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anonymous
 4 years ago
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 xaxis
d) The line y=9
So  I convert;
y = x+6
x = y6
y = x^2
x = sqrt(y)
Rest coming in an equation post 
anonymous
 4 years ago
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 xaxis d) The line y=9 So  I convert; y = x+6 x = y6 y = x^2 x = sqrt(y) Rest coming in an equation post 

This Question is Closed

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[2\pi \int\limits_{0}^{9} (3x)(\sqrt(y)  (y6))\] Is this the correct setup?

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.1well I would set it up the other way, in terms of x...

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0That was a typo, I meant to have (3y)(sqrt(y) etc, etc...

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.1dw:1327780091605:dw\[2\pi\int_{2}^{3}(3x)[(x+6)x^2]xdx\]is the way I see it easiest...

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0what letter are you working on now?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0A (Turings post above)

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.1doing 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<y<2

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.1sorry I had an extra x earlier\[2\pi\int_{2}^{3}(3x)[(x+6)x^2]dx\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Yeah, 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
 4 years ago
Best ResponseYou've already chosen the best response.0Going to try working it out, brb!

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.1dw:1327780376562:dwlook at the picture and you can see that the cylinders have radius r=(3x) and height h=x+6x^2 I see no reason to integrate with respect to y.

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.1dw:1327780500851:dw

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Yeah, 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
 4 years ago
Best ResponseYou've already chosen the best response.1Practice, practice :D

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.1as for part b and c you will just change the radius... part d I have to look at though.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Bam! 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
 4 years ago
Best ResponseYou've already chosen the best response.0Yeah. Cool, I'll get b/c and then work on d and hopefully be back with the right answer :) Thanks again, mate!

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Hrm, 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
 4 years ago
Best ResponseYou've already chosen the best response.1Yes 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\[h=2\sqrt y,0\le y\le4\]\[h=\sqrt y(y6),4\le y\le9\]\[r=9y\]dw:1327780922685:dw\[\int_{0}^{4}(9y)(2\sqrt y)dy+\int_{4}^{9}(9y)[\sqrt y(y6)]dy\]

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.1oh sorry, "no" to your last question archetype you could do the same with x for the radius if we were going about the yaxis

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Ah, thanks Turing. That's what i was thinking, but then I saw you say yes and I was, well.. confused. So for C its; \[2\pi \int\limits_{0}^{9} y(\sqrt(y)  (y6))\]

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.1@phi what do you mean the part to the left is double counted ? you mean the negative x portion?

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.1@archetype but between\[0\le y\le4\]the height will not be what you have

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.1ok, but do what radius do you mean? along the y axis the radius has no 'left' portion, just positive (up)dw:1327781599409:dwIf you are referring to an earlier problem I think we have confirmed the answer already, right archetype?

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.1that is supposed to be about y=9 above*

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.1that should be symmetric, but the drawing sucksdw:1327781872267:dw

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I've got an answer for A and B (both 625Pi/6). I do not yet have C or D.

phi
 4 years ago
Best ResponseYou've already chosen the best response.0sorry I was doing part C @Tur you are doing D correct

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0When looking at the graphs do you ALWAYS use the original equation? For some reason I'm looking at the graph of x=y6 and x=sqrt(y) to solve part C, but something is telling me I'm looking at the wrong graph.

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.1cool, 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\[h=\sqrt y(y6)\]for the interval \[4\le y\le9\]look at the picture carefully to see this

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.1dw:1327782089205:dwsee how the line y=4 divides our region in two? the height below it is 2sqrty, above it is sqrty(y6)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Awesome! 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
 4 years ago
Best ResponseYou've already chosen the best response.1good so the integral will be divided by the interval as well:\[2\pi\int_{0}^{4}y(2\sqrt y)dy+2\pi\int_{4}^{9}y[\sqrt y(y6)]dy\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I 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
 4 years ago
Best ResponseYou've already chosen the best response.1I'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
 4 years ago
Best ResponseYou've already chosen the best response.0Yeah  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
 4 years ago
Best ResponseYou've already chosen the best response.0Drawing that graph and putting the picture in like you did above really helped me.

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.1these 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
 4 years ago
Best ResponseYou've already chosen the best response.0Cool! Thanks! Well I really appreciate all of the help. Going to try to get my final answer on these guys.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Sweeet, 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
 4 years ago
Best ResponseYou've already chosen the best response.1My pleasure :) congrads
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