Find the volume of the solid obtained by rotating the region bounded by the given curves about the specified axis?

Question

anonymous · Answer

$$y=\frac{ 1 }{ x^3 }, y=0, x=3, x=6$$about the y-axis

anonymous · Answer

I haven't done this in a while, but the point of intersection is (3,1) so it should be: 

 pi*Integral( [3y^2]^2,y,0,1)=1.8*pi=5.654 

 since the rule is pi*integral( upper^2-lower^2,variable,lower x or y bound, upper x or y bound)

anonymous · Answer

Shouldn't the integration be from 3 to 6?

campbell_st · Answer

no... you substiute x = 3 and x = 6 to find the corresponding y values

so you are integrating from y = 1/216 to 1/27

you will also need to make x the subject... 

then it

$$V = \pi \int\limits_{\frac{1}{216}}^{\frac{1}{27}} x^2 dy$$

campbell_st · Answer

the diagram is 

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