#Volume with Disc Method Revolving Around Other Axes

#Table of Contents

1. Introduction
2. Volume of Revolution Around Axes Parallel to the x-axis
   • Key Concepts
   • Worked Example
3. Volume of Revolution Around Axes Parallel to the y-axis
   • Key Concepts
   • Worked Example
4. Practice Questions
5. Glossary
6. Summary and Key Takeaways
7. Exam Strategy

#Introduction

The disc method is a powerful technique used in calculus to find the volume of a solid of revolution. When a region in the plane is revolved around a line that is parallel to either the x-axis or the y-axis, we can use the disc method to calculate the volume of the resulting solid.

#Volume of Revolution Around Axes Parallel to the x-axis

#Key Concepts

• For a continuous function $f$, if the region bounded by:
  • The curve $y=f(x)$ and the line $y=k$
  • Between $x=a$ and $x=b$
• is rotated around the line $y=k$, then the volume of revolution is:

$$V = \pi \int_{a}^{b} (y - k)^{2} , dx$$

Note that $x$ here is a function of $y$.

• Thinking of this as an accumulation of change:
  • $\pi (y - k)^{2} \Delta x$ is the volume of a disc with:
    • Circular cross section of radius $\left|y - k\right|$
    • Length $\Delta x$
  • $\pi (y - k)^{2} , dx$ is the limit of this volume element as $\Delta x \to 0$
  • The integral is evaluated from $x=a$ to $x=b$

#Worked Example

Let $R$ be the region enclosed by the graph of $f(x) = 1 + e^{-x}$, the lines $x=1$ and $x=3$, and the line $y=-2$. The region $R$ is rotated about the horizontal line $y=-2$.

Solution: To find the volume, we use the formula: $$V = \pi \int_{1}^{3} \left[(1 + e^{-x}) - (-2)\right]^{2} , dx$$

Breaking it down: $$V = \pi \int_{1}^{3} (3 + e^{-x})^{2} , dx$$

Expanding the integrand: $$V = \pi \int_{1}^{3} (9 + 6e^{-x} + e^{-2x}) , dx$$

Integrating term by term: $$V = \pi \left[ 9x - 6e^{-x} - \frac{1}{2} e^{-2x} \right]_{1}^{3}$$

Evaluating the definite integral: $$V = \pi \left( \left[ 9 \cdot 3 - 6e^{-3} - \frac{1}{2} e^{-6} \right] - \left[ 9 \cdot 1 - 6e^{-1} - \frac{1}{2} e^{-2} \right] \right)$$

Simplifying the expression: $$V = \pi \left( 18 - 6e^{-3} - \frac{1}{2} e^{-6} + 6e^{-1} + \frac{1}{2} e^{-2} \right)$$

Final volume: $$V \approx 62.753 , \text{units}^3 , (\text{to 3 decimal places})$$

#Volume of Revolution Around Axes Parallel to the y-axis

#Key Concepts

• For a continuous function $f$, if the region bounded by:
  • The curve $y=f(x)$ and the line $x=k$
  • Between $y=a$ and $y=b$
• is rotated $360^\circ$ around the line $x=k$, then the volume of revolution is:

$$V = \pi \int_{a}^{b} (x - k)^{2} , dy$$

Note that $x$ here is a function of $y$. This will mean rewriting $y=f(x)$ in the form $x=g(y)$. Also note that the integration is done with respect to $y$.

#Worked Example

Let $R$ be the region enclosed by the graph of $y = \ln(x-3)$, the lines $x=4$ and $x=6$, and the line $y=-2$. The region $R$ is rotated about the vertical line $x=1$.

Solution: First, rewrite the function as a function of $y$: $$y = \ln(x-3) \implies e^{y} = x-3 \implies x = 3 + e^{y}$$

To find the volume, we use the formula: $$V = \pi \int_{-2}^{1} \left[(3 + e^{y}) - 1\right]^{2} , dy$$

Breaking it down: $$V = \pi \int_{-2}^{1} (2 + e^{y})^{2} , dy$$

Expanding the integrand: $$V = \pi \int_{-2}^{1} (4 + 4e^{y} + e^{2y}) , dy$$

Integrating term by term: $$V = \pi \left[ 4y + 4e^{y} + \frac{1}{2} e^{2y} \right]_{-2}^{1}$$

Evaluating the definite integral: $$V = \pi \left( \left[ 4 \cdot 1 + 4e^{1} + \frac{1}{2} e^{2} \right] - \left[ 4 \cdot (-2) + 4e^{-2} + \frac{1}{2} e^{-4} \right] \right)$$

Simplifying the expression: $$V = \pi \left( 4 + 4e + \frac{1}{2} e^{2} - (-8 + 4e^{-2} + \frac{1}{2} e^{-4}) \right)$$

Final volume: $$V \approx 81.735 , \text{units}^3 , (\text{to 3 decimal places})$$

#Practice Questions

1. Practice Question

Find the volume of the solid obtained by rotating the region bounded by $y = x^2$ and $y = 4$ about the line $y = 4$ from $x = -2$ to $x = 2. </practice_question> 2. <practice_question> Determine the volume of the solid formed by rotating the region bounded by$y = \sqrt{x}$,$x=4$, and$y=0$about the line$x=5$. </practice_question> 3. <practice_question> Calculate the volume of the solid obtained by rotating the region enclosed by$x = y^2$and$x = 4$about the line$x=4$. </practice_question>$

#Glossary

• Disc Method: A technique for calculating the volume of a solid of revolution by slicing the solid into discs.
• Volume of Revolution: The volume of a solid formed by rotating a region in the plane around a specified line.
• Definite Integral: An integral with upper and lower limits, representing the signed area under a curve.

#Summary and Key Takeaways

• The disc method involves slicing the solid into thin discs, calculating the volume of each disc, and summing these volumes using integration.
• When revolving around a line parallel to the x-axis, the volume is given byV = \pi \int_{a}^{b} (y - k)^{2} , dx$.$
• When revolving around a line parallel to the y-axis, the volume is given byV = \pi \int_{a}^{b} (x - k)^{2} , dy$.
• Carefully reframe the function as necessary to match the variable of integration.

#Exam Strategy

These tips and strategies will help you efficiently and accurately apply the disc method to calculate volumes of revolution in exam scenarios.