Math CentralQuandaries & Queries


find the volume of the solid formed when region bounded by y=x/3, y=2
and the y-axis. it is revolved about the x-axis.

the assignment was to use both the washer method and the shell method
but when i solved for the volume, i got different answers.
i think my shell method is wrong because i know i'm having difficulties
with using "dy" instead of "dx"
here's my work so far:

washer method:
R = 2, r = x/3
V=pi*integral from 0 to 6*4-(x2/9)dx
it becomes 16pi because after the integral it is
pi (4x-(x3/27)) and then i evaluated it

R = y, H = x = 3y
V = 2pi*integral from 0 to 6*3y2dy
it becomes 2pi(y3), evaluating it at 6
which turns out to be

but there's a huge difference between my two answers..
could you please help me find my mistake?

Hi Christina,

I agree with your washer method but there is one error in your shell method. In the shell method integral you are integrating with respect to y so the limits on the integral must be from y = 0 to y = 2. It looks like you have the limits as
x = 0 to x = 6.


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