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dessyj1

  • one year ago

Volume of revolution. How would I find the volume of revolution for a space between two curves y=-x^2 +9 y=-x from x=0 to x=2

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  1. Loser66
    • one year ago
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    graph it first

  2. dessyj1
    • one year ago
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    Already did, i would apply the formula but the space between the two curves is crossing the x-axis. What if the question asked for the vol of rev about the x-axis? How would i do that?

  3. dessyj1
    • one year ago
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    |dw:1434655395426:dw|

  4. Loser66
    • one year ago
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    Hey, volume or area?

  5. Loser66
    • one year ago
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    if it is area, then easy but if it is volume, then you need a rotation axis.

  6. dessyj1
    • one year ago
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    First off, this is a hypothetical question i made up. And i want to find the volume of rev

  7. Loser66
    • one year ago
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    What?? you made it up?? wow... @IrishBoy123

  8. dessyj1
    • one year ago
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    Usually, when we are asked to do a volume of rev in class, the space between the two curves are always in the 1st quadrant. I want to know if the procedure would change if the space were to be in the the 1st and the 4th simultaneously.

  9. dessyj1
    • one year ago
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    should i just shift both functions upwards to avoid complications?

  10. IrishBoy123
    • one year ago
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    you can only revolve this around y-axis whilst retaining any sense to the whole thing, ie whilst getting what might be described as a volume as @Loser66 said, you should draw it. that's the best advice by far. i can see only arrows on what you posted. not the curves and their intersections. when you see the actual **area** that you are revolving to form a **volume**, i am sure you will see the conundrum. you could revolve your area about any axis you want and get a mathematical answer but it wouldn't be a volume, would it?!

  11. dessyj1
    • one year ago
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    |dw:1434656246052:dw|

  12. dessyj1
    • one year ago
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    hmm

  13. dessyj1
    • one year ago
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    I do not know what to do from here on.

  14. IrishBoy123
    • one year ago
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    you do get the point, though? "solid" you can spin your enclosed area round the y axis and you will have a "solid" that has some real physical meaning. spin it around the x axis and it is just mathematical navel gazing. there's absolutely nothing wrong with that BTW but the answer is physically meaningless. so in your "design your own problem" mode, shift something up a bit so that the rev about x means something too |dw:1434656911930:dw| use a transform that gets that intercept onto the x axis.

  15. dessyj1
    • one year ago
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    Yes i do get your point.

  16. dessyj1
    • one year ago
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    I will shift both functions up by 2

  17. dessyj1
    • one year ago
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    then evaluate it.

  18. dessyj1
    • one year ago
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    would i use|dw:1434657740862:dw|

  19. IrishBoy123
    • one year ago
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    \(- x = - x^2 + 9 \) gives some really ugly arithmetic: \( x^2- x - 9= 0 \) the roots are: \(\frac {1 \pm \sqrt{1 - 4(-9) }}{2} = \frac{1\pm \sqrt{37}}{2} \) so the root in Q4 must be \( ( \frac{1+ \sqrt{37}}{2}, \frac{-1- \sqrt{37}}{2} )\) ouch, ouch and triple ouch.

  20. dessyj1
    • one year ago
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    i restricted the space to 0 to 2

  21. dessyj1
    • one year ago
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    so i wouldnt have to deal with that.

  22. IrishBoy123
    • one year ago
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    in terms of solving it, that is yet another matter. people here talk about wings and nut and washers. that is again totally fine if that is how you are learning it. i was shown how to do this in a very different way, building little integration elements and going from there to a full integration. that approach makes it impossible to miss the cone at the base of your shape. you only really need to integration one curve.

  23. dessyj1
    • one year ago
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    I understand the method of finding the intersections ans using that as your limits but i really just wanted to understand the concept. This way saves more time.

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