- #1

ziddy83

- 87

- 0

[tex] y= \frac{1}{x}, y=0, x=1, x=3[/tex]

rotated about [tex] y=-1 [/tex]

ok so, i drew out the shape, and for the radius i came up with:

[tex] r= \frac{1}{x} + 1 [/tex]

I think that's correct...so assuming that my radius is right, then the volume would be...

[tex] \pi \int_{1}^{3} ( \frac{1}{x} +1)^2 dx [/tex]

did i set this up right? Thanks..