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anonymous

  • one year ago

Use the shell method to find the volume of the solid generated by revolving the region bounded by the line y=x+2 and the parabola y=x^2 about the following line. y=4

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  1. Loser66
    • one year ago
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    draw it out, please

  2. Loser66
    • one year ago
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    |dw:1436218798400:dw|

  3. Loser66
    • one year ago
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    The intersection points:\(x^2 = x+2\) gives ust x =-1 and x =2 then the integral is defined as \[\int_{-1}^2( area~outer -area~inner)dx\\\pi* \int_{-1}^2 ((4-x^2)^2 -(4-(x+2))^2dx\]

  4. anonymous
    • one year ago
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    The answer is correct, but that is the washer method.

  5. Loser66
    • one year ago
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    oh, shell method!! ooooooooook

  6. Loser66
    • one year ago
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    I think we have to break it into 2 parts, the first one from -1 to 0 and the second one from 0 to 2 (w.r.t.x)

  7. dan815
    • one year ago
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    :)

  8. anonymous
    • one year ago
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    Clearly I don't know or I wouldn't be asking, but with the shell method aren't we taking it with respect to y. With that said, I've broken it down from 4-1 1-0 and still am not getting the correct answer

  9. Loser66
    • one year ago
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    |dw:1436221417680:dw|

  10. anonymous
    • one year ago
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    Got it

  11. xapproachesinfinity
    • one year ago
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    |dw:1436222834363:dw| the shape is like this i have some good drawing skills don't yo think hahah

  12. xapproachesinfinity
    • one year ago
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    |dw:1436222936259:dw|

  13. anonymous
    • one year ago
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    Thats why I don't bother with attempting to draw it online, I'm terrible. When I was taking the second integral between 1-0 I was doing y(2sqrty) I was leaving off the 4- for the radius of that one stupid simple mistakes

  14. anonymous
    • one year ago
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    Thank you though

  15. xapproachesinfinity
    • one year ago
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    you integrated with respect to x yes?

  16. anonymous
    • one year ago
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    dy

  17. xapproachesinfinity
    • one year ago
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    hmm why? the height of the cylinder goes in x direction

  18. anonymous
    • one year ago
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    correct this is the shell method

  19. xapproachesinfinity
    • one year ago
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    the shell method is the cylinder method can you show your work, how you integrated with dy

  20. anonymous
    • one year ago
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    \[\int\limits_{1}^{4}2\pi(4-y)(\sqrt{y}-(y-2)dy +\int\limits_{0}^{1}2\pi(4-y)(2\sqrt{y})dy\]

  21. xapproachesinfinity
    • one year ago
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    let me draw that nicely with a program! this drawing confuses me lol

  22. anonymous
    • one year ago
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    You can use the washer method with respect to X to get the same answer \[\int\limits_{-1}^{2}\pi((4-x)^2-(4-(x+2)))^2dx\]

  23. xapproachesinfinity
    • one year ago
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    why do you have 0 to 1?

  24. anonymous
    • one year ago
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    because you actually have 3 functions, but its "hidden" If you look at the graph at the point -1,1 you can no longer take the height using y-2. It changes to \[-\sqrt{y}. So \int\limits_{0}^{1} is taken with positive \sqrt{y} - -\sqrt{y}\]

  25. xapproachesinfinity
    • one year ago
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    sorry can't focus, i will check this later :)

  26. anonymous
    • one year ago
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    its correct the correct answer is 108pi/5

  27. freckles
    • one year ago
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    |dw:1436225174941:dw|

  28. freckles
    • one year ago
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    \[y=x+2 \\ x=y-2 \\ \text{ and } y=x^2 \\ \text{ gives } x=\sqrt{y} \text{ or } x=-\sqrt{y}\]

  29. freckles
    • one year ago
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    |dw:1436225286527:dw|

  30. freckles
    • one year ago
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    |dw:1436225325924:dw|

  31. freckles
    • one year ago
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    we have two difference things to look at because the left function switches at y=1

  32. anonymous
    • one year ago
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    we've gone through I'd like to close

  33. freckles
    • one year ago
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    like from y=0 to y=1 we have right function is x=sqrt(y) and left function x=-sqrt(y) and from y=1 to y=4 we have right function x=sqrt(y) and left function x=y-2

  34. freckles
    • one year ago
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    you can close it just trying to help anyone who is still trying to understand it

  35. anonymous
    • one year ago
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    Clicking on close and nothing is happening. Continue on if you want

  36. ybarrap
    • one year ago
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    Since this question is closed, I wanted to document for those interested the details of @Loser66's washer solution. Move the graphs down by 4 to make symmetric about the x-axis: |dw:1436229534071:dw| Setup and evaluate your integral: $$ \large{ \int_{-1}^2\int_{x-2}^{x^2-4}2\pi y~dy~dx\\ =\left.\begin{matrix}\pi\int_{-1}^2y^2 \\ \end{matrix}\right|^{x^2-4}_{x-2}~dx\\ =\pi\int_{-1}^2\left((x^2-4)^2-(x-2)^2\right)\\ =\pi\int_{-1}^2\left(x^4-9 x^2+4 x+12\right)~dx\\ =\pi \left(\frac{x^5}{5}-3 x^3+2 x^2+12 x\right)|^2_{-1}\\ } $$ Evaluation using wolfram gives (control+click ->): \(\href{http:///www.wolframalpha.com/input/?i=pi+*+integral++%28%28x%5E2-4%29%5E2+-+%28x-2%29%5E2%29+dx%2Cx%3D-1+to+x%3D2}{\cfrac{108\pi}{5}}\).

  37. ybarrap
    • one year ago
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    And... here is the solution using the shell method:\( \href{http:///www.wolframalpha.com/input/?i=2*pi*+%28integral+y*%28sqrt%284-y%29+-+%282-y%29+%29+dy%2Cy%3D0+to+y%3D3%29+%2B+%284*pi*integral+y*sqrt%284-y%29+++dy%2C+y%3D3+to+y%3D4%29}{\cfrac{108\pi}{5}\text{ (control+click me)}}. \) $$ ~~\!\bf\color{blue}{o}~\color{blue}{o}\\ ~~~\!\bullet \\ ~\smile \\ ~~\!\!/||\text{\\} $$

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