Question

Find the volume of a region bounded by the functions y = 2x + 6, the y-axis, and the x-axis as it revolves

around the x-axis.
GRAPH !
V=
units3
(round to the nearest tenth)

Answer 1

The bounded region in question is a triangle in the second quadrant. The line y = 2x + 6 has intercepts (0, 6) and (-3, 0), so revolving the triangle about the x-axis forms a cone with height 3 and whose circular base has radius 6. Its volume would then π/3 • 6² • 3 = 36π ≈ 113.1 units³.

But you're probably supposed to use calculus. Using the disk method, the integral for the volume is

`\displaystyle V=\pi\int_(-3)^0 (2x+6)^2\,\mathrm dx`

`\displaystyle V=\pi\int_(-3)^0(4x^2+24x+36)\,\mathrm dx`

`V=\pi\left(\frac43x^3+12x^2+36x\right)\bigg|_(-3)^0=\boxed{36\pi}`

as expected.