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Volume of rotation - my answer seems off a bit

  • #1
Find the area of the given function, rotated about the y axis

The Area below the line y=2 and above y=sin(x) from [tex]0-\pi[/tex]

I did this rotated around the X AXIS no problem,

I found the area of the disk to be [tex]\pi(4-sin^{2}[/tex](x))

The volume around the X AXIS to be[tex]\frac{7\pi^{2}}{2}[/tex]



The Attempt at a Solution



For the Y axis I have used this:[tex]\int^{\pi}_{0} 2\pi xf(x)dx[/tex]
Plugging in:

[tex]\int^{\pi}_{0} 2\pi x(2-sin(x))dx[/tex]

I solved this to be [tex]2\pi^{3}-2\pi^{2}[/tex]


Does this look right?
 

Answers and Replies

  • #2
33,181
4,860
Yes.
 
  • #3
Really? That is awesome!

Any second opinions?
 
  • #4
33,181
4,860
OK.
No.:smile:
 

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