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  1. TuringTest
    • 5 years ago
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    First we need the points A and B, know how to find that?

  2. ijlal
    • 5 years ago
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    yes i know that

  3. ijlal
    • 5 years ago
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    A has (1,5) and B(4,5)

  4. ijlal
    • 5 years ago
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    M(2,4)

  5. TuringTest
    • 5 years ago
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    right, so our integration will be from 1 to 4 what is the outer radius of each disk we will use in our method? what is the radius of the inner disk?

  6. ijlal
    • 5 years ago
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    i dont know how a cylinder is formed in this curve

  7. TuringTest
    • 5 years ago
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    we are making rings out of each section of area|dw:1327072712405:dw|the outer radius of each ring will be\[r_0=5\]the inner radius will be\[r_i=x+\frac{4}{x}\]the area of each disk wile therefor be\[\pi r_0^2-\pi r_i^2\to\pi(r_0^2-r_i^2)\]

  8. TuringTest
    • 5 years ago
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    |dw:1327073020023:dw|here is a cross-section of each ring|dw:1327073083149:dw|therefor adding up these rings we will integrate the areas of the rings from x=1 to x=4\[\pi\int_{1}^{4}r_0^2+r_i^2dx=\pi\int_{1}^{4}5^2-(x+\frac{4}{x})^2dx\]make sense?

  9. TuringTest
    • 5 years ago
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    *slight typo above, should be minus in the first integral (because we are subtracting the area of the inner disk from that of the outer)\[\pi\int_{1}^{4}r_0^2-r_i^2dx=\pi\int_{1}^{4}5^2-(x+\frac{4}{x})^2dx\]

  10. ijlal
    • 5 years ago
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    Yes the Volume of the ring is i guess 57pie

  11. TuringTest
    • 5 years ago
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    I guess, I haven't checked. I'll leave the integration for you to do. But your final answer is a volume of a shape, each section is a ring. (Just pointing out the discrepancy in calling the whole thing a ring)

  12. ijlal
    • 5 years ago
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    the answer of the question is 18pie i dont know how the cylinders height is 4units can you tell me that i will be done with my question done

  13. TuringTest
    • 5 years ago
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    which cylinder has a height of 4 units?

  14. ijlal
    • 5 years ago
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    there is only one cylinder formed

  15. TuringTest
    • 5 years ago
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    \[\pi\int_{1}^{4}r_0^2+r_i^2dx=\pi\int_{1}^{4}5^2-(x+\frac{4}{x})^2dx=\pi\int_{1}^{4}25-x^2-8-\frac{16}{x^2}dx\]\[=\pi(17x-\frac{x^3}{3}+\frac{16}{x})|_{1}^{4}\]\[=\pi[(68-\frac{16}{3}+4)-(17-\frac{1}{3}+16)]\]whatever that simplifies to... The whole shape will have a length of 3 units (from x=1 to x=4) if that is what you mean....

  16. TuringTest
    • 5 years ago
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    I got\[\pi\int_{1}^{4}r_0^2-r_i^2dx=\pi\int_{1}^{4}5^2-(x+\frac{4}{x})^2dx=\pi\int_{1}^{4}25-x^2-8-\frac{16}{x^2}dx\]\[=\pi(17x-\frac{x^3}{3}+\frac{16}{x})|_{1}^{4}\]\[=\pi[(68-\frac{16}{3}+4)-(17-\frac{1}{3}+16)]=32\pi\]and you say that you know the answer to be 18pi ?

  17. ijlal
    • 5 years ago
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    yes the answer is 18 pi it is given in the mark scheme of this question

  18. ijlal
    • 5 years ago
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    and this solution simplifies to 34pi not 32pi

  19. TuringTest
    • 5 years ago
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    Oh, I just made some silly arithmetic mistake apparently (this is why I left the integral for you to do) As you can see by wolfram it is set up right\[\pi\int_{1}^{4}r_0^2-r_i^2dx=\pi\int_{1}^{4}5^2-(x+\frac{4}{x})^2dx\] http://www.wolframalpha.com/input/?i=int%20from%201%20to%204%20(25-(x%2B4%2Fx)%5E2)dx&t=crmtb01 so you figure out where I messed up...

  20. ijlal
    • 5 years ago
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    the solution of this integral gives 90pi

  21. TuringTest
    • 5 years ago
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    no, look at the link I posted. Wolfram does no lie: http://www.wolframalpha.com/input/?i=pi*int+from+1+to+4+%285%5E2-%28x%2B4%2Fx%29%5E2%29dx there it is explicitly, clearly the answer to the integral I wrote is 18pi

  22. ijlal
    • 5 years ago
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    oh sorry i neglected the minus sign by mistake :P thank you :)

  23. TuringTest
    • 5 years ago
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    very welcome :) I apparently made a similar mistake trying to do it by eye, so it happens.

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