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

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

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

$V=\pi\left(\dfrac43x^3+12x^2+36x\right)\bigg|_{-3}^0=\boxed{36\pi}$

as expected.