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anonymous

  • one year ago

Find the volume of the solid formed by rotating the region inside the first quadrant enclosed by y=x^3,y=16x about the x-axis.

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  1. SolomonZelman
    • one year ago
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    your boundaries are \(y=x^3\) and \(y=16x\) and it is rotated about the x,axis correct?

  2. SolomonZelman
    • one year ago
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    x-axis *

  3. anonymous
    • one year ago
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    yes

  4. SolomonZelman
    • one year ago
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    that will look like a washer method|dw:1437943945723:dw|

  5. SolomonZelman
    • one year ago
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    do you know the washer method?

  6. SolomonZelman
    • one year ago
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    (don't be ahsamed not to know, if you don't know it is ok. the only person i can not help is the silent one)

  7. anonymous
    • one year ago
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    i do but have trouble with the set up

  8. SolomonZelman
    • one year ago
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    ok, lets figure the limits of integration together I will name them differently for our convinience \(f(x)=16x\) \(g(x)=x^3\)

  9. SolomonZelman
    • one year ago
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    lets find the points where these two functions intersect

  10. SolomonZelman
    • one year ago
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    16x=x³ 16=x² (and disregard negative x solutions because we are only in the 1st quadrant)

  11. SolomonZelman
    • one year ago
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    x=?

  12. anonymous
    • one year ago
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    4

  13. SolomonZelman
    • one year ago
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    yes, and one more?

  14. anonymous
    • one year ago
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    0

  15. SolomonZelman
    • one year ago
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    yes

  16. SolomonZelman
    • one year ago
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    So our limits of integration (for the region bounded by 16x and x³) is (x=0 and x=4) \ (i.e. the intersection points after which the region stops)

  17. SolomonZelman
    • one year ago
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    So far we know \(\large\color{slate}{\displaystyle\int\limits_{0}^{4}}\)

  18. anonymous
    • one year ago
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    okay got it and hw to determine inner vs outer for the formula

  19. SolomonZelman
    • one year ago
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    yes, and if you were to graph the two functions \(f(x)=16x\) and \(g(x)=x^3\) then for all values on the interval [0,4] (which is the interval we need) it is true that f(x)≥g(x)

  20. SolomonZelman
    • one year ago
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    So, g(x) is always below the f(x) in our case. right?

  21. SolomonZelman
    • one year ago
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    (you can test it using any value between 0 and 4)

  22. anonymous
    • one year ago
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    okay great thanks , makes more sense now , you were a big help!!

  23. SolomonZelman
    • one year ago
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    you are thinking me already, we aren;t done yet, aren we?

  24. SolomonZelman
    • one year ago
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    thanking*

  25. anonymous
    • one year ago
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    i just struggle with the set up I just worked out that practice problem and solved it form there

  26. SolomonZelman
    • one year ago
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    ok, but, just to make sure can you show me your integral that you will set up?

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