Calculating Volume of Rotated Curve: y=1/x, y=-1

[ziddy83]

Hey everyone...I need to find the volume of the following shape...

$$y= \frac{1}{x}, y=0, x=1, x=3$$

rotated about $$y=-1$$

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

$$r= \frac{1}{x} + 1$$

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

$$\pi \int_{1}^{3} ( \frac{1}{x} +1)^2 dx$$

did i set this up right? Thanks..

 

[whozum]

http://tutorial.math.lamar.edu/AllBrowsers/2413/VolumeWithRings.asp

Check out examples 3 and 4.

 

[OlderDan]

Hey everyone...I need to find the volume of the following shape...

$$y= \frac{1}{x}, y=0, x=1, x=3$$

rotated about $$y=-1$$

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

$$r= \frac{1}{x} + 1$$

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

$$\pi \int_{1}^{3} ( \frac{1}{x} +1)^2 dx$$

did i set this up right? Thanks..

Your solid has a hole in the middle. The y = 0 boundary excludes a cylinder of radius 1 from the solid. You need the difference of the squares of the outer radius and inner radius.