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  • 4 years ago

Calculus II - Volume of Solids - Setup Question \[y = 4x-x^2\] \[y=2-x\] Let D be the region bound by the curves above. Setup an integral representing the volume of the solid obtained by revolving R about each of the following lines; a) the x-axis b) the line y=5 c) the line y = -1 d) the line x = -1 e) the line x = 3 I have worked all five using the "washer method" and would like to get some feedback on the answers; a) \[\pi \int\limits_{-1}^{2} (4-x^2)^2 - (2-x)^2 \] b) \[\pi \int\limits_{-1}^{2} (2-x)^2 - (5-(4-x^2))^2 \] c) \[\pi \int\limits_{-1}^{2} (2+x)^2 - (1+(4-x^2))^2 \]

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  1. anonymous
    • 4 years ago
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    d) \[\pi \int\limits_{-1}^{2} (2-x)^2 - (1+(4-x^2))^2 \] e) \[\pi \int\limits_{-1}^{2} (2-x)^2 - (3-(4-x^2))^2 \]

  2. amistre64
    • 4 years ago
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    dx's

  3. amistre64
    • 4 years ago
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    im assuming its the area bound by the intersection of these functions

  4. amistre64
    • 4 years ago
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    washer method you have to make sure your integrating and "adjusting" according to the axis of roataion as well

  5. amistre64
    • 4 years ago
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    a function viewed from the x axis doesnt look the same as a function viewed from the y axis; it becomes its inverse function

  6. anonymous
    • 4 years ago
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    Yes, bound by the intersection. Shoot, adjusting. I also tried this for D (shell method); d) \[2\pi \int\limits_{-1}^{2} (x+1)[(4-x^2)-(2-x)]\]

  7. anonymous
    • 4 years ago
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    So anytime I am rotating about the Y-axis I need to adjust my equations to be in terms of Y?

  8. anonymous
    • 4 years ago
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    Using the washer method, that is.

  9. amistre64
    • 4 years ago
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    adjust your equations according to axis and method of ...sheel, disc, washer etc

  10. amistre64
    • 4 years ago
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    |dw:1327958145327:dw|

  11. amistre64
    • 4 years ago
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    R = f(x) r = g(x) pi (R^2 - r^2) = washer area integrated with the x axis

  12. amistre64
    • 4 years ago
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    washer is just a double disc

  13. TuringTest
    • 4 years ago
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    I meant shell*

  14. TuringTest
    • 4 years ago
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    shell about x integrate w/respect to x disk about x integrate w/respect to y and vice-versa

  15. amistre64
    • 4 years ago
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    thats better :)

  16. amistre64
    • 4 years ago
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    and look into your function from the way the method is looking at it; if its a radius define it from the radial axis

  17. anonymous
    • 4 years ago
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    Ok, so if you rotate around y axis R = f(y) r = g(y)

  18. amistre64
    • 4 years ago
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    if its a heaight, define it from the base axis

  19. amistre64
    • 4 years ago
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    f-1(y) but yes

  20. amistre64
    • 4 years ago
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    inverse when you are looking at it from the other axis that the equation is defined for

  21. amistre64
    • 4 years ago
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    y=x^2 looking at it from the yaxis is x=sqrt(y)

  22. amistre64
    • 4 years ago
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    |dw:1327958415125:dw|

  23. anonymous
    • 4 years ago
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    So I am assuming A is correct, but for B; \[\pi \int\limits_{-1}^{2}(y-2)^2 - (5-(\sqrt(4-y)))^2\] ?

  24. amistre64
    • 4 years ago
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    dont forget the dy if thats the case

  25. anonymous
    • 4 years ago
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    But tacking a dy onto the end its correct? :)

  26. amistre64
    • 4 years ago
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    y=4x−x^2 y=2−x ---------- -2+5x-x^2 = 0 x^2 -5x +2 = 0 x = 5/2 +- sqrt(17)/2 you sure your intercepts are right? Let D be the region bound by the curves above. Setup an integral representing the volume of the solid obtained by revolving R about each of the following lines; a) the x-axis

  27. amistre64
    • 4 years ago
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    http://www.wolframalpha.com/input/?i=4x%E2%88%92x%5E2%3D2-x

  28. anonymous
    • 4 years ago
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    We're actually given a picture to go along with the assignment; |dw:1327958708347:dw|

  29. amistre64
    • 4 years ago
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    that picture doesnt match your equations that you posted

  30. anonymous
    • 4 years ago
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    Ugh. I attached an X. y = 4 + x^2 Really sorry about that, but it shouldn't have effected much other then making everything wrong :)

  31. amistre64
    • 4 years ago
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    lol, its right everywhere except where its wrong :)

  32. anonymous
    • 4 years ago
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    Hahah

  33. amistre64
    • 4 years ago
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    better, -1,3 and 2,0 are going to be our "limits" or at least we can keep them in mind

  34. amistre64
    • 4 years ago
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    you want washer rotate about the x axis for a \[\pi\int_{-1}^{2}\ (4-x^2)^2-(2-x)^2 dx\] do you know what happens when you get the top and bottom in the wrong place?

  35. anonymous
    • 4 years ago
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    Negative answer?

  36. amistre64
    • 4 years ago
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    yep, so does position matter too much?

  37. anonymous
    • 4 years ago
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    Nope! :) Of course, some of the grade will be setting the equation up properly, I'm sure. But I see your point!

  38. amistre64
    • 4 years ago
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    good, cause all the texts I see waste so much time trying to tell you to set up top on top and bottom on bottom or else the sun will explode or some nonesense. the difference between a 2 and a -2 is just drop the negative :)

  39. anonymous
    • 4 years ago
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    Hahaha, exploding sun.

  40. anonymous
    • 4 years ago
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    So A and B are looking good, except for dx and dy respectively? Then for C I need to change to F(y) and D E should look good?

  41. amistre64
    • 4 years ago
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    b) the line y=5 this is the same as the xaxis; xaxis IS y=0 y=5 is just another parallel xaxis; what I do is subtract 5 from the eqs to set it back up to y=0

  42. amistre64
    • 4 years ago
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    y = 4-x^2 -5 -5 ---------- y = -1-x^2 same eq, different position

  43. amistre64
    • 4 years ago
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    y =2-x -5 -5 ------- y = -3-x

  44. anonymous
    • 4 years ago
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    Damn, that's a great idea. So that prevents you from having to subtract it from 5. If I had not done that it is correct, though?

  45. anonymous
    • 4 years ago
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    Can you do that with any Y or X = ? question. Just get the equation to F(y) or F(x) and add the x = or y = ?

  46. amistre64
    • 4 years ago
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    you swapped top and bottom on your B and forgot to -5 from one of the eqs

  47. amistre64
    • 4 years ago
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    yes, but with left and right shifts you have to modify the (x) perse say x^2 is measuered from x=5 instead of x=0 to move x^2 to the left by 5 we do: (x+5)^2

  48. amistre64
    • 4 years ago
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    it just about knowing how to move the graph around to a good vantage point

  49. anonymous
    • 4 years ago
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    Yeah, I am TERRIBLE at visualizing/setting up these graphs. Trying to get better, which is why I am on here constantly asking for setup help. Everytime I leave this website though I feel more and more confident.

  50. amistre64
    • 4 years ago
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    so B, washered about y=5; our integration hasnt shifted left and right so x to x is still good; -1 to 2 \[\pi \int_{-1}^{2} (4-x^2-5)^2-(2-x-5)^2 dx\]

  51. amistre64
    • 4 years ago
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    c) the line y = -1 this is the same thing; but +1 to move things to the right spot

  52. amistre64
    • 4 years ago
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    \pi \int_{-1}^{2} (4-x^2+1)^2-(2-x+1)^2 dx

  53. anonymous
    • 4 years ago
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    Now when it is X = -1 we will need to get it in terms of Y and then add one?

  54. amistre64
    • 4 years ago
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    forgot me delimiters :)

  55. amistre64
    • 4 years ago
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    one thing at time ...

  56. anonymous
    • 4 years ago
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    Haha, k. Trying to apply that principle to the next questions.

  57. amistre64
    • 4 years ago
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    \[B=\pi \int_{-1}^{2} (4-x^2+1)^2-(2-x+1)^2 dx\]

  58. amistre64
    • 4 years ago
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    that was C lol, but its a fancy looking mistake so its ok

  59. anonymous
    • 4 years ago
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    Beautiful, got the same thing written on my notepad in front of me.

  60. amistre64
    • 4 years ago
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    d) the line x = -1 our radiuseses are now look from the y axis out so we have to invert the equations to proper perspectives

  61. anonymous
    • 4 years ago
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    I've got to go afk for 15 minutes (moving from Library to Work) but I will be right back! If you are not here, thank you SO MUCH! You have given me a MUCH greater understanding of these concepts and I truly feel like i could tackle almost any washer problem with minimal difficulty!

  62. anonymous
    • 4 years ago
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    BRB, sir!

  63. amistre64
    • 4 years ago
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    y = 4 - x^2 y-4 =x^2 4-y = x^2 sqrt(4-y) = x y = 2-x y-2 = -x 2-y = x ok; and im sure they want inner outer set up "correctly" as well

  64. amistre64
    • 4 years ago
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    but yes, to adjust the eqs for x=-1; we +1 to our eqs now

  65. anonymous
    • 4 years ago
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    And then make sure the inner and outer equations are right. When doing inner/outer with respect to Y I assume we have to regraph the new equations and evaluate which is top and bottom?

  66. anonymous
    • 4 years ago
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    ahhh, now brb! Going to be late :)

  67. amistre64
    • 4 years ago
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    left and right; but yes

  68. amistre64
    • 4 years ago
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    the graphs dont change; we just see them from the y axis instead of the x axis

  69. amistre64
    • 4 years ago
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    of cours ethe math doesnt care if you use and x or y; just as long as your consistent

  70. amistre64
    • 4 years ago
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    for D about x=-1 \[\pi\int_{y=0}^{y=3}(\sqrt{4-y}+1)^2-(2-y+1)^2dy\]

  71. amistre64
    • 4 years ago
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    my ride is ready to leave, so thanks you for the time and good luck :)

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