# volume problem

• Aug 23rd 2009, 02:15 PM
volume 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 PM
Calculus26
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 PM
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 PM
HallsofIvy
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 PM
Calculus26
Canadian is using the shell method for rotating about the y axis

thats why the limits are x= 1 to 4