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anonymous

  • 5 years ago

if you are given the terms y = sqrt(x), y = 0, and x = 6, how do you find the volume of the solid generated by revolving it about the y-axis?

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  1. amistre64
    • 5 years ago
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    you can shell it

  2. amistre64
    • 5 years ago
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    2pi {S} x[sqrt(x)] dx ; [0,6]

  3. anonymous
    • 5 years ago
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    yeah, that's how i did it

  4. anonymous
    • 5 years ago
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    but the answer is a decimal: 35.27...

  5. amistre64
    • 5 years ago
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    x[sqrt(x)] can be re written as: sqrt(x^3) = x^(3/2) and integrate

  6. amistre64
    • 5 years ago
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    it gonna be a decimal regardless; pi is irrational/transcendental

  7. anonymous
    • 5 years ago
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    I am using a math homework software that won't take decimals for this answer. Should I just do it by hand?

  8. anonymous
    • 5 years ago
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    like have the fractional, un-evaluated part as and answer?

  9. amistre64
    • 5 years ago
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    if it dont take decimals, then use the exact version of it...

  10. anonymous
    • 5 years ago
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    I will try right now

  11. amistre64
    • 5 years ago
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    2x^(5/2) -------- * 2pi at x = 6 5

  12. anonymous
    • 5 years ago
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    that's the answer, but I need to evaluate at 6. and the computer won't take my decimal answer

  13. amistre64
    • 5 years ago
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    4.pi. sqrt(6.6.6.6.6) = 4.pi.36 sqrt(6) ---------------- ------------ 5 5

  14. amistre64
    • 5 years ago
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    144pi sqrt(6) ----------- should be an exact answer then 5

  15. anonymous
    • 5 years ago
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    you can evalutate it that way???

  16. amistre64
    • 5 years ago
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    yeah; thats what it equate to right?

  17. anonymous
    • 5 years ago
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    yes, and it takes that answer! Thanks everyone!

  18. amistre64
    • 5 years ago
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    :) just glad I was right lol :)

  19. watchmath
    • 5 years ago
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    Let me just use the washer method for comparison \(\int_0^{\sqrt{6}}\pi(36-y^4)\,dy=\pi(36y-\frac{y^5}{5})\mid_0^\sqrt{6}=\pi(36\sqrt{6}-\frac{36\sqrt{6}}{5})=\frac{144\pi\sqrt{6}}{5} \)

  20. anonymous
    • 5 years ago
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    okay, so I can see how you can do that a little more mathematically

  21. anonymous
    • 5 years ago
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    :)

  22. anonymous
    • 5 years ago
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    Thanks!

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