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anonymous
 5 years ago
Find the volume of the solid generated by revolving the region bound by the graphs of the equations around the line x=5.
y=5x, y=0, y=4, x=0
anonymous
 5 years ago
Find the volume of the solid generated by revolving the region bound by the graphs of the equations around the line x=5. y=5x, y=0, y=4, x=0

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amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1dw:1326820967395:dw

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0What integral do I take for using discs though?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1disc is just using circles instead of cylindars;\[\int pi f(x)^2dx\]

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1but it might be best to set it up better, our radius is in the form \(f^{1}(x)\) and there is an inner and outer radius to account for

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1y=5x x=5y x5=y x+5=y would be our radius function: or y+5=x might be a better notation

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1the outer radius is just "5" the inner radius is what we have to remove, right?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1dw:1326821336211:dw

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1\[piR^2pir^2\=pi(R^2r^2)=Area. of. ring/]

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1\[pi R^2pi r^2=pi(R^2r^2)=Area. of. ring/]

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1dint format on my screen hmmm

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1since the radius is in terms of "x" values pi int(5^2  (y+5)^2) dy, from 0 to 4 looks right
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