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MasterChief

  • one year ago

Could someone tell me what I did wrong in this volume problem? (Washer)

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  1. MASTERCHIEF
    • one year ago
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  2. phi
    • one year ago
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    it looks like you get 32+ 32/5 = 38.4 what makes you think that is wrong?

  3. MASTERCHIEF
    • one year ago
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    Well did I set up the integral according to the graph correctly?

  4. phi
    • one year ago
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    I would have to see the original question to be sure of that.

  5. MASTERCHIEF
    • one year ago
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    It's at the top of the image

  6. MASTERCHIEF
    • one year ago
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    x=y^2, x=4, about x=6

  7. phi
    • one year ago
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    is the lower bound known to be the x-axis ?

  8. MASTERCHIEF
    • one year ago
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    Yes

  9. phi
    • one year ago
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    yes, you set it up correctly. as a double check, if we find the volume of the whole region with center at x=6, and radius (6-y^2) \[\pi \int_0^2 (6-y^2)^2 dy = \pi\int_0^2 36+y^4 -12y^2 \ dy \\ = \pi(36y + \frac{y^5}{5} - 4y^3) \bigg|_0^2 = \pi(72 + \frac{32}{5}- 32)= \pi(40+ 6.4) \] now subtract off the volume of the "center cylinder" that has r=2 and h=2. Its volume is \( \pi r^2 h = \pi\cdot 4 \cdot 2= 8 \pi\) we get \[ 32\pi+6.4\pi = 38.4 \pi\] that is consistent with what you did.

  10. MASTERCHIEF
    • one year ago
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    Hmm the answer in the book is 384π/5

  11. phi
    • one year ago
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    that is twice as large as what we found. it implies they were finding the volume of |dw:1441656377364:dw|

  12. phi
    • one year ago
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    the lower limit appears to be \( -\sqrt{x} \) rather than 0

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