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Dido525

  • 2 years 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
    • 2 years ago
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    Anyone want to help me set up the integral?

  2. Dido525
    • 2 years ago
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    Anyone? :( .

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

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

  5. Dido525
    • 2 years ago
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    No.

  6. tkhunny
    • 2 years 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
    • 2 years ago
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    Yeah, I realized that.

  8. tkhunny
    • 2 years ago
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    Perfect. What are the two definitions?

  9. Dido525
    • 2 years ago
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    Wait you lost me. Never mind.

  10. Dido525
    • 2 years ago
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    @tkhunny

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

  12. Dido525
    • 2 years ago
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    yeah y=y-5 and y=5-y

  13. Dido525
    • 2 years ago
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    Wait. WHat am I doing...

  14. Dido525
    • 2 years ago
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    Braid dead lol.

  15. tkhunny
    • 2 years 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
    • 2 years ago
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    |dw:1358491765588:dw|

  17. Dido525
    • 2 years ago
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    Sorry. I had a stupid moment.

  18. tkhunny
    • 2 years 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
    • 2 years ago
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    Top function is \[\sqrt{x}+5\] Bottom function is \[5-\sqrt{x}\]

  20. tkhunny
    • 2 years 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
    • 2 years ago
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    Woudn't it be from 0 to 4?

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

  23. Dido525
    • 2 years ago
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    Haha thanks :) .

  24. tkhunny
    • 2 years ago
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    And the answer is????????

  25. Dido525
    • 2 years ago
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    136/3 which is wrong :( .

  26. Dido525
    • 2 years ago
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    I forgot pi whoops.

  27. tkhunny
    • 2 years 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
    • 2 years 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
    • 2 years ago
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    That should do it. Or, you can do them as separate integrals.

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

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

  32. Dido525
    • 2 years ago
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    I know. I did it out on maple.

  33. tkhunny
    • 2 years 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
    • 2 years 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
    • 2 years ago
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    When I learn it XD .

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