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

- anonymous

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- schrodinger

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- anonymous

@freckles @ganeshie8 @Hero @nincompoop @Robert136 @satellite73

- anonymous

2, I think.

- anonymous

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

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## More answers

- triciaal

##### 1 Attachment

- 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 cansâ€™ worth of water are in the body of a 160-pound runner? What unprovided information do you need to answer this question?

- anonymous

wingspan move man ... dick move.

- triciaal

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

- 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

|dw:1439477813129:dw|

- anonymous

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

- triciaal

when y = 0 then x = -1

- anonymous

i don't really see how that's useful

- triciaal

did you see the graph?

- 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

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

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

- anonymous

@ParthKohli @OregonDuck any ideas ?

- 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

it says that the cross sections perpendicular to the x axis are squares, only 3d objects have cross sections

- triciaal

so we need the integral between -1 and 1

- anonymous

already did that 1.8856

- anonymous

@pooja195

- triciaal

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

- freckles

|dw:1439482917191:dw|

- freckles

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

- freckles

sorta

- freckles

|dw:1439483183181:dw|

- freckles

@jdosio do you see how to setup the integral
let me know when you get back on

- freckles

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

- freckles

and use that as the integrand

- 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

damn now your offline, ill be on for the remainder of the day @freckles tell me when you get back on

- 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

\[V=\int\limits_{-1}^1 A(x) dx \\ \text{ where } A(x)=\text{ area of square }\]
|dw:1439494770565:dw|

- freckles

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

- freckles

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

- freckles

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

- anonymous

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

- 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

|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

|dw:1439495315876:dw|

- freckles

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

- anonymous

!! i get it !!

- freckles

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

- 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

okay im getting 2

- freckles

sounds great
that is what @IrishBoy123 got using double integrals

- anonymous

thanks for your help freckles , cant thank you enough

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