anonymous
  • anonymous
Calculus II - Volume By Shells Hello, having some problems with a volume by shells problem. I believe my issue is the setup. Question: Use the shells method to find the volume of a solid generated by revolving the region bounded by the lines y=x+6 and y=x^2 about the following lines; a) The line x=3 b) The line x=-2 c) The x-axis d) The line y=9 So - I convert; y = x+6 x = y-6 y = x^2 x = sqrt(y) Rest coming in an equation post -
Mathematics
  • Stacey Warren - Expert brainly.com
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katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
\[2\pi \int\limits_{0}^{9} (3-x)(\sqrt(y) - (y-6))\] Is this the correct setup?
TuringTest
  • TuringTest
for a ?
anonymous
  • anonymous
Yeah, sorry.

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TuringTest
  • TuringTest
well I would set it up the other way, in terms of x...
anonymous
  • anonymous
That was a typo, I meant to have (3-y)(sqrt(y) etc, etc...
TuringTest
  • TuringTest
|dw:1327780091605:dw|\[2\pi\int_{-2}^{3}(3-x)[(x+6)-x^2]xdx\]is the way I see it easiest...
anonymous
  • anonymous
what letter are you working on now?
anonymous
  • anonymous
A (Turings post above)
TuringTest
  • TuringTest
doing shell method here in terms of y seems strange to me, because the area is bonded by y=x^2 on both sides between 0
TuringTest
  • TuringTest
bounded*
TuringTest
  • TuringTest
sorry I had an extra x earlier\[2\pi\int_{-2}^{3}(3-x)[(x+6)-x^2]dx\]
anonymous
  • anonymous
Yeah, that sure does look a lot easier. Every example I have seen if you are rotating around X (axis or line) then you get the equation in terms of Y. That is the reason I convert them... I have a feeling I'm telling myself incorrect rules.
anonymous
  • anonymous
Going to try working it out, brb!
TuringTest
  • TuringTest
|dw:1327780376562:dw|look at the picture and you can see that the cylinders have radius r=(3-x) and height h=x+6-x^2 I see no reason to integrate with respect to y.
TuringTest
  • TuringTest
|dw:1327780500851:dw|
anonymous
  • anonymous
Yeah, you're right. I really need to get better at drawing/interpreting a graph on every problem. I've been sort of lazy with them up to this point.
TuringTest
  • TuringTest
Practice, practice :D
TuringTest
  • TuringTest
as for part b and c you will just change the radius... part d I have to look at though.
anonymous
  • anonymous
Bam! Got it! Thanks a lot for the help! I'm going to try to do the second one by myself, although I already know it will be insanely easy since we already have the setup.
anonymous
  • anonymous
Yeah. Cool, I'll get b/c and then work on d and hopefully be back with the right answer :) Thanks again, mate!
anonymous
  • anonymous
Hrm, rotating about the X axis everything is the same, but the radius is X, right? I got B no problem, but I'm having problems with C.
TuringTest
  • TuringTest
Yes to your last question^ I think we have to split the integral in part d up because as I mentioned, the area is bound by y=x^2 on both sides between y=0 and y=4, so\[h=2\sqrt y,0\le y\le4\]\[h=\sqrt y-(y-6),4\le y\le9\]\[r=9-y\]|dw:1327780922685:dw|\[\int_{0}^{4}(9-y)(2\sqrt y)dy+\int_{4}^{9}(9-y)[\sqrt y-(y-6)]dy\]
TuringTest
  • TuringTest
oh sorry, "no" to your last question archetype you could do the same with x for the radius if we were going about the y-axis
anonymous
  • anonymous
Ah, thanks Turing. That's what i was thinking, but then I saw you say yes and I was, well.. confused. So for C its; \[2\pi \int\limits_{0}^{9} y(\sqrt(y) - (y-6))\]
TuringTest
  • TuringTest
@phi what do you mean the part to the left is double counted ? you mean the negative x portion?
TuringTest
  • TuringTest
@archetype but between\[0\le y\le4\]the height will not be what you have
TuringTest
  • TuringTest
ok, but do what radius do you mean? along the y axis the radius has no 'left' portion, just positive (up)|dw:1327781599409:dw|If you are referring to an earlier problem I think we have confirmed the answer already, right archetype?
TuringTest
  • TuringTest
that is supposed to be about y=9 above*
TuringTest
  • TuringTest
that should be symmetric, but the drawing sucks|dw:1327781872267:dw|
anonymous
  • anonymous
I've got an answer for A and B (both 625Pi/6). I do not yet have C or D.
phi
  • phi
sorry I was doing part C @Tur you are doing D correct
anonymous
  • anonymous
When looking at the graphs do you ALWAYS use the original equation? For some reason I'm looking at the graph of x=y-6 and x=sqrt(y) to solve part C, but something is telling me I'm looking at the wrong graph.
TuringTest
  • TuringTest
cool, ok so for C the only thing we are missing is that fact that the height formula for each shell depends which interval you are in\[h=2\sqrt y\]for the interval\[0\le y\le4\]and\[h=\sqrt y-(y-6)\]for the interval \[4\le y\le9\]look at the picture carefully to see this
TuringTest
  • TuringTest
|dw:1327782089205:dw|see how the line y=4 divides our region in two? the height below it is 2sqrty, above it is sqrty-(y-6)
anonymous
  • anonymous
Awesome! I graphed the two things out myself and was going to ask if it was 2sqrt(y) because of the other line intersecting at that point.
TuringTest
  • TuringTest
good so the integral will be divided by the interval as well:\[2\pi\int_{0}^{4}y(2\sqrt y)dy+2\pi\int_{4}^{9}y[\sqrt y-(y-6)]dy\]
anonymous
  • anonymous
I find this stuff so insanely difficult, haha. I guess like you said, practice, practice, practice. This is one of those days I wish I was a savant. Or just good at math. I'd take good at math.
TuringTest
  • TuringTest
I'd like to think I'm good at math, but really it's just a matter of familiarity with the ways to visualize things
anonymous
  • anonymous
Yeah - and I think that is a lot of my problem. Not much else in school requires visualization like Volumes of Solids. At least nothing that I have taken.
anonymous
  • anonymous
Drawing that graph and putting the picture in like you did above really helped me.
TuringTest
  • TuringTest
these are really good notes http://tutorial.math.lamar.edu/Classes/CalcI/VolumeWithCylinder.aspx they even teach you how to do things other than just revolutions here if you read more than just this section
anonymous
  • anonymous
Cool! Thanks! Well I really appreciate all of the help. Going to try to get my final answer on these guys.
TuringTest
  • TuringTest
no prob good luck!
anonymous
  • anonymous
Sweeet, got them both. And I feel like I came away with a far better understanding of the setup. Thanks again, man, it was really awesome of you to help me out (mainly since we've been working together for over an hour).
TuringTest
  • TuringTest
My pleasure :) congrads

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