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Find the volume of the solid obtained by rotating the region bounded by the given curve about the specified axis. x^2+(y-5)^2=16 about the y axis. Volume = __?__
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so we have a donut pretty much
radius of 4; and we are 5 away from the center so yeah; a donut
shell method might be appicable
|dw:1333341918298:dw|
\[\int 2pix\ f(x)\ dx\] \[f(x)=\sqrt{16-x^2}+5\] \[4pi\int_{2}^{10}x\sqrt{16-x^2}+5x\ dx\]
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one idea that might work, and be simpler is this: |dw:1333342330845:dw|
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