The volume of a solid formed by rotating the curve y = f(x) about the x-axis through a full rotation is given by which integral?

The volume of a solid formed by rotating the curve y = f(x) about the x-axis through a full rotation is given by which integral?

• A. 2π ∫ y dx
• B. π ∫ f(x)^2 dx
• C. π ∫ y^2 dx
• D. π ∫ x^2 dy

Answer: (C)

When a solid is formed by rotating a curve about the x-axis, use the disk method. A tiny slice perpendicular to the x-axis becomes a disk with radius equal to the y-value at that x. The area of each disk is π times the radius squared, so π y^2, and its thickness is dx. Let x range from a to b; the volume is the integral of these disk volumes: V = ∫_a^b π y^2 dx. Since y = f(x), this is V = π ∫_a^b [f(x)]^2 dx, which you can also write as π ∫ y^2 dx. This form captures the radius being y and the integration being with respect to x. The other form with 2π ∫ y dx would not give the volume in this setup.

When a solid is formed by rotating a curve about the x-axis, use the disk method. A tiny slice perpendicular to the x-axis becomes a disk with radius equal to the y-value at that x. The area of each disk is π times the radius squared, so π y^2, and its thickness is dx. Let x range from a to b; the volume is the integral of these disk volumes: V = ∫_a^b π y^2 dx. Since y = f(x), this is V = π ∫_a^b [f(x)]^2 dx, which you can also write as π ∫ y^2 dx. This form captures the radius being y and the integration being with respect to x. The other form with 2π ∫ y dx would not give the volume in this setup.