anonymous
  • anonymous
How to come up with the integral for the volume of the rotation of a region bounded by curves with reference to a vertical line?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
I understand the process of rotating a region with the x-axis and other lines parallel to it, but I keep having trouble formulating the integral of the equation representing the volume whenever the rotation is in reference with the y-axis and its meridians.
anonymous
  • anonymous
What I usually do in such case is that I integrate the function with respect to y
anonymous
  • anonymous
The region is bounded by the curves of:\[y=x ^{2/3}\]\[y=0\]\[x=0\]The rotation is in reference with the y-axis

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TuringTest
  • TuringTest
draw the picture
anonymous
  • anonymous
@TuringTest yes that's a very good idea :D
anonymous
  • anonymous
It helps a lot
TuringTest
  • TuringTest
thanks, it's the only one I ever use
anonymous
  • anonymous
then you identify the crosssectional area
anonymous
  • anonymous
|dw:1333616173853:dw| This a rough sketch of how the graphs should look
anonymous
  • anonymous
excuse be, may I ask is it really y=0 and x=0?
anonymous
  • anonymous
So my parameters for the integral would be the min and max of the intersection?
anonymous
  • anonymous
I'm sorry, the rotation is in reference with the line x=0 but the region is above the line y=0.
TuringTest
  • TuringTest
|dw:1333616352713:dw|can't be just those three bounds, what's the fourth? there is no intersection besides at the origin, unless you have omitted an eqn.
TuringTest
  • TuringTest
what about on the right side though?
anonymous
  • anonymous
I did once again I'm sorry the line I drew in the first drawing should be the line x=1 and the intersection is at (1,1).
anonymous
  • anonymous
ah ok
TuringTest
  • TuringTest
|dw:1333616595171:dw|now which axis are we going around?
anonymous
  • anonymous
so the cross sectional area A is pie(x^(2/3))^2
anonymous
  • anonymous
I understand that I need to rewrite all the equations in terms of x, that is having the x as dependent on the y var.
anonymous
  • anonymous
then we know we'll integrate from 0 to 1 so \[\int\limits_{0}^{1}\pi(x^{2/3})^2dx\]
anonymous
  • anonymous
The axis would be the y-axis.
anonymous
  • anonymous
that is if we are to rotate about the x axis
anonymous
  • anonymous
Don't I need to subtract that volume by the volume of the cylinder formed by the line x=1 with radius 1?
anonymous
  • anonymous
yes
TuringTest
  • TuringTest
that is one way the other is shell method
TuringTest
  • TuringTest
...but then you would integrate along y
anonymous
  • anonymous
@TuringTest hmmm.. can you tell something about this?
TuringTest
  • TuringTest
oh no, srry, thinking of going around x-axis again
anonymous
  • anonymous
I think I'm getting it after all, what I had to do since the very beginning was change the equation y=x^(2/3) to x=y^(3/2) and use that as the radius for the inner disk and 1 as the radius for the outer disk.
TuringTest
  • TuringTest
yeah that would work fine
TuringTest
  • TuringTest
actually, that's probably the easiest way as I said last time, more than one way to skin a cat
anonymous
  • anonymous
and from there proceed to integrate from 0 to to 1 since I'm 'slicing' this figure in horizontal sections.
anonymous
  • anonymous
Yeah, I'm having a test over this tomorrow and I need to know this by heart.
TuringTest
  • TuringTest
well, vertical sections|dw:1333617106925:dw|here's a visual for you of two of the rings you are talking about (the best I can do with what I'm given)
TuringTest
  • TuringTest
the inner radius is as you said: \(r_i=x=y^{3/2}\) and the outer is\(r_o=1\) the intesrsection is still \((1,1)\) so you can look at it as subtracting from the cylinder of radius 1 if you wish...
TuringTest
  • TuringTest
|dw:1333617404061:dw|
anonymous
  • anonymous
Yes I meant to say that I was slicing this 'cylinder' from left to right. Thanks for the clarification.
TuringTest
  • TuringTest
I think you'll do fine on your test I'm tired... going to sleep good night :)
anonymous
  • anonymous
Yeah thanks for all the help, good night.

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