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Dido525

  • one year ago

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.

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  1. Dido525
    • one year ago
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    Anyone want to help me set up the integral?

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

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

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

  5. Dido525
    • one year ago
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    No.

  6. tkhunny
    • one year ago
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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!

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

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

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

  10. Dido525
    • one year ago
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    @tkhunny

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

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

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

  14. Dido525
    • one year ago
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    Braid dead lol.

  15. tkhunny
    • one year ago
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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!

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

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

  18. tkhunny
    • one year ago
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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.

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

  20. tkhunny
    • one year ago
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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!

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

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

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

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

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

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

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

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

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

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

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

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

  33. tkhunny
    • one year ago
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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 \(8\sqrt{x}\).

  34. tkhunny
    • one year ago
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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!

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

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