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CoconutJJ

  • one year ago

The bounded area by the graph y=sqrt(x), y=x-4 and the x axis is rotated along the x axis to form a solid. Use Integration By Shells to find the Volume of the Solid.

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  1. misssunshinexxoxo
    • one year ago
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    This might help utilize the "Shell" method; https://www.khanacademy.org/math/integral-calculus/solid_revolution_topic/solid_of_revolution/v/solid-of-revolution-part-5

  2. CoconutJJ
    • one year ago
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    |dw:1435377786774:dw| This is the bounded area

  3. CoconutJJ
    • one year ago
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    Wait so how does the integral change when you rotate it around the x axis instead of y?

  4. misssunshinexxoxo
    • one year ago
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    https://www.wyzant.com/resources/lessons/math/calculus/integration/finding_volume

  5. misssunshinexxoxo
    • one year ago
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    With "washer method"

  6. misssunshinexxoxo
    • one year ago
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    To rotate about the y-axis, swap the roles of x and y throughout the above discussion. To rotate about a different axis, change coordinates so that the axis of rotation is one of the coordinate axes, and then do the computation.

  7. misssunshinexxoxo
    • one year ago
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    Follow rules of "shell" and "washer" method to guide throughout finding volume

  8. anonymous
    • one year ago
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    been a while, and I was never great at these so take this with a grain of salt. I believe you have to change the formulas around and integrate in terms of y. |dw:1435378239112:dw|

  9. anonymous
    • one year ago
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    and then the limits will be the y coordinates of the intersection points

  10. CoconutJJ
    • one year ago
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    so then the thickness of the cylinder is dy?

  11. CoconutJJ
    • one year ago
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    |dw:1435378522268:dw|

  12. UsukiDoll
    • one year ago
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    @peachpi it's been a while for me too.

  13. misssunshinexxoxo
    • one year ago
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    Peachpi seems absolutely right; this is a tough area

  14. anonymous
    • one year ago
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    this is a solid. so there's no hole, right?

  15. anonymous
    • one year ago
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    the area lies on the axis it's being rotated on

  16. UsukiDoll
    • one year ago
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    but eventually we have to integrate this thing... who is going to be a and who is going to be b on the integral symbol ?

  17. CoconutJJ
    • one year ago
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    well they intersect when x=6.562

  18. anonymous
    • one year ago
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    yeah, but the integral is in y. a is 0. b you solve y² = y + 4 to get

  19. CoconutJJ
    • one year ago
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    How is the integral with respect to y ?

  20. UsukiDoll
    • one year ago
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    maybe it has something to do with the x axis being rotated so we can't use that? I have no idea x.x

  21. UsukiDoll
    • one year ago
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    oh I see where the a and b for the integral is |dw:1435378968903:dw|

  22. UsukiDoll
    • one year ago
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    those intersection points are the values for a and b

  23. CoconutJJ
    • one year ago
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    |dw:1435378996664:dw|

  24. UsukiDoll
    • one year ago
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    well a = 0 is correct.. but for the b we need to find the value of that intersection point but as @peachpi mentioned we have to solve " y² = y + 4 to get"

  25. anonymous
    • one year ago
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    I think it is simpler to do it horizontally than vertically, which would require two integrals.

  26. UsukiDoll
    • one year ago
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    hmmm y^2-y-4 = 0 and then factor ? OH NO! I DO NOT WANT TO DO 2 INTEGRALS

  27. UsukiDoll
    • one year ago
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    (-1)^2-4(1)(-4) = 1-4(-4) = 1+ 16 = 17 and it's not a perfect square.. dahhhh

  28. anonymous
    • one year ago
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    What is the point of intersection?

  29. CoconutJJ
    • one year ago
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    |dw:1435379165728:dw|

  30. UsukiDoll
    • one year ago
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    you graph the two equations first. and then they should have common points.

  31. CoconutJJ
    • one year ago
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    |dw:1435379200509:dw|

  32. UsukiDoll
    • one year ago
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    with 0 being one of them

  33. anonymous
    • one year ago
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    they're not nice numbers. about (6.56, 2.56)

  34. anonymous
    • one year ago
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    |dw:1435379227291:dw|

  35. CoconutJJ
    • one year ago
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    So if we use the shell method from 0 to 4, we are basically using the disc method?

  36. UsukiDoll
    • one year ago
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    double integrals.. double the anti derivatives :S

  37. anonymous
    • one year ago
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    |dw:1435379315446:dw|

  38. anonymous
    • one year ago
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    \[ \int dV = 2\pi\int rh~dy = 2\pi\int \limits_0^{y_1} y(f(y)-g(y))~dy \]

  39. CoconutJJ
    • one year ago
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    THANK YOU! I think I can do it from here now :)

  40. anonymous
    • one year ago
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    The volume of a cylinder is: \[ V = \pi r^2h \]When you differentiate, it is: \[ \frac{dV}{dr} = 2\pi r h \implies \int dV = 2\pi\int rh~dr \]That's how I think about it.

  41. CoconutJJ
    • one year ago
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    Thankks

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