The base of a solid is bounded by the curve y = sqrt(x + 1), the x-axis and the line x = 1. The cross sections, taken perpendicular to the x-axis, are squares. Find the volume of the solid 1 2 2.333 None of these

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The base of a solid is bounded by the curve y = sqrt(x + 1), the x-axis and the line x = 1. The cross sections, taken perpendicular to the x-axis, are squares. Find the volume of the solid 1 2 2.333 None of these

Mathematics
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2, I think.
dont really want the answer as much as i want to know how to do it :P

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PLEASE HELP WILL FAN AND MEDAL According to an article in Runners' World magazine: On average the human body is more than 50 percent water [by weight]. Runners and other endurance athletes average around 60 percent. This equals about 120 soda cans' worth of water in a 160-pound runner! Investigate their calculation. Approximately how many soda cans’ worth of water are in the body of a 160-pound runner? What unprovided information do you need to answer this question?
wingspan move man ... dick move.
I think we need the area under the curve between 0 and 1 times the height
Is the region bounded by sqrt(x+1), y=0, and x=1 rotated around some line? Because without that, so far, you only have a 2D object....
|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
Never claimed to be good at math, but to me the question is missing, because as it is the very region you are dealing with is 2D.
pretty sure that its a complete question and that the object described is 3d
It is just bound by some (explicit) functions, and there is no reason for it to be 3D. Is it rotated about some line? You said it is not.
it says that the cross sections perpendicular to the x axis are squares, only 3d objects have cross sections
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|
@jdosio do you see how to setup the integral let me know when you get back on
in other words you need to find the area of the square
and use that as the integrand
+1 @freckles http://www.wolframalpha.com/input/?i=%5Cint_%7By%3D0%7D%5E%7B%5Csqrt%7B2%7D%7D+%5Cint_%7Bx%3Dy%5E2+-+1%7D%5E%7B1%7D+%28%5Csqrt%7Bx%2B1%7D%29+dx+dy
damn now your offline, ill be on for the remainder of the day @freckles tell me when you get back on
also thats the only part i have a problem with,(setting up the integral ) i know what the solid looked like and all just not sure how to calculate its volume
\[V=\int\limits_{-1}^1 A(x) dx \\ \text{ where } A(x)=\text{ area of square }\] |dw:1439494770565:dw|
\[A(x)=\text{ area of square}=\text{ height } ^2\]
the integrand is just x+1 that is A(x)=x+1
\[V=\int\limits_{-1}^1 (x+1) dx\]
isint it \[\sqrt{x+1}\]
that is the height you also have the base is that too the area of a square=base*height or height^2 since base=height
|dw:1439495278485:dw| if that length here is sqrt(x+1) then the base length of that shape is also sqrt(x+1) since the shape is a square
|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
\[V=\int\limits _{-1}^1 A(x) dx \\ A(x)=\sqrt{x+1} \cdot \sqrt{x+1}=x+1 \\ V=\int\limits_{-1}^{1} (x+1) dx \] V represents the volume by the way
okay im getting 2
sounds great that is what @IrishBoy123 got using double integrals
thanks for your help freckles , cant thank you enough

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