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Dido525

The region bounded by the given curves is rotated about the specified axis. Find the volume V of the resulting solid by any method. x = (y − 5)^2, x = 4; about y = 3 I am trying to use the disc method but I am not sure how I would.

  • one year ago
  • one year ago

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  1. Dido525
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    Anyone want to help me set up the integral?

    • one year ago
  2. Dido525
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    Anyone? :( .

    • one year ago
  3. Dido525
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    Well I found the intersection points for y which are y=3 and y=7.

    • one year ago
  4. tkhunny
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    Good. This is a nice problem. Have you heard of Pappus's Theorem?

    • one year ago
  5. Dido525
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    No.

    • one year ago
  6. tkhunny
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    Too bad. It is much simpler with the Centroid. Another lesson on another day, perhaps. The Whole thing: \(\pi\int\limits_{0}^{4}(y_{1}-3)^{2}\;dx\) The Extra thing: \(\pi\int\limits_{0}^{4}(y_{2}-3)^{2}\;dx\) Do you see the nature of the problem that requires two definitions for 'y'? It's NOT a function!

    • one year ago
  7. Dido525
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    Yeah, I realized that.

    • one year ago
  8. tkhunny
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    Perfect. What are the two definitions?

    • one year ago
  9. Dido525
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    Wait you lost me. Never mind.

    • one year ago
  10. Dido525
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    @tkhunny

    • one year ago
  11. tkhunny
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    \(x = (y-5)^{2}\) Solve for \(y\). You should get TWO results.

    • one year ago
  12. Dido525
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    yeah y=y-5 and y=5-y

    • one year ago
  13. Dido525
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    Wait. WHat am I doing...

    • one year ago
  14. Dido525
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    Braid dead lol.

    • one year ago
  15. tkhunny
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    ?? Where did the \(x\) go? \(\sqrt{x} = y-5\;for\;y>5\) \(\sqrt{x} = 5-y\;for\;y<5\) Keep solving!

    • one year ago
  16. Dido525
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    |dw:1358491765588:dw|

    • one year ago
  17. Dido525
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    Sorry. I had a stupid moment.

    • one year ago
  18. tkhunny
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    Super. You almost beat me to it. Now, which one is on top and which one is on the bottom? These are my little y-functions.

    • one year ago
  19. Dido525
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    Top function is \[\sqrt{x}+5\] Bottom function is \[5-\sqrt{x}\]

    • one year ago
  20. tkhunny
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    The Whole thing: \(\pi\int\limits_{0}^{5}((\sqrt{x}+5) −3)^{2}\;dx\) The Extra part: Oh, you do that one!

    • one year ago
  21. Dido525
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    Woudn't it be from 0 to 4?

    • one year ago
  22. tkhunny
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    Yes. I just spotted that typo. Sorry about that. Good eye!

    • one year ago
  23. Dido525
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    Haha thanks :) .

    • one year ago
  24. tkhunny
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    And the answer is????????

    • one year ago
  25. Dido525
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    136/3 which is wrong :( .

    • one year ago
  26. Dido525
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    I forgot pi whoops.

    • one year ago
  27. tkhunny
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    That's the whole thing. You have to subtract the small piece. That's why we had two integrals.

    • one year ago
  28. Dido525
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    So it would be:\[\pi \int\limits_{0}^{4}((\sqrt{x}+5)-3)^2-((5-\sqrt{x})-3)^2dx\]

    • one year ago
  29. tkhunny
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    That should do it. Or, you can do them as separate integrals.

    • one year ago
  30. Dido525
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    THanks so much! Lets see if I get the right answer...

    • one year ago
  31. Dido525
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    Yeah it's right :) . Thanks so much!

    • one year ago
  32. Dido525
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    I know. I did it out on maple.

    • one year ago
  33. tkhunny
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    Interestingly, if you do a little algebra, you can do a whole lot less calculus. That wonderful integrand you created simplifies to \(8\sqrt{x}\).

    • one year ago
  34. tkhunny
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    Well, for future reference, if the ENTIRE area is on one side of the axis of rotation, you can pull Pappus' Theorem out of your hat! Good luck when the time comes!

    • one year ago
  35. Dido525
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    When I learn it XD .

    • one year ago
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