Volumes of Revolution Math Example 3

Follow the full solution, then compare it with the other examples linked below.

Example 3

easy
Find the volume of the solid formed by rotating y=2xy = 2x from x=0x=0 to x=3x=3 around the xx-axis.

Solution

  1. 1
    V=ฯ€โˆซ03(2x)2โ€‰dx=4ฯ€โˆซ03x2โ€‰dx=4ฯ€[x33]03=4ฯ€โ‹…9=36ฯ€V = \pi\int_0^3(2x)^2\,dx = 4\pi\int_0^3 x^2\,dx = 4\pi\left[\frac{x^3}{3}\right]_0^3 = 4\pi \cdot 9 = 36\pi.

Answer

36ฯ€36\pi
This cone has base radius 6 and height 3. The cone volume formula 13ฯ€r2h=36ฯ€\frac{1}{3}\pi r^2 h = 36\pi agrees, confirming the disc-method result.

About Volumes of Revolution

Finding the volume of a three-dimensional solid formed by rotating a two-dimensional region around an axis. The disc/washer method uses circular cross-sections perpendicular to the axis; the shell method uses cylindrical shells parallel to the axis.

Learn more about Volumes of Revolution โ†’

More Volumes of Revolution Examples

Example 1 easy

Find the volume of the solid formed by rotating [formula] around the [formula]-axis from [formula] t

Example 2 hard

Find the volume when the region between [formula] and [formula] ([formula]) is rotated around the [f

Example 4 medium

Use the shell method to find the volume when the region bounded by [formula], [formula], [formula] i

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