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anonymous

  • 4 years ago

Calculus II - Volume By Shells Hello, having some problems with a volume by shells problem. I believe my issue is the setup. Question: Use the shells method to find the volume of a solid generated by revolving the region bounded by the lines y=x+6 and y=x^2 about the following lines; a) The line x=3 b) The line x=-2 c) The x-axis d) The line y=9 So - I convert; y = x+6 x = y-6 y = x^2 x = sqrt(y) Rest coming in an equation post -

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  1. anonymous
    • 4 years ago
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    \[2\pi \int\limits_{0}^{9} (3-x)(\sqrt(y) - (y-6))\] Is this the correct setup?

  2. TuringTest
    • 4 years ago
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    for a ?

  3. anonymous
    • 4 years ago
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    Yeah, sorry.

  4. TuringTest
    • 4 years ago
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    well I would set it up the other way, in terms of x...

  5. anonymous
    • 4 years ago
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    That was a typo, I meant to have (3-y)(sqrt(y) etc, etc...

  6. TuringTest
    • 4 years ago
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    |dw:1327780091605:dw|\[2\pi\int_{-2}^{3}(3-x)[(x+6)-x^2]xdx\]is the way I see it easiest...

  7. anonymous
    • 4 years ago
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    what letter are you working on now?

  8. anonymous
    • 4 years ago
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    A (Turings post above)

  9. TuringTest
    • 4 years ago
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    doing shell method here in terms of y seems strange to me, because the area is bonded by y=x^2 on both sides between 0<y<2

  10. TuringTest
    • 4 years ago
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    bounded*

  11. TuringTest
    • 4 years ago
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    sorry I had an extra x earlier\[2\pi\int_{-2}^{3}(3-x)[(x+6)-x^2]dx\]

  12. anonymous
    • 4 years ago
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    Yeah, that sure does look a lot easier. Every example I have seen if you are rotating around X (axis or line) then you get the equation in terms of Y. That is the reason I convert them... I have a feeling I'm telling myself incorrect rules.

  13. anonymous
    • 4 years ago
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    Going to try working it out, brb!

  14. TuringTest
    • 4 years ago
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    |dw:1327780376562:dw|look at the picture and you can see that the cylinders have radius r=(3-x) and height h=x+6-x^2 I see no reason to integrate with respect to y.

  15. TuringTest
    • 4 years ago
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    |dw:1327780500851:dw|

  16. anonymous
    • 4 years ago
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    Yeah, you're right. I really need to get better at drawing/interpreting a graph on every problem. I've been sort of lazy with them up to this point.

  17. TuringTest
    • 4 years ago
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    Practice, practice :D

  18. TuringTest
    • 4 years ago
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    as for part b and c you will just change the radius... part d I have to look at though.

  19. anonymous
    • 4 years ago
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    Bam! Got it! Thanks a lot for the help! I'm going to try to do the second one by myself, although I already know it will be insanely easy since we already have the setup.

  20. anonymous
    • 4 years ago
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    Yeah. Cool, I'll get b/c and then work on d and hopefully be back with the right answer :) Thanks again, mate!

  21. anonymous
    • 4 years ago
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    Hrm, rotating about the X axis everything is the same, but the radius is X, right? I got B no problem, but I'm having problems with C.

  22. TuringTest
    • 4 years ago
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    Yes to your last question^ I think we have to split the integral in part d up because as I mentioned, the area is bound by y=x^2 on both sides between y=0 and y=4, so\[h=2\sqrt y,0\le y\le4\]\[h=\sqrt y-(y-6),4\le y\le9\]\[r=9-y\]|dw:1327780922685:dw|\[\int_{0}^{4}(9-y)(2\sqrt y)dy+\int_{4}^{9}(9-y)[\sqrt y-(y-6)]dy\]

  23. TuringTest
    • 4 years ago
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    oh sorry, "no" to your last question archetype you could do the same with x for the radius if we were going about the y-axis

  24. anonymous
    • 4 years ago
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    Ah, thanks Turing. That's what i was thinking, but then I saw you say yes and I was, well.. confused. So for C its; \[2\pi \int\limits_{0}^{9} y(\sqrt(y) - (y-6))\]

  25. TuringTest
    • 4 years ago
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    @phi what do you mean the part to the left is double counted ? you mean the negative x portion?

  26. TuringTest
    • 4 years ago
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    @archetype but between\[0\le y\le4\]the height will not be what you have

  27. TuringTest
    • 4 years ago
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    ok, but do what radius do you mean? along the y axis the radius has no 'left' portion, just positive (up)|dw:1327781599409:dw|If you are referring to an earlier problem I think we have confirmed the answer already, right archetype?

  28. TuringTest
    • 4 years ago
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    that is supposed to be about y=9 above*

  29. TuringTest
    • 4 years ago
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    that should be symmetric, but the drawing sucks|dw:1327781872267:dw|

  30. anonymous
    • 4 years ago
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    I've got an answer for A and B (both 625Pi/6). I do not yet have C or D.

  31. phi
    • 4 years ago
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    sorry I was doing part C @Tur you are doing D correct

  32. anonymous
    • 4 years ago
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    When looking at the graphs do you ALWAYS use the original equation? For some reason I'm looking at the graph of x=y-6 and x=sqrt(y) to solve part C, but something is telling me I'm looking at the wrong graph.

  33. TuringTest
    • 4 years ago
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    cool, ok so for C the only thing we are missing is that fact that the height formula for each shell depends which interval you are in\[h=2\sqrt y\]for the interval\[0\le y\le4\]and\[h=\sqrt y-(y-6)\]for the interval \[4\le y\le9\]look at the picture carefully to see this

  34. TuringTest
    • 4 years ago
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    |dw:1327782089205:dw|see how the line y=4 divides our region in two? the height below it is 2sqrty, above it is sqrty-(y-6)

  35. anonymous
    • 4 years ago
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    Awesome! I graphed the two things out myself and was going to ask if it was 2sqrt(y) because of the other line intersecting at that point.

  36. TuringTest
    • 4 years ago
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    good so the integral will be divided by the interval as well:\[2\pi\int_{0}^{4}y(2\sqrt y)dy+2\pi\int_{4}^{9}y[\sqrt y-(y-6)]dy\]

  37. anonymous
    • 4 years ago
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    I find this stuff so insanely difficult, haha. I guess like you said, practice, practice, practice. This is one of those days I wish I was a savant. Or just good at math. I'd take good at math.

  38. TuringTest
    • 4 years ago
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    I'd like to think I'm good at math, but really it's just a matter of familiarity with the ways to visualize things

  39. anonymous
    • 4 years ago
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    Yeah - and I think that is a lot of my problem. Not much else in school requires visualization like Volumes of Solids. At least nothing that I have taken.

  40. anonymous
    • 4 years ago
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    Drawing that graph and putting the picture in like you did above really helped me.

  41. TuringTest
    • 4 years ago
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    these are really good notes http://tutorial.math.lamar.edu/Classes/CalcI/VolumeWithCylinder.aspx they even teach you how to do things other than just revolutions here if you read more than just this section

  42. anonymous
    • 4 years ago
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    Cool! Thanks! Well I really appreciate all of the help. Going to try to get my final answer on these guys.

  43. TuringTest
    • 4 years ago
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    no prob good luck!

  44. anonymous
    • 4 years ago
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    Sweeet, got them both. And I feel like I came away with a far better understanding of the setup. Thanks again, man, it was really awesome of you to help me out (mainly since we've been working together for over an hour).

  45. TuringTest
    • 4 years ago
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    My pleasure :) congrads

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