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 -

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- anonymous

\[2\pi \int\limits_{0}^{9} (3-x)(\sqrt(y) - (y-6))\]
Is this the correct setup?

- TuringTest

for a ?

- anonymous

Yeah, sorry.

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- TuringTest

well I would set it up the other way, in terms of x...

- anonymous

That was a typo, I meant to have (3-y)(sqrt(y) etc, etc...

- TuringTest

|dw:1327780091605:dw|\[2\pi\int_{-2}^{3}(3-x)[(x+6)-x^2]xdx\]is the way I see it easiest...

- anonymous

what letter are you working on now?

- anonymous

A (Turings post above)

- 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

bounded*

- TuringTest

sorry I had an extra x earlier\[2\pi\int_{-2}^{3}(3-x)[(x+6)-x^2]dx\]

- 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

Going to try working it out, brb!

- 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

|dw:1327780500851:dw|

- 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

Practice, practice :D

- TuringTest

as for part b and c you will just change the radius...
part d I have to look at though.

- 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

Yeah. Cool, I'll get b/c and then work on d and hopefully be back with the right answer :) Thanks again, mate!

- 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

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

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

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

@phi what do you mean the part to the left is double counted ?
you mean the negative x portion?

- TuringTest

@archetype
but between\[0\le y\le4\]the height will not be what you have

- 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

that is supposed to be about y=9 above*

- TuringTest

that should be symmetric, but the drawing sucks|dw:1327781872267:dw|

- anonymous

I've got an answer for A and B (both 625Pi/6). I do not yet have C or D.

- phi

sorry I was doing part C
@Tur you are doing D correct

- 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

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

|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

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

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

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

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

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

Drawing that graph and putting the picture in like you did above really helped me.

- 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

Cool! Thanks! Well I really appreciate all of the help. Going to try to get my final answer on these guys.

- TuringTest

no prob
good luck!

- 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

My pleasure :)
congrads

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