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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=5-x, y=0, y=4, x=0

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  1. amistre64
    • 5 years ago
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    |dw:1326820967395:dw|

  2. amistre64
    • 5 years ago
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    id shell it

  3. anonymous
    • 5 years ago
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    What integral do I take for using discs though?

  4. amistre64
    • 5 years ago
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    disc is just using circles instead of cylindars;\[\int pi f(x)^2dx\]

  5. amistre64
    • 5 years ago
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    but 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

  6. amistre64
    • 5 years ago
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    y=5-x x=5-y x-5=-y x+5=y would be our radius function: or y+5=x might be a better notation

  7. amistre64
    • 5 years ago
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    the outer radius is just "5" the inner radius is what we have to remove, right?

  8. amistre64
    • 5 years ago
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    |dw:1326821336211:dw|

  9. amistre64
    • 5 years ago
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    \[piR^2-pir^2\=pi(R^2-r^2)=Area. of. ring/]

  10. amistre64
    • 5 years ago
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    \[pi R^2-pi r^2=pi(R^2-r^2)=Area. of. ring/]

  11. amistre64
    • 5 years ago
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    dint format on my screen hmmm

  12. amistre64
    • 5 years ago
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    since 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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