amistre64
  • amistre64
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).
Mathematics
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SOLVED
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chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
Ah this question still o.O
amistre64
  • amistre64
Yeah :) but I found out what I was doing wrong.... other than trying to drink and derive :)
anonymous
  • 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
  • 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
  • anonymous
A cross section of the torus.
amistre64
  • amistre64
yep...
anonymous
  • anonymous
Can't you just integrate it?
amistre64
  • 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
  • amistre64
math: yes, but couldnt figure out how to last night :)
anonymous
  • anonymous
\[x = y - y^2 + 4\] \[x = y^2 + 1\]
anonymous
  • anonymous
Okay so the bounds are y = -3, y = 1/2
anonymous
  • anonymous
No...
anonymous
  • anonymous
What?
amistre64
  • amistre64
Krb: goog good. the bounds of integration are -1 and 3/2
anonymous
  • anonymous
Yeah those are the bounds ^
anonymous
  • 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
  • amistre64
you got the equation right; now square them individually before you subtract them..
anonymous
  • anonymous
So \[\int\limits_{(-1)}^{(3/2)}(y - y^2 + 4) - (y^2 + 1)\]
anonymous
  • anonymous
That's what i would think
anonymous
  • 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
  • amistre64
think; the rotation is around the "y axis" so to speak
anonymous
  • anonymous
yeah..thats what i mean, so its like integrating the difference of squares of the equations right? and these equations signify -----?
amistre64
  • amistre64
Krb: getting closer. we need to integrate [F(y)]^2 - [G(x)]^2 in the bounds
anonymous
  • 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
  • amistre64
Think; exactly :)
anonymous
  • anonymous
I'm confused as to why it would be squared
amistre64
  • amistre64
typo...[G(y)]^2
anonymous
  • anonymous
i guess it should be squared...shouldn't it?
amistre64
  • 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
  • amistre64
pi* (S) F(y)^2 dy - (S) G(y)^2 dy The equations ARE our Radiusessess
anonymous
  • anonymous
OH
anonymous
  • 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
  • anonymous
multiplied by a pi
anonymous
  • anonymous
SO we are looking at it from a top down view, and then you need pi * r^2 for both "circles"
amistre64
  • amistre64
exactly
anonymous
  • anonymous
on efor inner and the other for outer 'circle' i guess
amistre64
  • amistre64
think: thats the gyst of it yeah
anonymous
  • anonymous
thnx!!
anonymous
  • anonymous
\[[\pi \int\limits_{-1}^{(3/2)}(y - y^2 + 4)^2] - [\pi \int\limits_{-1}^{(3/2)}(y^2 + 1)^2]\]
amistre64
  • amistre64
Krb: your a genius :)
anonymous
  • anonymous
No, you are :)
anonymous
  • anonymous
u both are :)
amistre64
  • amistre64
we should start a mensa club right here lol
anonymous
  • anonymous
the open study math mensa
anonymous
  • anonymous
:) :) so krb ur'e pursuing your masters too?
amistre64
  • amistre64
C1 - C2 is what we were missing last night. 2pi(R) - 2pi(r) does not equal 2pi(R-r)
amistre64
  • amistre64
it does, but it doesnt when you integrate lol
anonymous
  • anonymous
wer those constants that u'll missed out yesterday?
anonymous
  • anonymous
Ends up being 85.9029
anonymous
  • anonymous
think: I'm honestly not sure
anonymous
  • anonymous
krb: that's alright
anonymous
  • anonymous
i dunno why it gets posted twice always, when i click only once!
anonymous
  • anonymous
I think I probably would like to get a Masters
anonymous
  • anonymous
good

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