- #1

Lo.Lee.Ta.

- 217

- 0

^{2}-6y =x, rotated around x=1.

2. R= y

^{2}-6y -1

r= y-10 -1

∫2 to 5 of [([itex]\pi[/itex](y

^{2}-6y -1)

^{2}) - [itex]\pi[/itex](y - 11)

^{2}]

∫[itex]\pi[/itex][y

^{4}- 12y

^{3}+ 4y

^{2}+ 12y + 1] - [itex]\pi[/itex][y

^{2}- 22y + 121]dy

= [itex]\pi[/itex][y

^{5}/5 - 12y

^{4}/4 + 4y

^{3}/3 +12

^{2}/2 +y] - [itex]\pi[/itex][y

^{3}/3 - 22y

^{2}/2 + 121y] |2 to 5

...LONG SUBSTITUTION...

= -928.33[itex]\pi[/itex] - 371.67[itex]\pi[/itex] + 4.93[itex]\pi[/itex] + 200.67[itex]\pi[/itex]

= -1094.4[itex]\pi[/itex]

...Is this the right answer...? I don't really think it's right because it's negative!

But what am I doing wrong here? :(

Thanks SO much for your help! :D