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NathanJHW

  • one year ago

The region R is bounded by the x-axis, y-axis, x = 3 and y = 1/sqrt(x+1).

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  1. NathanJHW
    • one year ago
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    A. Find the area of region R. B.Find the volume of the solid formed when the region R is revolved about the x-axis C. The solid formed in part B is divided into two solids of equal volume by a plane perpendicular to the x-axis. Find the x-value where this plane intersects the x-axis.

  2. perl
    • one year ago
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    first lets graph the region https://www.desmos.com/calculator/2bz6ctgloa

  3. NathanJHW
    • one year ago
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    Yeah I was able to do that much.

  4. perl
    • one year ago
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    so the area of the region R , we take the integral $$ \Large \int_{0}^{3} \frac{1}{\sqrt{x+1}}~dx$$

  5. NathanJHW
    • one year ago
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    I got 2

  6. perl
    • one year ago
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    thats correct so far

  7. NathanJHW
    • one year ago
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    So 2 is the answer to A then right?

  8. perl
    • one year ago
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    right

  9. NathanJHW
    • one year ago
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    Alright, then how do we do part B?

  10. perl
    • one year ago
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    part B we revolve about x axis $$ \Large \int_{0}^{3} \pi \left( \frac{1}{\sqrt{x+1}} \right)^2~dx$$

  11. NathanJHW
    • one year ago
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    pi ln (4)

  12. perl
    • one year ago
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    thats correct

  13. NathanJHW
    • one year ago
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    Now how do we do part c?

  14. perl
    • one year ago
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    we need to find the x point that divides the volume in half

  15. perl
    • one year ago
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    |dw:1433367290427:dw|

  16. perl
    • one year ago
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    you can't just cut it in half, because the region is not symmetric

  17. NathanJHW
    • one year ago
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    So then where do we cut it?

  18. perl
    • one year ago
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    solve for c $$\LARGE { \int_{0}^{\color{red}c} \pi \left( \frac{1}{\sqrt{x+1}} \right)^2~dx = \frac{1}{2}\cdot \pi \ln 4 } $$

  19. NathanJHW
    • one year ago
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    pi ln (c+1)

  20. NathanJHW
    • one year ago
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    Wait sorry I'm solving for c not integrating.

  21. NathanJHW
    • one year ago
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    I got ln|x+1| - ln|1| = ln|4| - ln|x+1|

  22. NathanJHW
    • one year ago
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    x=1

  23. perl
    • one year ago
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    right, so we need to solve ln| c + 1 | - ln| 0 + 1| = 1/2 * Pi * ln(4)

  24. NathanJHW
    • one year ago
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    is it not x/c =1?

  25. NathanJHW
    • one year ago
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    x/c means x or c as I put x when I solved.

  26. perl
    • one year ago
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    $$ \Large { \int_{0}^{c} \pi \left( \frac{1}{\sqrt{x+1}} \right)^2~dx \\~\\=\int_{0}^{c} \pi \left( \frac{1}{x+1} \right)~dx \\~\\ =\pi \left[ \ln |x + 1| \right]_{0}^{c} \\~\\ =\pi ( \ln|c+1| - \ln|0+1|) \\~\\ =\pi ( \ln|c+1| - \ln|1|) \\~\\ =\pi \ln|c+1| } $$

  27. perl
    • one year ago
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    ok so far?

  28. NathanJHW
    • one year ago
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    Yeah

  29. perl
    • one year ago
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    so we need to solve pi ln ( c + 1) = 1/2 * pi * ln(4) ln(c+1) = 1/2 * ln (4) ln(c+1) = ln(4 ^(1/2) ) ln(c+1) = ln(2) so c = 1

  30. NathanJHW
    • one year ago
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    Alright so I guess I was right, just needed to see the work.

  31. NathanJHW
    • one year ago
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    Thanks for the help.

  32. perl
    • one year ago
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    your welcome :)

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