Volume from rotating function about y axis

[nameVoid]

y=1/x, y=0, x=1,x=1, rotate around y axis;

curious how to set this up using washers

 

[Saladsamurai]

It's been a long time since I have done this, but I think this should be done with disks & I believe you meant x=1 and y=1 ?

 

[HallsofIvy]

You have "x= 1, x= 1". Surely, that's not what you meant! I am going to assume the limits are x= a, x= b, with a< b.

Because you have constant x limits (I assume), and are rotating around the y-axis, "washers" is not the best way to do this. The simplest method would be shells.

But since you ask:

There is no graph below y= 1/b. But you have y= 0 as the lower limit for the figure. So, for y= 0 to 1/a, this is a cylinder. its volume is $\pi(b^2- a^2)(1/a)$, the area between the two circles, of radii a and b, times it height, 1/a.

For y= 1/a to 1/b, draw a horizontal line from line from x= (a, y) to (1/y, y) (Since y= 1/x, x= 1/y). Rotated about the y-axis that gives a "washer". Its area is the difference between the areas of the two circles: $\pi((1/y)^2- a^2)$. The volume of that "washer" is that area times its thickness, "dy": $\pi\int_{y= 1/a}^{1/b} ((1/y)^2- a^2)dy$. Don't forget to add that first volume to get the volume of the entire thing.

 

[tiny-tim]

y=1/x, y=0, x=1,x=1, rotate around y axis;

curious how to set this up using washers

Hi nameVoid!

I'm not sure what you mean , but the general technique is to cut it into slices with thickness dx … then the volume of each slice is (area)dx, and so the total volume is ∫(area)dx.