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agentx5

  • 2 years ago

Is this setup correct? Q: "Use the method of cylindrical shells to find the volume V of the solid obtained by rotating the region bounded by the given curves about the x-axis." x = 1 + y\(^2\), x = 0, y = 1, y = 3 \[\huge\int\limits_0^{\sqrt{2}} (2 \pi)(y)((3) - (1+y^2)) dy\] y=3 is the right function x=1+y\(^2\) is the left function right - left Graph below:

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  1. agentx5
    • 2 years ago
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    |dw:1341360986379:dw| Look good @KingGeorge ? :-)

  2. KingGeorge
    • 2 years ago
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    I think you switched some things around. First of all, the lines you drew for y=1,3 are vertical lines when they should be horizontal. In short, check the area you want integrate again.

  3. agentx5
    • 2 years ago
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    Wait a minute I'm getting zero when I evaluate...

  4. agentx5
    • 2 years ago
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    Oh... I think believe I see what I did there, let's try that graph again: |dw:1341361482202:dw|

  5. KingGeorge
    • 2 years ago
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    That looks better. So your new integrand and bounds would be...?

  6. agentx5
    • 2 years ago
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    \[\huge\int\limits_1^3 (2 \pi)(y)((3) - (1+y^2)) dy = -\frac{26}{3}\] The shells are growing out from the x-axis but I'm still missing something, a radius change perhaps? But which one? +1 or -1?

  7. agentx5
    • 2 years ago
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    Seems like 1-y is logical for a shift upward, the others evaluate negative

  8. KingGeorge
    • 2 years ago
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    The only change I would make is in the integrand again. The right side is now \((1+y)^2\) and the left is 0. Hence, you should have \[\huge\int\limits_1^3 (2 \pi)(y)( (1+y^2)-0) dy \]

  9. agentx5
    • 2 years ago
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    \[\large \int\limits_1^3 (2 \pi)(y)( (1+y^2)-0) dy =^? \int\limits_1^3 (2 \pi)(1-y)((3) - (1+y^2)) dy = \frac{44}{3}\]

  10. agentx5
    • 2 years ago
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    No wait the first is 44/3 and the second is 44\(\pi\)/3

  11. KingGeorge
    • 2 years ago
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    I think I forgot a close parentheses. \[\huge\int\limits_1^3 (2 \pi)(y)( (1+y)^2) dy\]

  12. agentx5
    • 2 years ago
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    Just noticed it ran off the page too, sry about breaking margins >_<

  13. agentx5
    • 2 years ago
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    \[\huge\int\limits_1^3 (2π)(y)((1+y)^2)dy = \frac{248\pi }{3}\] Seems logical, let me give it a try...

  14. agentx5
    • 2 years ago
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    Alas no, it is incorrect

  15. agentx5
    • 2 years ago
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    Sheesh these particular problems are being, well, problems tonight :D

  16. agentx5
    • 2 years ago
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    *looks back over steps*

  17. KingGeorge
    • 2 years ago
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    Wait, I misread the question. There shouldn't be a close parentheses there XD Have you already tried it without the close parentheses that I added?

  18. agentx5
    • 2 years ago
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    (2π)(y)((1+y)^2)dy <-- literally what I started working with

  19. agentx5
    • 2 years ago
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    Oh it's on the wrong side!

  20. KingGeorge
    • 2 years ago
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    How about \[\large\int\limits_1^3 (2 \pi)(y)( 1+y^2) dy?\]

  21. agentx5
    • 2 years ago
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    That's a quadratic now by mistake

  22. agentx5
    • 2 years ago
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    And 48π is perfectly correct, awesome!!!

  23. agentx5
    • 2 years ago
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    Comes out very nice in the end, no fractions

  24. KingGeorge
    • 2 years ago
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    Awesome. Now I just need to read the problems correctly.

  25. agentx5
    • 2 years ago
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    Me too lol

  26. agentx5
    • 2 years ago
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    What am I thinking, making vertical lines when it asked for horizontal lol

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