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anonymous

  • one year ago

Use the shell method to set up and evaluate the integral that gives the volume of the solid generated by revolving the plane region about the y-axis for y=x y=0 and x=2

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  1. Zale101
    • one year ago
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    First, draw the graphs and identify the enclosed area. Have you done that already?

  2. anonymous
    • one year ago
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    Yes

  3. Zale101
    • one year ago
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    |dw:1434058993462:dw|

  4. Zale101
    • one year ago
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    Since we are revolving the shell around the y-axis, we will be using the example i circled. |dw:1434059042077:dw|

  5. Zale101
    • one year ago
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    Makes sense?

  6. anonymous
    • one year ago
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    yeah! except your axis kind of confuse me I thought the y and x were switched

  7. anonymous
    • one year ago
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    But yeah I understand.

  8. Zale101
    • one year ago
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    But before anything else. You will analyze the rectangle and see what function along the x-axis is higher and what functions where the x-axis is lower. |dw:1434059232453:dw|

  9. anonymous
    • one year ago
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    ok I think I get it. The one thing im really confused on is the integral. do you use the numbers on the x-axis for the integral?

  10. anonymous
    • one year ago
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    \[\int\limits_{0}^{3} 2\pi (x ^{2})\] so like this ?

  11. anonymous
    • one year ago
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    i used that and then plugged the integral in so i got \[18\pi \] as my answer is that right. I think we may have learned it a different way because we never used that Big X little x thing we use a set equation that combines circumference and height

  12. Zale101
    • one year ago
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    |dw:1434059661157:dw|

  13. Zale101
    • one year ago
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    I forgot the mention, the intervals are based on where the spot is enclosed on the y-axis because we have dy and everything should be based on Y.

  14. Zale101
    • one year ago
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    First step: Let's identify the intervals. What do you think the intervals are?

  15. anonymous
    • one year ago
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    i gave you the wrong one by accident the problem says y=0 to x=3 so i think those are the intervals

  16. anonymous
    • one year ago
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    i told you 2 but its 3 sorry about that

  17. Zale101
    • one year ago
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    okay

  18. Zale101
    • one year ago
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    So the intervals were right. But what did you get for the big x and the small x?

  19. anonymous
    • one year ago
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    I don't know because the formula we learned dose not have that anywhere in there so i don't understand that.

  20. anonymous
    • one year ago
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    I honestly thought i had it right and just wanted to double check my answers and make sure but now im super confused

  21. Zale101
    • one year ago
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    \(\large a(x)=2\pi(radius)(height)\)

  22. anonymous
    • one year ago
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    Yeah!

  23. Zale101
    • one year ago
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    |dw:1434060538490:dw| \(\large a(x)=2pi(y)(X-x)\)

  24. Zale101
    • one year ago
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    \[\Large \int\limits_{0}^{3}2 \pi y(X-x)dy\]

  25. anonymous
    • one year ago
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    ok i still don't get the X-x thing i thought it was just the equation which is in this case just x but then u times that by x from 2pi move the 2 pi out front because its a constant take the anti derivative of the x^2 so then you end up with \[2\pi \int\limits_{0}^{3} 1/3 x^{3}\]

  26. Zale101
    • one year ago
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    |dw:1434060977099:dw| The radius is y, and the height is (X-x)

  27. anonymous
    • one year ago
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    thanks for helping btw i know its a pain

  28. Zale101
    • one year ago
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    |dw:1434061121877:dw| The big X is x=3 and the small x is y=x

  29. Zale101
    • one year ago
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    Makes sense? Hope I didn't confused you...

  30. anonymous
    • one year ago
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    ok yeah that makes sense.

  31. anonymous
    • one year ago
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    ok! so then the next step is the anti derivative of that?

  32. Zale101
    • one year ago
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    Yes.

  33. Zale101
    • one year ago
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    Use the fundamental theorem of calculus to find the definite integral. \[\Large \int\limits_{a}^{b}f(y)dy=F(b)-F(a)\]

  34. anonymous
    • one year ago
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    so 9pi or am i still not getting it?

  35. anonymous
    • one year ago
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    omg! ok thank you so much!!

  36. Zale101
    • one year ago
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    Np!

  37. courtneygraley009
    • one year ago
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    This calculator is designed to make the calculation of volumes easier without remembering the difficult formulas of volume. Try http://www.acalculator.com/volume-conversion-calculator.html

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