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rainbow22 Group Title

find the volume of the solid formed by revolving the region bounded by y=x^2, y=0, and x=2 about the y axis

  • one year ago
  • one year ago

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  1. Thomas9 Group Title
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    Do you know about multivariable calculus perhaps? Or do you want the easy-not-understanding-what-you're-doing way

    • one year ago
  2. rainbow22 Group Title
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    The answer is 8pi

    • one year ago
  3. rainbow22 Group Title
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    No, I am supposed to use the Disk Washer method. I can do it with antiderive(2-y) to get the right answer but I'm pretty sure that's not following th formula I'm given which is outer radius minus inner radius.

    • one year ago
  4. Thomas9 Group Title
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    Disk Washer?

    • one year ago
  5. rainbow22 Group Title
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    There's a hole in the solid.. so it's called Washer Method.

    • one year ago
  6. Thomas9 Group Title
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    So how does this method work?

    • one year ago
  7. rainbow22 Group Title
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    I do antiderivative from a-b in terms of x or y (y in this case because I rotate about y axis) pi (outer radius)^2 - (inner radius)^2

    • one year ago
  8. rainbow22 Group Title
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    \[\int\limits_{a}^{b} \pi (outer radius)^2 -(innerradius)^2 dx\]

    • one year ago
  9. Thomas9 Group Title
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    I don't how that works, I'm afraid.

    • one year ago
  10. zepdrix Group Title
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    |dw:1359838432884:dw|

    • one year ago
  11. zepdrix Group Title
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    |dw:1359838561883:dw|So let's look at one slice.

    • one year ago
  12. zepdrix Group Title
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    Oh you're doing the washer/disk method, my mistake.. Lemme slice that differently.

    • one year ago
  13. zepdrix Group Title
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    |dw:1359838731527:dw|Ok this type of slice will give us a Disk.

    • one year ago
  14. zepdrix Group Title
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    |dw:1359838764034:dw|So we want to get the Volume of this disk.

    • one year ago
  15. zepdrix Group Title
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    The outer radius appears to be the line x=2. And the inner radius our function which is in terms of x, we'll need it in terms of y to integrate since we sliced in the y direction (dy thickness)

    • one year ago
  16. zepdrix Group Title
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    \[\large y=x^2 \qquad \rightarrow \qquad x=\sqrt y\]

    • one year ago
  17. zepdrix Group Title
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    \[\large V=\pi\left[(2)^2-(\sqrt y)^2\right]dy\]

    • one year ago
  18. zepdrix Group Title
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    Then to find the total volume in this enclosed area, we Integrate (add up all the slices) from one intersecting point to another. Where do they intersect? Ummm looks like.. y=0 and y=4?

    • one year ago
  19. zepdrix Group Title
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    Simplifying things down gives us, \[\large \pi \int\limits_0^4 4-y \;dy\]

    • one year ago
  20. zepdrix Group Title
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    Imma check my work real quick to make sure I didn't make a mistake somewhere. A little tired, it's possible. lol

    • one year ago
  21. zepdrix Group Title
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    Yah I think that's right. Confused about any of that? I went through it a little sloppy D:

    • one year ago
  22. rainbow22 Group Title
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    It's right. I realize what I did wrong! I had the wrong intersections/points. I did instead from -2 to 2.. not realizing that it was based on terms of y . thank you so much!

    • one year ago
  23. zepdrix Group Title
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    Oh cool c:

    • one year ago
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