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Dido525

  • 2 years ago

Use the method of cylindrical shells to find the volume V generated by rotating the region bounded by the given curves about the specified axis. y = 7e^x, y = 7e−x, x = 1; about the y-axis

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  1. Dido525
    • 2 years ago
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    I don't get what I did wrong in my work. Let me post it.

  2. Dido525
    • 2 years ago
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    |dw:1358999196019:dw|

  3. Dido525
    • 2 years ago
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    Did I set up the integral properly?

  4. Dido525
    • 2 years ago
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    @Mimi_x3 @satellite73

  5. tkhunny
    • 2 years ago
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    What did you get for your points of intersection? y = 7e^x vs x = 1 is obviously (1,7e) y = 7e−x vs x = 1 is obviously (1,7e-1) But what about y = 7e^x vs y = 7e−x ? Since this cannot be solved by direct methods, we get (0.9488478,18.0791250) Maybe the problem statement is not written correctly?

  6. Dido525
    • 2 years ago
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    Let me send a pic.

  7. Dido525
    • 2 years ago
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  8. Dido525
    • 2 years ago
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    I used the e points of intersection so I used 0 to 1.

  9. Dido525
    • 2 years ago
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    x not e.

  10. tkhunny
    • 2 years ago
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    \(y = 7e^{x}\) -- You wrote this one okay. \(y = 7e^{-x}\) -- This is quite a different story. Please learn to code LaTeX or to use the Equation Editor. It will be a great benefit to you now and in the future. You're f(x) is not correct. You just picked one of the curves. You need the distance between them.

  11. Dido525
    • 2 years ago
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    Ohh I can write in latex. It just takes too long lol.

  12. tkhunny
    • 2 years ago
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    It would have avoided much time and error. 7e-x is VERY different from 7e^(-x) or \(7e^{-x}\)

  13. Dido525
    • 2 years ago
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    So I wrote my integral as: \[14\pi \int\limits_{}^{}xe^x dx\]

  14. Dido525
    • 2 years ago
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    How do I incorporate the e^-x then?

  15. tkhunny
    • 2 years ago
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    \(e^{x}\) is no good. You need \(e^{x} - e^{-x}\).

  16. Dido525
    • 2 years ago
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    Huh? Why?

  17. Dido525
    • 2 years ago
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    Nevermind. I see why.

  18. Dido525
    • 2 years ago
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    The limits of intergation should still be the same right?

  19. tkhunny
    • 2 years ago
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    Did you draw a picture? The height of your cyllinders is the difference of the two curves.

  20. Dido525
    • 2 years ago
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    Yeah I did, One moment.

  21. Dido525
    • 2 years ago
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    |dw:1359002030111:dw|

  22. Dido525
    • 2 years ago
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    Something like that.

  23. tkhunny
    • 2 years ago
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    That's it. And the final answer is?

  24. Dido525
    • 2 years ago
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    I see why it's the difference though.

  25. Dido525
    • 2 years ago
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    Sec.

  26. Dido525
    • 2 years ago
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    \[\frac{ 28\pi }{ e}\] is what I got.

  27. tkhunny
    • 2 years ago
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    I like it. Now, your punishment for writing one of the equations incorrectly in the beginning is to find the point of intersection for what you wrote. Find the value of x such that \(7e^{x} = 7e - x\). It's a wonderful exploration.

  28. Dido525
    • 2 years ago
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    Eww... Natural logs...

  29. tkhunny
    • 2 years ago
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    No, no. It's far worse than that. :-) When you are ready...

  30. Dido525
    • 2 years ago
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    Ahh I see... Hehe...

  31. Dido525
    • 2 years ago
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    Yeah... I have to use a calculator lol.

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