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Use the shell method to find the volume of the solid generated by revolving  about the y-axis. x=sqrt(9-x^2) and x=0

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The equation should be\[y =\sqrt{9-x^2}\]?
Assume the equation of the line is \(y=\sqrt{9-x^2}\) Domain of the function: [-3, 3] So, integrate from x=0 to x=3 \[V=\int_0^3 2\pi xf(x) dx = 2\pi \int_0^3 x\sqrt{9-x^2} dx=...\]
if revolving about the y-axis, the integration should forward dy so, v=pi*int((f(y))^2 dy [y1,y2] with y1 and y2 are under interval and up interval respectively

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Other answers:

Shell method.... about y-axis => dx See this:
@RadEn Yours is disc method, I think
yea, sorry i knew just new about the sell methode, @Callisto... sorry again :)
It's okay, don't worry :)

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