I assume that the line x=0 is also part of the boundary. If so, you just have a cone with radius 24 and height 3, with a volume of
v = π/3 * 24^2 * 3 = 576π
Oh, ok -- if you want to use calculus, then using discs of thickness dx,
v = ∫[-3,0] πr^2 dx
where r = y = 8x+24
v = π∫[-3,0] (8x+24)^2 dx
= 64π∫[-3,0] (x+3)^2 dx
= 64π∫[-3,0] x^2+6x+9 dx
= 64π (1/3 x^3 + 3x^2 + 9x) [-3,0]
= 576π
using shells of thickness dy,
v = ∫[0,24] 2πrh dy
where r = y and h = x = -(y-24)/8
v = π/4 ∫[0,24] -y(y-24) dy
= -π/4 (1/3 y^3 - 12y^2) [0,24]
= 576π
Find the volume of the solid obtained by rotating the region bounded by
y=8x+24, y=0
about the y-axis.
2 answers
Rats. I did it about the x-axis. Still, you see the method. Redo the rotation, and see whether you come up with 72π
2 answers