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).

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- amistre64

sounds like a torus...

- amistre64

need to figure out the solution to the system of equations to know what our "bounds" are gonna be...

- amistre64

i got an idea, but its just to early in the morning to get it right in me head.....

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## More answers

- amistre64

the bounds for the integral are gonna be [-1,(3/2)]

- anonymous

That is an odd region...is it the odd oval shape?

- amistre64

(S) y-y^2 dy = (1/2)y^2 -(1/3) y^3 |-1-3/2

- amistre64

its a funky shape fer sure :)

- anonymous

use circular ring method..

- amistre64

F(-1) = 5/6

- anonymous

That shape isn't circular...

- anonymous

i know the answer but i dont know the solution..

- amistre64

F(3/2) = 9/8

- anonymous

v=85.902 cubic units

- amistre64

your one step ahead of me then :) I am just trying to find the area of the shape

- anonymous

.plz help me..

- amistre64

for F(x) + get an area of 7/24....which could be totally wrong, but ill see :)

- anonymous

I don't understand exactly which region he is asking for...

- amistre64

G(x) = (S) y^2-3 dy = (1/3)y^3 -3y | -1,(3/2)

- 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

Why does this site suck at showing links...

- amistre64

the region for the area is between y=-1 and y=(3/2)

- 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

i solved the system of equations and got y=-1 and y=(3/2) as solutions

- anonymous

Yes i see those bounds

- amistre64

If I did it right; the area of F(x) = 7/24. Now I need to subtract area of G(x) right?

- amistre64

G(-1) = 8/3

- anonymous

So where exactly does the line x = -4 come in then

- amistre64

you spin the area around x=-4 to find the area of the torus shape

- anonymous

Oh that's the center

- amistre64

i hope its the center :)

- amistre64

G(3/2) = -27/8 if Im doing it right :)

- anonymous

where should i put the strip??is it horizontal or vertical?

- amistre64

area of G(x) = -145/24 ?
Horizontal I beleive

- anonymous

well it can't be

- amistre64

total area of shape is F(x) - G(x) = 152/24

- amistre64

how do we spin it? multiply by circumference of the circle?

- anonymous

integrate from 0 to 2*pi

- anonymous

i thought

- amistre64

I can find the area between curves alright, but after that I get lost :)

- amistre64

Area = 38/6 = 19/3...thats as far as I can reduce it; if its right

- amistre64

how do we integrate the spin? Do we need to find the domain of the area are anything?

- anonymous

I'm not exactly sure

- amistre64

lol.....comeon..your smarter than me for sure :)

- anonymous

i thought it was the integral from 0 to 2*pi, but that wouldn't make much sense in this case

- anonymous

What about using the ring method

- anonymous

That would probably just be much more difficult

- amistre64

not that versed in the ring method...

- amistre64

sounds like we'd integrate with respect to the radius....

- 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

gotta go back and find the bends then :)

- amistre64

one of those bounds is x=1/4

- amistre64

is that right? y-y^2 x'=1-2y=0 y=1/2. .5-.5^2 = .25 x=1/4 right?

- anonymous

My brain hurts lol. Okay so you're deriving y-y^2, solving for y, and plugging it back in?

- amistre64

y^2 -3 x'=2y=0 y=0 other bound is x=-3

- anonymous

Yeah that seems right

- amistre64

yep, only thing I can do :)

- amistre64

sp we got our bends at x=-3 and x=1/4....now what?

- amistre64

spin it at x=-4 somehow

- amistre64

each slice is spin from -3 to 1/4 in dx increments and then add them all up

- anonymous

Im confused on the area

- 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

Yeah that's what i did actually :)

- amistre64

which area :) the area between the curves?

- 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

Or i should say...to the right of the parabola

- 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

Ok i see

- amistre64

back in algebra here :) when y-y^2 = y^2-3 we get solutions to the equations :)

- amistre64

2y^2 -y -3 =0
(y-...)(y+....)

- 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

now, since we are talking about spinning a circle; do we (S) 2pi(x)(area)? dx

- 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

cant hurt to try:
(S) 2pi(x)(19/3) dx = 19pi(x^2)/3 | 1,4.25

- amistre64

114.39583333(pi) - 6.3333333333(pi)

- amistre64

I get 108.0625 pi....but am I right?

- amistre64

ill need alot more practice for this I guess :)

- amistre64

v=85.902 cubic units is what the person said to begin with; and that doesnt match mine

- 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

the key lies in the y axis. distance from the y axis bounds i think..not the x-axis

- anonymous

what is the integral of
1/((x+2)√(x+3))

- amistre64

im wrong somewhere :)

- amistre64

i need more practice on integrals... .

- amistre64

have you intergrated by parts?

- anonymous

yeah...but dunno if its correct

- amistre64

did you do it more than once?

- amistre64

(S) 1/x+2 dx = ln(x+2) that much works out

- amistre64

i gotta get some sleep :)....chat with ya later think...

- anonymous

okay amistre...btw u have taken u=1/x+2?

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