Find the volume of the solid created by the bounded region of (x=y-y^2) and (x=y^2 -3) when it is spun around the line (x=-4).

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- amistre64

- katieb

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- anonymous

Ah this question still o.O

- amistre64

Yeah :) but I found out what I was doing wrong.... other than trying to drink and derive :)

- anonymous

So we at least know you can add 4 to all terms to change the bounds and have it revolved around x = 0

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- anonymous

And it's a torus, right? So you need to integrate the area of the torus around the line in a full circle.

- anonymous

A cross section of the torus.

- amistre64

yep...

- anonymous

Can't you just integrate it?

- amistre64

its pi (S) [4+F(y)]^2 - [4+G(y)]^2 | -1, 3/2
I kept trying:
pi (S) [(4+F(y)) - (4+G(y))]^2 | -1, 3/2

- amistre64

math: yes, but couldnt figure out how to last night :)

- anonymous

\[x = y - y^2 + 4\]
\[x = y^2 + 1\]

- anonymous

Okay so the bounds are y = -3, y = 1/2

- anonymous

No...

- anonymous

What?

- amistre64

Krb: goog good. the bounds of integration are -1 and 3/2

- anonymous

Yeah those are the bounds ^

- anonymous

i need to figure out how to use the equations to find the solids of revolution..if this is x=-4, then the line we get is a vertical one right? so its been revolved upwards?

- amistre64

you got the equation right; now square them individually before you subtract them..

- anonymous

So \[\int\limits_{(-1)}^{(3/2)}(y - y^2 + 4) - (y^2 + 1)\]

- anonymous

That's what i would think

- anonymous

Krb you're always on this site. I like how when I look at paul's notes, you were the first one on open study.

- amistre64

think; the rotation is around the "y axis" so to speak

- anonymous

yeah..thats what i mean, so its like integrating the difference of squares of the equations right? and these equations signify -----?

- amistre64

Krb: getting closer. we need to integrate [F(y)]^2 - [G(x)]^2 in the bounds

- anonymous

Yes, normally when integrating with respect to x you are finding the are underneath the curve that drops to the x axis. So in this case you are finding the area "to the side of the curve" that drops to the y axis

- amistre64

Think; exactly :)

- anonymous

I'm confused as to why it would be squared

- amistre64

typo...[G(y)]^2

- anonymous

i guess it should be squared...shouldn't it?

- amistre64

Krb: if we do it seperately for each on, there is no real "area" so to speak until we combine them. Does that make sense?

- amistre64

pi* (S) F(y)^2 dy - (S) G(y)^2 dy
The equations ARE our Radiusessess

- anonymous

OH

- anonymous

if y1 & y2 are the curves then ,
\[\int\limits_{}^{}[y1^{2}-y2^{2}]\] gives the volume of the solid? where y1=upper curve & y2 the lower

- anonymous

multiplied by a pi

- anonymous

SO we are looking at it from a top down view, and then you need pi * r^2 for both "circles"

- amistre64

exactly

- anonymous

on efor inner and the other for outer 'circle' i guess

- amistre64

think: thats the gyst of it yeah

- anonymous

thnx!!

- anonymous

\[[\pi \int\limits_{-1}^{(3/2)}(y - y^2 + 4)^2] - [\pi \int\limits_{-1}^{(3/2)}(y^2 + 1)^2]\]

- amistre64

Krb: your a genius :)

- anonymous

No, you are :)

- anonymous

u both are :)

- amistre64

we should start a mensa club right here lol

- anonymous

the open study math mensa

- anonymous

:) :) so krb ur'e pursuing your masters too?

- amistre64

C1 - C2 is what we were missing last night.
2pi(R) - 2pi(r) does not equal 2pi(R-r)

- amistre64

it does, but it doesnt when you integrate lol

- anonymous

wer those constants that u'll missed out yesterday?

- anonymous

Ends up being 85.9029

- anonymous

think: I'm honestly not sure

- anonymous

krb: that's alright

- anonymous

i dunno why it gets posted twice always, when i click only once!

- anonymous

I think I probably would like to get a Masters

- anonymous

good

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