Areas and volumes of revolution (HL)

Introduction to Areas and Volumes of Revolution

In your previous calculus studies, you learned how to find the area between a curve and the \(x\)-axis. In this chapter, we take that knowledge further into the Higher Level (HL) territory. We will explore how to find the area between a curve and the \(y\)-axis, and how to "spin" these 2D shapes to create 3D solids, calculating their volumes. This technique is used by engineers and designers to calculate the capacity of everything from soda bottles to rocket nose cones!

1. Area Enclosed by a Curve and the \(y\)-axis

Usually, we integrate with respect to \(x\) to find the area under a curve. However, sometimes the region we are interested in is tucked between the curve and the vertical \(y\)-axis. To find this, we simply "switch our perspective."

The Formula

If we have a curve defined as \(x = f(y)\), the area \(A\) between the curve and the \(y\)-axis from \(y = c\) to \(y = d\) is given by:

\(A = \int_{c}^{d} x \, dy\)

How to approach these problems:

1. Rearrange the equation: If your equation is in the form \(y = f(x)\), you must rearrange it to make \(x\) the subject (i.e., \(x = g(y)\)).
2. Identify the bounds: Look for the lower limit (\(c\)) and upper limit (\(d\)) on the \(y\)-axis.
3. Integrate: Use your standard integration rules, but treat \(y\) as the variable.

Example: Find the area between \(y = \ln(x)\), the \(y\)-axis, and the lines \(y = 0\) and \(y = 1\).
First, rearrange \(y = \ln(x)\) to get \(x = e^y\).
Then, calculate \(A = \int_{0}^{1} e^y \, dy\).
Evaluating the integral: \([e^y]_{0}^{1} = e^1 - e^0 = e - 1\).

Quick Review: When integrating along the \(y\)-axis, your boundaries and your differential (\(dy\)) must both be in terms of \(y\)!

2. Volumes of Revolution about the \(x\)-axis

Imagine taking a 2D area trapped under a curve and spinning it 360 degrees around the \(x\)-axis. This "sweep" creates a solid 3D shape. Because we are spinning in a circle, the cross-sections of this solid are circles (or discs), which is why we use the area of a circle formula (\(\pi r^2\)) inside our integral.

The Formula

The volume \(V\) of the solid generated by rotating the curve \(y = f(x)\) around the \(x\)-axis from \(x = a\) to \(x = b\) is:

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

Step-by-Step Process:

1. Square the function: Take your expression for \(y\) and square it (\(y^2\)).
2. Set up the integral: Place the \(\pi\) outside the integral so you don't forget it.
3. Integrate and evaluate: Find the anti-derivative and plug in your bounds \(a\) and \(b\).
4. Final Touch: Multiply the result by \(\pi\).

Common Mistake: Many students forget to square the function or forget to include \(\pi\). A good mnemonic is: "Volume is a Pie ( \(\pi\) ) Square ( \(r^2\) ) Meal."

3. Volumes of Revolution about the \(y\)-axis

This is the vertical version of the previous concept. We take an area and spin it around the vertical \(y\)-axis.

The Formula

The volume \(V\) of the solid generated by rotating the curve \(x = g(y)\) around the \(y\)-axis from \(y = c\) to \(y = d\) is:

\(V = \pi \int_{c}^{d} x^2 \, dy\)

Key Strategy:

Since we are rotating around the \(y\)-axis, everything in the integral must be in terms of \(y\). If you are given \(y = \sqrt{x}\), you must square both sides to get \(x = y^2\). Then, for the volume formula, you square it again to get \(x^2 = (y^2)^2 = y^4\).

Example: Find the volume when \(y = x^3\) is rotated around the \(y\)-axis from \(y = 1\) to \(y = 8\).
1. Rearrange for \(x\): \(x = y^{1/3}\).
2. Square it for the formula: \(x^2 = (y^{1/3})^2 = y^{2/3}\).
3. Set up integral: \(V = \pi \int_{1}^{8} y^{2/3} \, dy\).
4. Integrate: \(V = \pi [\frac{3}{5}y^{5/3}]_{1}^{8}\).
5. Evaluate: \(V = \pi (\frac{3}{5}(8)^{5/3} - \frac{3}{5}(1)^{5/3}) = \pi (\frac{3}{5}(32) - \frac{3}{5}(1)) = \frac{93\pi}{5}\).

4. Summary Table for Volumes

Use this table to keep your formulas straight during revision:

Rotation Axis: \(x\)-axis
Integral Type: \(\int \dots dx\)
What to square: \(y^2\)
Formula: \(V = \pi \int_{a}^{b} [f(x)]^2 \, dx\)

Rotation Axis: \(y\)-axis
Integral Type: \(\int \dots dy\)
What to square: \(x^2\)
Formula: \(V = \pi \int_{c}^{d} [g(y)]^2 \, dy\)

5. Important Tips for Success

Did you know? If you rotate a straight diagonal line passing through the origin, you create a cone! If you rotate a horizontal line, you create a cylinder. You can actually use these calculus formulas to prove the volume formulas you learned in middle school.

Common Pitfalls to Avoid:

• The "Washers" Problem: If you are finding the volume between two curves, the formula becomes \(V = \pi \int (y_{top}^2 - y_{bottom}^2) \, dx\). Do not subtract the functions and then square them; you must square them individually!
• Calculator Usage: On Paper 2 or 3, you can use your GDC to evaluate these integrals. Ensure you know how to input "Numerical Integral" accurately.
• Bounds: Always check if the question gives you \(x\)-limits or \(y\)-limits. If you are rotating around the \(y\)-axis but given \(x\)-limits, you must convert them by plugging them into the original equation.

Key Takeaway: Whether you are finding area or volume, the most important step is identifying the axis of revolution. This tells you whether to integrate with respect to \(x\) or \(y\). Once you have that, just remember to "Square the radius and add the \(\pi\)" for volumes!