^3}{3} + 2x^2 + 4x \right]_{0}^{2} \)
\( V = \pi \left( \dfrac{8}{3} + 8 + 8 \right) \)
\( V = \pi \cdot \dfrac{56}{3} \)
\( V = \dfrac{56\pi}{3} \ \text{cubic units} \)