What if the axis of revolution is a line like y = k instead of the x-axis?

Adjust the radius function to measure the distance from the curve to the new axis. The radius becomes R(x) = |f(x) - k|.

When should the washer method be used instead of the disk method?

Use the washer method when there is a gap between the planar region and the axis of revolution, requiring the subtraction of an inner volume.

Can the bounds of integration be negative?

Yes, the limits of integration represent coordinates on the axis and can be negative, but the resulting volume must always be positive.

How to find the volume of a solid of revolution with the disk method

The disk method calculates the volume of a solid generated by revolving a continuous curve around a coordinate axis. It applies when the axis of revolution forms a boundary of the planar region, yielding solid circular cross-sections with no inner cavity.

The setup

Identify the axis of revolution and determine the bounds of integration. Express the bounding curve as a radius function perpendicular to the axis of revolution: R(x) for the x-axis or R(y) for the y-axis.

The steps

Define the radius function R(x) or R(y) measuring the distance from the axis of revolution to the curve. 2. Set up the volume integral V = \pi \int_a^b [R(x)]^2 dx for horizontal axes, or V = \pi \int_c^d [R(y)]^2 dy for vertical axes. 3. Expand the squared term algebraically. 4. Evaluate the definite integral using the Fundamental Theorem of Calculus.

Checking the result

Confirm that the final volume is strictly positive. Check that the integration limits correspond exactly to the geometric intersections of the bounding curves.

Common errors

Failing to square the radius function inside the integral. Integrating with respect to the wrong variable, such as using dx for a y-axis revolution. Treating \pi as a variable rather than a constant that can be factored out.

Worked example

Find the volume of the solid generated by revolving the region bounded by y = \sqrt{x}, y = 0, and x = 4 around the x-axis.

The radius function is R(x) = \sqrt{x}. The region extends from x = 0 to x = 4. Set up the integral: V = \pi \int_{0}^{4} (\sqrt{x})^2 dx. Simplify the integrand: V = \pi \int_{0}^{4} x dx. Find the antiderivative: V = \pi [\frac{1}{2}x^2]_{0}^{4}. Evaluate the bounds: V = \pi (\frac{1}{2}(16) - 0). The final volume is V = 8\pi.

FAQ

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Answer form

References: Stewart Calculus, 8th Edition, Chapter 6 · OpenStax Calculus Volume 1, Chapter 2 · Thomas Calculus, 14th Edition, Chapter 6