"Find the volume of the solid obtained by rotating the region bounded by the given curves about the given axis:"

y = x^4, y = 1; about y = 2

I set up my integral as π∫(1 - x^4)^2 - (1)^2 dx from -1 to 1.
I calculated this, several times, to get -(26π)/45. (I realize you take the absolute value of a negative area)
The answer in the book is (208π)/45.

Question

"Find the volume of the solid obtained by rotating the region bounded by the given curves about the given axis:"

y = x^4, y = 1; about y = 2

I set up my integral as π∫(1 - x^4)^2 - (1)^2 dx from -1 to 1.
I calculated this, several times, to get -(26π)/45. (I realize you take the absolute value of a negative area)
The answer in the book is (208π)/45.

anonymous · Answer

why did u take (1-x^4)^2??

anonymous · Answer

and what is the limit in x direction??

anonymous · Answer

You are doing shell yet you are using washing equation

anonymous · Answer

I ended up figuring it out.  I had to take (2 - x^4) as the radius since y = 2 was the middle of the whole big thing and x^4 was the bottom.  The top minus the bottom then is the radius.

Since y = x^4 and y = 1 bound the object, the limits in x direction are -1 and 1.

You can use shells and washers interchangeably, just like you can solve for area in respect to both x and y to get the same value.  This problem was in a section that dealt with washers as well.  Shells won't be introduced till the next chapter.