anonymous
  • anonymous
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
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
@freckles @ganeshie8 @Hero @nincompoop @Robert136 @satellite73
anonymous
  • anonymous
2, I think.
anonymous
  • anonymous
dont really want the answer as much as i want to know how to do it :P

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triciaal
  • triciaal
1 Attachment
rds8701
  • rds8701
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?
anonymous
  • anonymous
wingspan move man ... dick move.
triciaal
  • triciaal
I think we need the area under the curve between 0 and 1 times the height
SolomonZelman
  • SolomonZelman
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....
triciaal
  • triciaal
|dw:1439477813129:dw|
anonymous
  • anonymous
its not rotated its |dw:1439477867812:dw| this is what it should loo like
triciaal
  • triciaal
when y = 0 then x = -1
anonymous
  • anonymous
i don't really see how that's useful
triciaal
  • triciaal
did you see the graph?
anonymous
  • anonymous
yes i have a ti-84 iv graphed it , i have a table of values, don't see the use to it though
SolomonZelman
  • SolomonZelman
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.
anonymous
  • anonymous
pretty sure that its a complete question and that the object described is 3d
anonymous
  • anonymous
@ParthKohli @OregonDuck any ideas ?
SolomonZelman
  • SolomonZelman
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.
anonymous
  • anonymous
it says that the cross sections perpendicular to the x axis are squares, only 3d objects have cross sections
triciaal
  • triciaal
so we need the integral between -1 and 1
anonymous
  • anonymous
already did that 1.8856
anonymous
  • anonymous
@pooja195
triciaal
  • triciaal
we have half a parabola that opens right guess none of the above
freckles
  • freckles
|dw:1439482917191:dw|
freckles
  • freckles
|dw:1439483111182:dw| I'm bad about drawing 3d objects but this is what the shape looks like
freckles
  • freckles
sorta
freckles
  • freckles
|dw:1439483183181:dw|
freckles
  • freckles
@jdosio do you see how to setup the integral let me know when you get back on
freckles
  • freckles
in other words you need to find the area of the square
freckles
  • freckles
and use that as the integrand
IrishBoy123
  • IrishBoy123
+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
anonymous
  • anonymous
damn now your offline, ill be on for the remainder of the day @freckles tell me when you get back on
anonymous
  • anonymous
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
freckles
  • freckles
\[V=\int\limits_{-1}^1 A(x) dx \\ \text{ where } A(x)=\text{ area of square }\] |dw:1439494770565:dw|
freckles
  • freckles
\[A(x)=\text{ area of square}=\text{ height } ^2\]
freckles
  • freckles
the integrand is just x+1 that is A(x)=x+1
freckles
  • freckles
\[V=\int\limits_{-1}^1 (x+1) dx\]
anonymous
  • anonymous
isint it \[\sqrt{x+1}\]
freckles
  • freckles
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
freckles
  • freckles
|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
freckles
  • freckles
|dw:1439495315876:dw|
freckles
  • freckles
\[A(x)=\sqrt{x+1} \cdot \sqrt{x+1}=(x+1)\]
anonymous
  • anonymous
!! i get it !!
freckles
  • freckles
notice the lower limit is x=-1 and the upper limit is x=1
freckles
  • freckles
\[V=\int\limits _{-1}^1 A(x) dx \\ A(x)=\sqrt{x+1} \cdot \sqrt{x+1}=x+1 \\ V=\int\limits_{-1}^{1} (x+1) dx \] V represents the volume by the way
anonymous
  • anonymous
okay im getting 2
freckles
  • freckles
sounds great that is what @IrishBoy123 got using double integrals
anonymous
  • anonymous
thanks for your help freckles , cant thank you enough

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