Solid Of Revolution Problem - Washer

[Bellwether]

Homework Statement

Find the volume of a solid created when the area between the function y=x^2+1 and the x-axis (for 0<x<2) is rotated about the line y=-2.

Homework Equations

Vs = ∏*r^2*h

The Attempt at a Solution

I can't seem to set this up correctly and am thrown by the inner radius being the x-axis. I originally set up the integral as : Int {0,2} ∏((x^2+1) + 3)^2 dx

I think this creates a full solid around y=-2 -- would I just want to add 1 to the radius instead of 3?

Many thanks!

 

[tiny-tim]

Welcome to PF!

Hi Bellwether! Welcome to PF!

(try using the X² button just above the Reply box )

I can't seem to set this up correctly and am thrown by the inner radius being the x-axis. I originally set up the integral as : Int {0,2} ∏((x^2+1) + 3)^2 dx

That's for the outer radius …

you now need to subtract the volume for the inner radius. ​

 

[QuarkCharmer]

For problems like this, I used to prefer adding 2 to the function to shift the whole thing up 2 units of y. Then you can simply revolve the function around the line y = 0, and the volume result will be the same.

 

[Bellwether]

Thanks for the welcome!

So it's, ∫∏((x²+1)+3)² - 2²) ?

 

[tiny-tim]

Hi Bellwether!

(just got up :zzz:)

Thanks for the welcome!

So it's, ∫∏((x²+1)+3)² - 2²) ?

∫ π((x²+1) + 2)² - 2²) dx

 

[Bellwether]

I'm confused -- why is the outer radius just +2 , instead of 3? Isn't (x² + 1) three units away from the axis of revolution?

 

[tiny-tim]

no, you've counted the "1" twice …

(x² + 1) - (-2 ) = x² + 3 ​

 

[Bellwether]

Of course, thank you.