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anonymous

  • 5 years ago

Use the disk or the shell method to find the volume of the solid generated by revolving the region bounded by the graphs of the equations about each given line. y = 6/x^2 y = 0 x = 1 x = 3 What method should I use?

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  1. amistre64
    • 5 years ago
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    all of them lol

  2. anonymous
    • 5 years ago
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    thanks :D

  3. anonymous
    • 5 years ago
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    I've tried all of them, and I still get the wrong answer

  4. anonymous
    • 5 years ago
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    The shell method gives me -8pi, and the disk method gives me a ridiculous number

  5. amistre64
    • 5 years ago
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    the question seems to be asking you to do it 4 different times...

  6. anonymous
    • 5 years ago
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    Oh, my question uses the y-axis.

  7. anonymous
    • 5 years ago
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    There's a list of them, and I have already done the one revolving around the x-axis

  8. amistre64
    • 5 years ago
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    is this our region?

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  9. amistre64
    • 5 years ago
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    if you wanna spin it round the y axis; shell it. 2pi 6 {S} x(1/x^2) dx ; [1,3]

  10. anonymous
    • 5 years ago
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    Yes...so I thought to use the shell method first....right., that's what I did, and got -8pi

  11. amistre64
    • 5 years ago
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    1/x ints up to ln(x) right? so 12pi ln(x)

  12. anonymous
    • 5 years ago
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    wait---you separated the 6 from the equation, and multiplied by x instead of (x + 1) or some variation of that?

  13. amistre64
    • 5 years ago
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    12pi ln(3) should be it ; since ln(1) = 0

  14. amistre64
    • 5 years ago
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    constants dont get inted; they get pulled out, thats why we dont int pi and 2pi and the like

  15. amistre64
    • 5 years ago
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    we dont derive constants either; so why bother inting them lol

  16. amistre64
    • 5 years ago
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    sine the y axis is alreay x = 0; no need to move anything

  17. anonymous
    • 5 years ago
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    so we don't add or subtract anything from the r(x) factor in [\int\limits_{a}^{b}r(x)h(x)dx\] if y = 0?

  18. anonymous
    • 5 years ago
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    \[2\pi \int\limits_{a}^{b}r(x)h(x)dx\]

  19. amistre64
    • 5 years ago
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    nope; the radius in the shell method is as you move from 1 to 3 in this and the height is 6/x^2

  20. anonymous
    • 5 years ago
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    the radius in the shell method is...? as you move from 1 to 3?

  21. amistre64
    • 5 years ago
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    what are we spinning around?

  22. anonymous
    • 5 years ago
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    the y-axis

  23. amistre64
    • 5 years ago
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    does the y axis the same as the x=0 axis?

  24. anonymous
    • 5 years ago
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    ?

  25. amistre64
    • 5 years ago
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    is the y axis that same as the x=0 line?

  26. anonymous
    • 5 years ago
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    y = 0 and x = 0 look the same....oh, okay

  27. amistre64
    • 5 years ago
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    since y axis IS x=0; then what do we move to get this to x=0? nothing right?

  28. amistre64
    • 5 years ago
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    so r(x) = x ; h(x) = 6/x^2 r(x)h(x) = 6/x from 1 to 3

  29. amistre64
    • 5 years ago
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    12pi/x ints to 12pi ln(x) which means the the answer is: 12pi ln(3)

  30. anonymous
    • 5 years ago
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    the answer is right! i get it----what your second to last post was. This makes sense now.

  31. amistre64
    • 5 years ago
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    :)

  32. anonymous
    • 5 years ago
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    Thanks! See yah next time

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