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shubhamsrg
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Consider a cylinder with radius = 5 cm and height = 20 cm. The cylinder is placed vertically on the ground and water is filled in to a height of 10 cm from the ground.
Now the cylinder is closed from the top and it is placed on the ground horizontally. Find the new height of water from the ground .
 one year ago
 one year ago
shubhamsrg Group Title
Consider a cylinder with radius = 5 cm and height = 20 cm. The cylinder is placed vertically on the ground and water is filled in to a height of 10 cm from the ground. Now the cylinder is closed from the top and it is placed on the ground horizontally. Find the new height of water from the ground .
 one year ago
 one year ago

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shubhamsrg Group TitleBest ResponseYou've already chosen the best response.0
dw:1359278725894:dw
 one year ago

Azteck Group TitleBest ResponseYou've already chosen the best response.0
Find the volume of the water when the cylinder is vertically upwards.
 one year ago

shubhamsrg Group TitleBest ResponseYou've already chosen the best response.0
I have tried all. Couldn't crack it.
 one year ago

Azteck Group TitleBest ResponseYou've already chosen the best response.0
Just do it.
 one year ago

Azteck Group TitleBest ResponseYou've already chosen the best response.0
I will tell you what to do next. I might have got it just by looking at the question.
 one year ago

shubhamsrg Group TitleBest ResponseYou've already chosen the best response.0
V = 250pi
 one year ago

shubhamsrg Group TitleBest ResponseYou've already chosen the best response.0
So, what next ?
 one year ago

frx Group TitleBest ResponseYou've already chosen the best response.0
All you need to know: http://mathworld.wolfram.com/CylindricalSegment.html
 one year ago

Azteck Group TitleBest ResponseYou've already chosen the best response.0
Do you know how to calculate the volume of apartial cylinder?
 one year ago

shubhamsrg Group TitleBest ResponseYou've already chosen the best response.0
Nopes I don't, am referring to the link @frx gave .
 one year ago

sirm3d Group TitleBest ResponseYou've already chosen the best response.1
dw:1359279395322:dw \[\Large{A=\int_{5}^h 2\sqrt{5^2y^2}} \;\mathrm dy\]
 one year ago

Azteck Group TitleBest ResponseYou've already chosen the best response.0
Found this. The first suggestion works for a *vertical* cylinder, but not for a horizontal one. Your first step will depend on the fluid level in the tank... You start the same way, . I'm assuming it's a circular cylinder, where If the cylinder is more than half full, you'll want to start with the volume of the full cylinder, V = r^2*length*pi then subtract the unfilled space at the top. The crosssection of this empty space takes the form of a circular segment, whose area is given by the formula A = 1/2 * r^2 * (x  sin(x)) where r is the circle's radius, and x is the angle (measured in radians) formed between the two radii drawn from the points on the tank wall where the fluid level ends. Multiply this area by the cylinder's length to obtain the volume of empty space. Subtract this number from the full cylinder's volume to arrive at your fluid volume. If the cylinder is less than half full, you'll do something similar...only this time you'll be finding the cross sectional area of the fluid sitting on the bottom of the tank and multiplying that by the tank's length to find the volume. You can also use this method if the tank is exactly half full...you'll see that the angle between the two radii is flat (pi radians), forming a diameter...and as sin(pi) = 0, the area formula reduces to A= (1/2) * r^2 * pi
 one year ago

sirm3d Group TitleBest ResponseYou've already chosen the best response.1
it's \[\Large A=\int_{5}^{5+h}2\sqrt{25y^2}\;\mathrm dy\]dw:1359279823727:dw
 one year ago

shubhamsrg Group TitleBest ResponseYou've already chosen the best response.0
@Azteck how do I find "x" ?
 one year ago

sirm3d Group TitleBest ResponseYou've already chosen the best response.1
that's \(5+h\) for the upper limit.
 one year ago

shubhamsrg Group TitleBest ResponseYou've already chosen the best response.0
I am unable to follow, @sirm3d
 one year ago

sirm3d Group TitleBest ResponseYou've already chosen the best response.1
i'm sorry. i assumed you know integral calculus.
 one year ago

shubhamsrg Group TitleBest ResponseYou've already chosen the best response.0
I do know integral calculus .
 one year ago

shubhamsrg Group TitleBest ResponseYou've already chosen the best response.0
Wait, am getting it, 2 mins.
 one year ago

sirm3d Group TitleBest ResponseYou've already chosen the best response.1
it's the area under the curve.
 one year ago

shubhamsrg Group TitleBest ResponseYou've already chosen the best response.0
How do you get 5^2  y^2 ?
 one year ago

shubhamsrg Group TitleBest ResponseYou've already chosen the best response.0
oh wait,
 one year ago

shubhamsrg Group TitleBest ResponseYou've already chosen the best response.0
So you're finding the length of an element at a distance y from origin.
 one year ago

sirm3d Group TitleBest ResponseYou've already chosen the best response.1
dw:1359280450587:dw
 one year ago

shubhamsrg Group TitleBest ResponseYou've already chosen the best response.0
Ah ok so our working eqn will be A(20) = 250pi Where A =integral (2 sqrt(25 y^2)) from y= 5 to (5+h)
 one year ago

sirm3d Group TitleBest ResponseYou've already chosen the best response.1
in the process, you will be deriving the area bounded by the arc and a chord. you can also derive the area as the difference between the area of the sector of a circle and the isosceles triangle.
 one year ago

shubhamsrg Group TitleBest ResponseYou've already chosen the best response.0
=>[ sqrt(25y^2) + 25sin^1( y/5) ] from y=5 to 5+h = 25pi/2 =>( sqrt( 25  (h5)^2) + 25sin^1((h5)/5) ) = 25pi am getting complex solutions, please someone confirm. http://www.wolframalpha.com/input/?i=%28+sqrt%28+25++%28h5%29%5E2%29+%2B+25sin%5E1%28%28h5%29%2F5%29+%29++%3D+25pi
 one year ago

shubhamsrg Group TitleBest ResponseYou've already chosen the best response.0
Oh wait, RHS will be 0 ./
 one year ago

shubhamsrg Group TitleBest ResponseYou've already chosen the best response.0
Seems fine thanks a lot @sirm3d
 one year ago

sauravshakya Group TitleBest ResponseYou've already chosen the best response.1
Isnt it simply 5 cm as volume of water =1/2 volume of cylinder
 one year ago

shubhamsrg Group TitleBest ResponseYou've already chosen the best response.0
I see your point, I guess it won't be 5 since the area does not increases uniformly as we move up from ground level when it is placed horizontally. Area around center is more than at the bottom; or at top.
 one year ago

shubhamsrg Group TitleBest ResponseYou've already chosen the best response.0
When cylinder was placed vertically, area was increasing uniformly,but not in the case of horizontal positioning.
 one year ago

sauravshakya Group TitleBest ResponseYou've already chosen the best response.1
dw:1359287336567:dw If I will cut the cylinder from the diameter in two parts won't their volume be equal
 one year ago

sauravshakya Group TitleBest ResponseYou've already chosen the best response.1
dw:1359287729887:dw
 one year ago

shubhamsrg Group TitleBest ResponseYou've already chosen the best response.0
Indeed! Rechecked my calculations! It is 5 only!
 one year ago

shubhamsrg Group TitleBest ResponseYou've already chosen the best response.0
It was real simple then, guess calculus would have been involved only if, say, radius was 4 cm. hm, thanks @sauravshakya and @sirm3d
 one year ago
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