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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?