dessyj1
  • dessyj1
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
Mathematics
katieb
  • katieb
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Loser66
  • Loser66
graph it first
dessyj1
  • dessyj1
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?
dessyj1
  • dessyj1
|dw:1434655395426:dw|

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Loser66
  • Loser66
Hey, volume or area?
Loser66
  • Loser66
if it is area, then easy but if it is volume, then you need a rotation axis.
dessyj1
  • dessyj1
First off, this is a hypothetical question i made up. And i want to find the volume of rev
Loser66
  • Loser66
What?? you made it up?? wow... @IrishBoy123
dessyj1
  • dessyj1
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.
dessyj1
  • dessyj1
should i just shift both functions upwards to avoid complications?
IrishBoy123
  • IrishBoy123
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?!
dessyj1
  • dessyj1
|dw:1434656246052:dw|
dessyj1
  • dessyj1
hmm
dessyj1
  • dessyj1
I do not know what to do from here on.
IrishBoy123
  • IrishBoy123
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.
dessyj1
  • dessyj1
Yes i do get your point.
dessyj1
  • dessyj1
I will shift both functions up by 2
dessyj1
  • dessyj1
then evaluate it.
dessyj1
  • dessyj1
would i use|dw:1434657740862:dw|
IrishBoy123
  • IrishBoy123
\(- x = - x^2 + 9 \) 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.
dessyj1
  • dessyj1
i restricted the space to 0 to 2
dessyj1
  • dessyj1
so i wouldnt have to deal with that.
IrishBoy123
  • IrishBoy123
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.
dessyj1
  • dessyj1
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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