How do you find the volume of the region bounded by #y=6x# #y=x# and #y=18# is revolved about the y axis?

1 Answer
Aug 21, 2015

Use washers. to get #V = 1890 pi#

Explanation:

The region is the bounded region in:

graph{(y-6x)(y-x)(y-0.0001x-18) sqrt(81-(x-9)^2)sqrt(85-(y-9)^2)/sqrt(81-(x-9)^2)sqrt(85-(y-9)^2) = 0 [-28.96, 44.06, -7.7, 28.83]}

Taking vertical slices and integrating over #x# would require two integrals, so take horizontal slices.

Rewrite the region: #x=1/6y#, #x=y# and #y=18#

As #y# goes from #0# to #18#, x goes from #x = 1/6y# on the left, to #x=y# on the right.
The greater radius is #R = y# and the lesser is #r = 1/6y#

Evaluate
#pi int_0^18 (R^2-r^2) dy = pi int_0^18 (y^2 - (y/6)^2) dy#

# = (35pi)/36 int_0^18 y^2 dy#

# = 1890pi#

(Steps omitted because once it is set up, I think this is a straightforward integration.)