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).

- anonymous

- Stacey Warren - Expert brainly.com

Hey! We 've verified this expert answer for you, click below to unlock the details :)

- schrodinger

I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!

- 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.....

Looking for something else?

Not the answer you are looking for? Search for more explanations.

## 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?

Looking for something else?

Not the answer you are looking for? Search for more explanations.