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Moyo30

  • one year ago

Find the volume of the torus (doughnut) formed by rotating the circle x2 + (y − 2)2 = 1 about the x-axis.

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  1. Moyo30
    • one year ago
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    Hint: \[\int\limits_{-1}^{1}\sqrt{1-x ^{2}}=\frac{ \pi }{ 2 }\]

  2. klimenkov
    • one year ago
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    |dw:1357858021843:dw|

  3. Moyo30
    • one year ago
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    ???

  4. klimenkov
    • one year ago
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    Do you know the formula for the volume of solid rotating \(y=f(x)\) about \(x\)-axis?

  5. Moyo30
    • one year ago
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    yes.

  6. klimenkov
    • one year ago
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    |dw:1357858353355:dw|

  7. Moyo30
    • one year ago
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    \[\pi \int\limits_{a}^{b}f(x)dx\]

  8. klimenkov
    • one year ago
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    You forgot 2.

  9. Moyo30
    • one year ago
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    ma bad

  10. klimenkov
    • one year ago
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    |dw:1357858535828:dw| Subtract the right volume from the left. Got it?

  11. Moyo30
    • one year ago
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    y subtracting?

  12. klimenkov
    • one year ago
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    |dw:1357858823945:dw| It is like subtracting this areas.

  13. klimenkov
    • one year ago
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    \(V=\pi\int_{-1}^1(\sqrt{1-x^2}+2)^2dx-\pi\int_{-1}^1(-\sqrt{1-x^2}+2)^2dx\)

  14. klimenkov
    • one year ago
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    If you don't get it now just sit and think about it very thoroughly.

  15. Moyo30
    • one year ago
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    |dw:1357858979556:dw| the graph of x^2+(y-2)^2=1 is something like that how do we get to the penultimate drawing

  16. Moyo30
    • one year ago
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    i kinda get it you reflect a portion of the circle in the x axis and find the region bounded by the two arcs|dw:1357859345394:dw|

  17. klimenkov
    • one year ago
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    if you try to write y depending on x, you will get 2 functions: \(y=\sqrt{1-x^2}+2\) and \(y=-\sqrt{1-x^2}+2\). Now think what will be if you rotate them about x-axes. And then use your imagination to unterstand how to get the needed volume. Really the plot is:|dw:1357859397346:dw|

  18. klimenkov
    • one year ago
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    But if you are lazy and dishonest student you can just use the formulas I've written above.

  19. Moyo30
    • one year ago
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    thx

  20. klimenkov
    • one year ago
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    Have you understood something?

  21. Moyo30
    • one year ago
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    yes it is basically a volume but it complicated by the fact that the function that we get after making y the subject of the formula is composed of two equations so we add the two. right?

  22. klimenkov
    • one year ago
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    It is very hard to tell if you got something or not. I don't know your level of knowledge. Tell me honestly: can you find the area between \(y=x^2\) and \(y=1-x^2\)?

  23. Moyo30
    • one year ago
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    subtract x^2 from 1-x^2 and integrate

  24. klimenkov
    • one year ago
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    Why square?

  25. Moyo30
    • one year ago
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    juss realised my prob no squaring

  26. Moyo30
    • one year ago
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    integrate, limits the intersection of the lines

  27. klimenkov
    • one year ago
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    Yes. The similar situation is here. But now we have a volume to subtract. Hope, you got something, if not - very bad.

  28. Moyo30
    • one year ago
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    so basically in this prob they ask us to find the volume rather than the area

  29. Moyo30
    • one year ago
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    can you pls draw the resultant volume if it is not too diff

  30. klimenkov
    • one year ago
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    Yes. The volume of the doughnut.|dw:1357861006645:dw|

  31. Moyo30
    • one year ago
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    thx for your help

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