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

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@freckles @ganeshie8 @Hero @nincompoop @Robert136 @satellite73

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dont really want the answer as much as i want to know how to do it :P

rds8701
 one year ago
Best ResponseYou've already chosen the best response.0PLEASE 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 160pound runner! Investigate their calculation. Approximately how many soda cans’ worth of water are in the body of a 160pound runner? What unprovided information do you need to answer this question?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0wingspan move man ... wingspan move.

triciaal
 one year ago
Best ResponseYou've already chosen the best response.0I think we need the area under the curve between 0 and 1 times the height

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.0Is 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
 one year ago
Best ResponseYou've already chosen the best response.0dw:1439477813129:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0its not rotated its dw:1439477867812:dw this is what it should loo like

triciaal
 one year ago
Best ResponseYou've already chosen the best response.0when y = 0 then x = 1

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i don't really see how that's useful

triciaal
 one year ago
Best ResponseYou've already chosen the best response.0did you see the graph?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yes i have a ti84 iv graphed it , i have a table of values, don't see the use to it though

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.0Never 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
 one year ago
Best ResponseYou've already chosen the best response.0pretty sure that its a complete question and that the object described is 3d

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@ParthKohli @OregonDuck any ideas ?

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.0It 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
 one year ago
Best ResponseYou've already chosen the best response.0it says that the cross sections perpendicular to the x axis are squares, only 3d objects have cross sections

triciaal
 one year ago
Best ResponseYou've already chosen the best response.0so we need the integral between 1 and 1

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0already did that 1.8856

triciaal
 one year ago
Best ResponseYou've already chosen the best response.0we have half a parabola that opens right guess none of the above

freckles
 one year ago
Best ResponseYou've already chosen the best response.2dw:1439482917191:dw

freckles
 one year ago
Best ResponseYou've already chosen the best response.2dw:1439483111182:dw I'm bad about drawing 3d objects but this is what the shape looks like

freckles
 one year ago
Best ResponseYou've already chosen the best response.2dw:1439483183181:dw

freckles
 one year ago
Best ResponseYou've already chosen the best response.2@jdosio do you see how to setup the integral let me know when you get back on

freckles
 one year ago
Best ResponseYou've already chosen the best response.2in other words you need to find the area of the square

freckles
 one year ago
Best ResponseYou've already chosen the best response.2and use that as the integrand

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0damn now your offline, ill be on for the remainder of the day @freckles tell me when you get back on

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0also 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
 one year ago
Best ResponseYou've already chosen the best response.2\[V=\int\limits_{1}^1 A(x) dx \\ \text{ where } A(x)=\text{ area of square }\] dw:1439494770565:dw

freckles
 one year ago
Best ResponseYou've already chosen the best response.2\[A(x)=\text{ area of square}=\text{ height } ^2\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.2the integrand is just x+1 that is A(x)=x+1

freckles
 one year ago
Best ResponseYou've already chosen the best response.2\[V=\int\limits_{1}^1 (x+1) dx\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0isint it \[\sqrt{x+1}\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.2that 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
 one year ago
Best ResponseYou've already chosen the best response.2dw: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
 one year ago
Best ResponseYou've already chosen the best response.2dw:1439495315876:dw

freckles
 one year ago
Best ResponseYou've already chosen the best response.2\[A(x)=\sqrt{x+1} \cdot \sqrt{x+1}=(x+1)\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.2notice the lower limit is x=1 and the upper limit is x=1

freckles
 one year ago
Best ResponseYou've already chosen the best response.2\[V=\int\limits _{1}^1 A(x) dx \\ A(x)=\sqrt{x+1} \cdot \sqrt{x+1}=x+1 \\ V=\int\limits_{1}^{1} (x+1) dx \] V represents the volume by the way

freckles
 one year ago
Best ResponseYou've already chosen the best response.2sounds great that is what @IrishBoy123 got using double integrals

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0thanks for your help freckles , cant thank you enough
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