anonymous
  • anonymous
find the volume of the solid generated by revolving the line x=-4.The region bounded by the two parabolas x=(y-y^2),x=(y^2-3).
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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amistre64
  • amistre64
sounds like a torus...
amistre64
  • amistre64
need to figure out the solution to the system of equations to know what our "bounds" are gonna be...
amistre64
  • amistre64
i got an idea, but its just to early in the morning to get it right in me head.....

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amistre64
  • amistre64
the bounds for the integral are gonna be [-1,(3/2)]
anonymous
  • anonymous
That is an odd region...is it the odd oval shape?
amistre64
  • amistre64
(S) y-y^2 dy = (1/2)y^2 -(1/3) y^3 |-1-3/2
amistre64
  • amistre64
its a funky shape fer sure :)
anonymous
  • anonymous
use circular ring method..
amistre64
  • amistre64
F(-1) = 5/6
anonymous
  • anonymous
That shape isn't circular...
anonymous
  • anonymous
i know the answer but i dont know the solution..
amistre64
  • amistre64
F(3/2) = 9/8
anonymous
  • anonymous
v=85.902 cubic units
amistre64
  • amistre64
your one step ahead of me then :) I am just trying to find the area of the shape
anonymous
  • anonymous
.plz help me..
amistre64
  • amistre64
for F(x) + get an area of 7/24....which could be totally wrong, but ill see :)
anonymous
  • anonymous
I don't understand exactly which region he is asking for...
amistre64
  • amistre64
G(x) = (S) y^2-3 dy = (1/3)y^3 -3y | -1,(3/2)
anonymous
  • anonymous
These are the graphs http://www.wolframalpha.com/input/?i=x+%3D+y-y^2%2C+x%3D+y^2-3%2C+x%3D-4
anonymous
  • anonymous
Why does this site suck at showing links...
amistre64
  • amistre64
the region for the area is between y=-1 and y=(3/2)
anonymous
  • anonymous
So it's the odd oval shape in the center http://www.wolframalpha.com/input/?i=x+%3D+y-y^2%2C+x%3D+y^2-3
amistre64
  • amistre64
i solved the system of equations and got y=-1 and y=(3/2) as solutions
anonymous
  • anonymous
Yes i see those bounds
amistre64
  • amistre64
If I did it right; the area of F(x) = 7/24. Now I need to subtract area of G(x) right?
amistre64
  • amistre64
G(-1) = 8/3
anonymous
  • anonymous
So where exactly does the line x = -4 come in then
amistre64
  • amistre64
you spin the area around x=-4 to find the area of the torus shape
anonymous
  • anonymous
Oh that's the center
amistre64
  • amistre64
i hope its the center :)
amistre64
  • amistre64
G(3/2) = -27/8 if Im doing it right :)
anonymous
  • anonymous
where should i put the strip??is it horizontal or vertical?
amistre64
  • amistre64
area of G(x) = -145/24 ? Horizontal I beleive
anonymous
  • anonymous
well it can't be
amistre64
  • amistre64
total area of shape is F(x) - G(x) = 152/24
amistre64
  • amistre64
how do we spin it? multiply by circumference of the circle?
anonymous
  • anonymous
integrate from 0 to 2*pi
anonymous
  • anonymous
i thought
amistre64
  • amistre64
I can find the area between curves alright, but after that I get lost :)
amistre64
  • amistre64
Area = 38/6 = 19/3...thats as far as I can reduce it; if its right
amistre64
  • amistre64
how do we integrate the spin? Do we need to find the domain of the area are anything?
anonymous
  • anonymous
I'm not exactly sure
amistre64
  • amistre64
lol.....comeon..your smarter than me for sure :)
anonymous
  • anonymous
i thought it was the integral from 0 to 2*pi, but that wouldn't make much sense in this case
anonymous
  • anonymous
What about using the ring method
anonymous
  • anonymous
That would probably just be much more difficult
amistre64
  • amistre64
not that versed in the ring method...
amistre64
  • amistre64
sounds like we'd integrate with respect to the radius....
anonymous
  • anonymous
yes the inner radius would be the maximum for the first equation, and the outer radius would be the maximum for the 2nd i believe
amistre64
  • amistre64
gotta go back and find the bends then :)
amistre64
  • amistre64
one of those bounds is x=1/4
amistre64
  • amistre64
is that right? y-y^2 x'=1-2y=0 y=1/2. .5-.5^2 = .25 x=1/4 right?
anonymous
  • anonymous
My brain hurts lol. Okay so you're deriving y-y^2, solving for y, and plugging it back in?
amistre64
  • amistre64
y^2 -3 x'=2y=0 y=0 other bound is x=-3
anonymous
  • anonymous
Yeah that seems right
amistre64
  • amistre64
yep, only thing I can do :)
amistre64
  • amistre64
sp we got our bends at x=-3 and x=1/4....now what?
amistre64
  • amistre64
spin it at x=-4 somehow
amistre64
  • amistre64
each slice is spin from -3 to 1/4 in dx increments and then add them all up
anonymous
  • anonymous
Im confused on the area
amistre64
  • amistre64
make it easier and spin add 4 to x so that we have the middle at 0 and our limits at 1 and the other at 4.25
anonymous
  • anonymous
Yeah that's what i did actually :)
amistre64
  • amistre64
which area :) the area between the curves?
anonymous
  • anonymous
yes, the integral of x = y-y^2 would give the area under the parabola, but how would you remove the sections on the sides to get the torus
anonymous
  • anonymous
Or i should say...to the right of the parabola
amistre64
  • amistre64
the two equations meet at the points -1 and 3/2; so these are the bounds we use. to find the area right?
anonymous
  • anonymous
Ok i see
amistre64
  • amistre64
back in algebra here :) when y-y^2 = y^2-3 we get solutions to the equations :)
amistre64
  • amistre64
2y^2 -y -3 =0 (y-...)(y+....)
amistre64
  • amistre64
I use that to find the area of F(x) and G(x) then: F(x)-G(x) gives us the area we need
amistre64
  • amistre64
now, since we are talking about spinning a circle; do we (S) 2pi(x)(area)? dx
amistre64
  • amistre64
it should be the same concept as finding the area between curves; use the lower and upper bounds of x and subtract them?
amistre64
  • amistre64
cant hurt to try: (S) 2pi(x)(19/3) dx = 19pi(x^2)/3 | 1,4.25
amistre64
  • amistre64
114.39583333(pi) - 6.3333333333(pi)
amistre64
  • amistre64
I get 108.0625 pi....but am I right?
amistre64
  • amistre64
ill need alot more practice for this I guess :)
amistre64
  • amistre64
v=85.902 cubic units is what the person said to begin with; and that doesnt match mine
anonymous
  • anonymous
Yeah it doesn't, my brain is too fried for this right now, I've got diff eq in 5 hours, best of luck though
amistre64
  • amistre64
the key lies in the y axis. distance from the y axis bounds i think..not the x-axis
anonymous
  • anonymous
what is the integral of 1/((x+2)√(x+3))
amistre64
  • amistre64
im wrong somewhere :)
amistre64
  • amistre64
i need more practice on integrals... .
amistre64
  • amistre64
have you intergrated by parts?
anonymous
  • anonymous
yeah...but dunno if its correct
amistre64
  • amistre64
did you do it more than once?
amistre64
  • amistre64
(S) 1/x+2 dx = ln(x+2) that much works out
amistre64
  • amistre64
i gotta get some sleep :)....chat with ya later think...
anonymous
  • anonymous
okay amistre...btw u have taken u=1/x+2?

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