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monroe17 Group Title

Find the volume formed by rotating the region enclosed by: y=x^2, x=y^2 about the line x=-5

  • one year ago
  • one year ago

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  1. tkhunny Group Title
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    Find the points of intersection. This will also be your integral limits with this lovely region.

    • one year ago
  2. monroe17 Group Title
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    how do i do that?

    • one year ago
  3. tkhunny Group Title
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    \(x^{2} = \sqrt{x}\) -- Solve!

    • one year ago
  4. monroe17 Group Title
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    0 and 1 :)

    • one year ago
  5. tkhunny Group Title
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    Perfect. Now, decide disks or shells. With this well-behaved region, it really doesn't much matter. Just pick one.

    • one year ago
  6. monroe17 Group Title
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    disks

    • one year ago
  7. tkhunny Group Title
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    Disks it is, then. \(\pi\int\limits_{0}^{1} OuterRadius^{2} - InnerRadius^{2} d(Height)\) The "Height" is the x-direction, so we have: \(\pi\int\limits_{0}^{1} OuterRadius^{2} - InnerRadius^{2} dx\) Now indentify that inner and outer radii. What say you?

    • one year ago
  8. monroe17 Group Title
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    x^2+5

    • one year ago
  9. tkhunny Group Title
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    Well, I would prefer the more literal (-5-x^2), but it will do since we are just going to square it. Is that the inner or the outer radius?

    • one year ago
  10. monroe17 Group Title
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    my bad, and R right?

    • one year ago
  11. tkhunny Group Title
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    No worries. x^2 is the inner radius. Are we calling that 'r'? What is the outer radius, "R"?

    • one year ago
  12. monroe17 Group Title
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    isn't (-5-x^2) the outer radius? and yes

    • one year ago
  13. tkhunny Group Title
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    No, \(x^{2}\) and \(\sqrt{x}\) are a little funny on [0,1] \(\sqrt{x} \ge x^2\;for\;x\in[0,1]\)

    • one year ago
  14. monroe17 Group Title
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    uhm.. im not sure ;/

    • one year ago
  15. tkhunny Group Title
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    r is \((-5 - x^{2})\) R is \((-5 - \sqrt{x})\) That's all. Did you draw a picture? You will see it at a glance.

    • one year ago
  16. monroe17 Group Title
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    I'm still confused toward the sqrt(x) where'd that come from?

    • one year ago
  17. tkhunny Group Title
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    x = y^2 ==> y = sqrt(x) We have integration "dx", so we need the representation in terms of x.

    • one year ago
  18. monroe17 Group Title
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    got it ;p lol just had to see the steps..

    • one year ago
  19. tkhunny Group Title
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    Well, there you have it. Let's see what you get.

    • one year ago
  20. monroe17 Group Title
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    I got 109pi/30?

    • one year ago
  21. tkhunny Group Title
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    THAT is the right answer. Good work! Wading through it is FUN, isn't it?!

    • one year ago
  22. monroe17 Group Title
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    haha.. totally ;p

    • one year ago
  23. tkhunny Group Title
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    Once we realized what the region was, and where the square root came from, you were all over it. This is encouraging! Move on to the next one!

    • one year ago
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