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