what is the volume of the region bound by

y=4/x and y=(x-3)^2 about the y-axis.

I was trying to use this:

2pi*INT[ x(4/x)-x(x-3)^2]

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- Aug 23rd 2009, 02:15 PMcanadinaicon25volume problem
what is the volume of the region bound by

y=4/x and y=(x-3)^2 about the y-axis.

I was trying to use this:

2pi*INT[ x(4/x)-x(x-3)^2] - Aug 23rd 2009, 02:24 PMCalculus26
Are you using x = 1 and 4 for your limits of integration?

If you are submit your work to see where the problem might be--your integrand is correct - Aug 23rd 2009, 02:30 PMcanadinaicon25
yeah, forgot the bounds. intersection is at x=1 and x=4. I must be making an algebraic mistake then.

4x-(x^4/4-2x^3+9x^2/2)

evaluated between 4 and 1? - Aug 23rd 2009, 02:38 PMHallsofIvy
**why**are you integrating between y= 1 and y= 4? If you draw a graph you will see that parabola $\displaystyle y= (x-3)^2$ extends down to y= 0. I suggest you do this as two parts: integrate what you have from y= 1 to y= 4 then set up a different integral for the part from y= 0 to y= 1. - Aug 23rd 2009, 02:41 PMCalculus26
Canadian is using the shell method for rotating about the y axis

thats why the limits are x= 1 to 4

Canadian -- 4x-(x^4/4-2x^3+9x^2/2) is correct - Aug 23rd 2009, 02:48 PMCalculus26
you should get 27pi/2