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2, I think.

dont really want the answer as much as i want to know how to do it :P

wingspan move man ... dick move.

I think we need the area under the curve between 0 and 1 times the height

|dw:1439477813129:dw|

its not rotated its |dw:1439477867812:dw| this is what it should loo like

when y = 0 then x = -1

i don't really see how that's useful

did you see the graph?

yes i have a ti-84 iv graphed it , i have a table of values, don't see the use to it though

pretty sure that its a complete question and that the object described is 3d

@ParthKohli @OregonDuck any ideas ?

so we need the integral between -1 and 1

already did that 1.8856

we have half a parabola that opens right
guess none of the above

|dw:1439482917191:dw|

|dw:1439483111182:dw|
I'm bad about drawing 3d objects but this is what the shape looks like

sorta

|dw:1439483183181:dw|

in other words you need to find the area of the square

and use that as the integrand

\[A(x)=\text{ area of square}=\text{ height } ^2\]

the integrand is just x+1
that is A(x)=x+1

\[V=\int\limits_{-1}^1 (x+1) dx\]

isint it \[\sqrt{x+1}\]

|dw:1439495315876:dw|

\[A(x)=\sqrt{x+1} \cdot \sqrt{x+1}=(x+1)\]

!! i get it !!

notice the lower limit is x=-1
and the upper limit is x=1

okay im getting 2

sounds great
that is what @IrishBoy123 got using double integrals

thanks for your help freckles , cant thank you enough