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AP® Calculus AB-BC

Disc Method Formula: AP® Calculus AB–BC Review

Turning a flat picture into a 3-D object feels like magic. Yet, on the AP® Calculus exams, this “magic” has a name: the disc method formula. Mastering it unlocks quick, error-free points whenever a volume of revolution question appears. This guide explains the idea, shows how to set it up for both axes, and walks through two fully solved examples. A handy vocabulary chart at the end ties everything together.

Disc Method Basics

What Is the Disc Method?

Imagine stacking identical coins into a neat column. Each coin is a thin disc, and together they form a solid cylinder. The disc method copies that idea. A 2-D region is spun around a line (the axis of revolution). This spin sweeps out many tiny circular slices, each with radius given by a function. Adding the volumes of these slices—using integration—gives the final 3-D volume.

When to Choose the Disc Method (vs. Washer or Shell)

Use the disc method formula when:

  • The cross sections are solid circles (no hole).
  • The region touches the axis of revolution.
  • Only one function sets the radius at every slice.

Conversely, select the washer method if there is a hole, and choose the shell method if cylindrical shells look simpler.

Deriving the Disc Method Formula

Revolving Around the x-Axis

  1. Consider a small horizontal slice of thickness \Delta x at position x.
  2. After revolving around the x-axis, the slice becomes a disc of radius r(x).
  3. The slice’s volume is approximately \pi \bigl(r(x)\bigr)^{2}\Delta x.
  4. Adding all slices from x=a to x=b and taking the limit as \Delta x \to 0 produces
\displaystyle V = \pi \int_{a}^{b} \bigl(r(x)\bigr)^{2}dx

Revolving Around the y-Axis

By switching the roles of x and y, the exact same reasoning gives

V = \pi \displaystyle\int_{c}^{d} \bigl(r(y)\bigr)^{2}dy

Key Conditions

Before integrating, always check:

  • The radius is measured straight to the axis of revolution.
  • Variables in the radius match the variable of integration.
  • Bounds correspond to the chosen variable.

Setting Up the Integral: A Graph-First Approach

Sketching prevents many mistakes. Follow this checklist:

  1. Draw the region and label the axis of revolution.
  2. Decide whether discs are horizontal (x-integration) or vertical (y-integration).
  3. Identify the radius function directly from the graph.
  4. Mark the correct lower and upper bounds.
  5. Confirm there is no inner “hole.” If a hole exists, switch to the washer method.

Because the disc method formula squares the radius, a sign error will not ruin the answer; however, wrong bounds or a missing \pi will.

Worked Example 1: Around the x-Axis

Problem

Find the volume of the solid generated when the region under y=\sqrt{x} from x=0 to x=4 is revolved about the x-axis.

Solution Steps

1. Graph and radius

The curve y=\sqrt{x} lies above the x-axis. Spinning it around the x-axis creates solid discs. The radius at position x equals the function value:

r(x) = \sqrt{x}

2. Integral setup

Lower bound a=0; upper bound b=4. V = \pi \displaystyle\int_{0}^{4} \bigl(\sqrt{x}\bigr)^{2}dx

3. Simplify the integrand

Since (\sqrt{x})^{2}=x, V = \pi \displaystyle\int_{0}^{4} xdx

4. Integrate

V = \pi \left[ \tfrac{1}{2}x^{2} \right]_{0}^{4} = \pi \left( \tfrac{1}{2}\cdot16 - 0\right)=8\pi

Interpretation

The solid has volume 8\pi\ \text{cubic units}.

Worked Example 2: Around the y-Axis

Problem

The region to the right of x = y^{2} for 0\le y\le 3 is revolved about the y-axis. Find the volume.

Solution Steps

1. Graph and radius

The parabola x = y^{2} opens rightward. The axis of revolution is the y-axis. A vertical slice at height y spins into a disc whose radius equals the x-distance to the axis:

r(y) = y^{2}

2. Integral setup

Lower bound c=0; upper bound d=3. V = \pi \displaystyle\int_{0}^{3} \bigl(y^{2}\bigr)^{2}dy

3. Simplify the integrand

(y^{2})^{2}=y^{4}, so V = \pi \displaystyle\int_{0}^{3} y^{4}dy

4. Integrate

V = \pi \left[ \tfrac{1}{5}y^{5} \right]_{0}^{3} = \pi\left( \tfrac{1}{5}\cdot243 - 0\right)=\tfrac{243}{5}\pi

Interpretation

The rotated region fills \tfrac{243}{5}\pi\ \text{cubic units}.

Common Pitfalls and Pro Tips

  • Forgetting \pi: Multiply the final integral by \pi every time.
  • Wrong bounds: Always match bounds to the axis’ variable.
  • Mixing variables: Do not place x inside an integral with dy.
  • Squared radius sign error: Negative values inside the square do not matter, but ensure the radius is positive by definition.

Quick Reference Chart: Key Vocabulary & Symbols

TermDefinition
DiscA solid, flat-faced cylinder slice formed by revolving a line segment
Radius function r(x) or r(y)Distance from curve to axis, used inside the square
Axis of revolutionLine (x-axis or y-axis) about which the region is spun
Bounds a,b,c,dStart and end values for the integral limits
Cross-sectional areaArea of one disc: \pi r^{2}
Disc method formulaV=\pi\displaystyle\int r^{2}d(\text{variable})

Conclusion

The disc method formula transforms a simple area into a tangible 3-D volume with one elegant integral:

V = \pi \displaystyle\int \bigl(\text{radius}\bigr)^{2}d(\text{variable})

Remember to sketch first, choose the correct variable, square the radius, and attach \pi. With steady practice, these steps will become automatic, ensuring quick, confident solutions on the AP® Calculus AB–BC exams. Integrate these tips into daily study, and watch volume problems turn from intimidating to routine.

Sharpen Your Skills for AP® Calculus AB-BC

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