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anonymous
 4 years ago
How to solve this.
A cylinder of length 2x is inscribed in a sphere of radius a . Between one end ot this cylinder & sphere, another cylinder is inscribed with one end on an end of the first cylinder so that the axes of the cylinders are collinear. Show that the sum of the volumes of the two cylinders is,
V= 2(Pi) ( x + y )^( a^2  x^2  4y)
Thanks
anonymous
 4 years ago
How to solve this. A cylinder of length 2x is inscribed in a sphere of radius a . Between one end ot this cylinder & sphere, another cylinder is inscribed with one end on an end of the first cylinder so that the axes of the cylinders are collinear. Show that the sum of the volumes of the two cylinders is, V= 2(Pi) ( x + y )^( a^2  x^2  4y) Thanks

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Shayaan_Mustafa
 4 years ago
Best ResponseYou've already chosen the best response.0Are you understanding question?

Mertsj
 4 years ago
Best ResponseYou've already chosen the best response.0This is a hard problem. I think we need Satellite.

Mertsj
 4 years ago
Best ResponseYou've already chosen the best response.0I think maybe. I tried to draw it and that is not easy.

Shayaan_Mustafa
 4 years ago
Best ResponseYou've already chosen the best response.0nothing is hard. we can do anything. just need of help.

Mertsj
 4 years ago
Best ResponseYou've already chosen the best response.0Also it would be helpful if we knew what y represents.

Shayaan_Mustafa
 4 years ago
Best ResponseYou've already chosen the best response.0i think coordinates. because he talk about axes. isn't it?

Shayaan_Mustafa
 4 years ago
Best ResponseYou've already chosen the best response.0first we need a diagram. otherwise it will take more time to understand.

Mertsj
 4 years ago
Best ResponseYou've already chosen the best response.0The axes of the cylinders. So I would assume that means the line down the middle of the cylinder that is perpendicular to both bases.

Mertsj
 4 years ago
Best ResponseYou've already chosen the best response.0Or perhaps I am wrong in my assumption that these are right circular cylinders.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Tried to figure out the diagram too

Mertsj
 4 years ago
Best ResponseYou've already chosen the best response.0Is there any information as to what y represents?

Shayaan_Mustafa
 4 years ago
Best ResponseYou've already chosen the best response.0The axes of the cylinders. So I would assume that means the line down the middle of the cylinder that is perpendicular to both bases. Yeah i agree.

Shayaan_Mustafa
 4 years ago
Best ResponseYou've already chosen the best response.0A cylinder of length 2x is inscribed in a sphere of radius a dw:1327783252116:dw Is this right? for the above mentioned line?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I think x & y represent the value of r^2

Mertsj
 4 years ago
Best ResponseYou've already chosen the best response.0So what we need to know now is the radius of the cylinder so that we can find the area of the bases of the cylinder.

Shayaan_Mustafa
 4 years ago
Best ResponseYou've already chosen the best response.0can you elaborate the wording from the word "between one end of this cylinder?" so that we could figure it out.

Mertsj
 4 years ago
Best ResponseYou've already chosen the best response.0So is there some relationship between a sphere and the radius of the circle whose center is ax units from the endpoint of the diameter?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I think one end of the cylinder is touching each other

Shayaan_Mustafa
 4 years ago
Best ResponseYou've already chosen the best response.0@Mertsj. Here is just a sphere not a circle.

Shayaan_Mustafa
 4 years ago
Best ResponseYou've already chosen the best response.0we need some more help. now i want to answer this question.

Mertsj
 4 years ago
Best ResponseYou've already chosen the best response.0But if the cylinder is inscribed in a sphere, isn't the base of the cylinder a circle of the sphere?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Trying to visualize it

Shayaan_Mustafa
 4 years ago
Best ResponseYou've already chosen the best response.0hmmm... yes you are right. but does this matter here?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0It can be mean that the cylinder is in lying position

Shayaan_Mustafa
 4 years ago
Best ResponseYou've already chosen the best response.0yes cbrsam you are right. cylinders are in lying position. therefore their axes are collinear otherwise this is not possible.

Shayaan_Mustafa
 4 years ago
Best ResponseYou've already chosen the best response.0@MertsJ. Are their axes collinear or not?

Mertsj
 4 years ago
Best ResponseYou've already chosen the best response.0The problem says they are. I need to put another cylinder in my drawing but I am not good at drawing. Do you want to try?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0If we see only one cylinders . 2a  2x is the part that protrude out at the end of cylinder

Mertsj
 4 years ago
Best ResponseYou've already chosen the best response.0Actually ax since there is an equal piece at both ends of the cylinder.

Shayaan_Mustafa
 4 years ago
Best ResponseYou've already chosen the best response.0remember guys. axes must be collinear. otherwise we could not be able to find volume.

Mertsj
 4 years ago
Best ResponseYou've already chosen the best response.0What class are you taking. Is this a calculus problem?

Shayaan_Mustafa
 4 years ago
Best ResponseYou've already chosen the best response.0ax for one side yes it is.

Mertsj
 4 years ago
Best ResponseYou've already chosen the best response.0Is it a calculus problem?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0okk .. I will approach this as .. the big cylinder (with length 2x) would have radius as sqrt (a^2  x^2) the smaller cylinder would have radius as sqrt(a^2  (x+y)^2) .. this is assuming y is the length of the smaller cylinder ... then if I solve .. i get sum of volume as pie*(a^2 + y^2  x^2)*(y + 2x)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0but the answer ain't matchin

Shayaan_Mustafa
 4 years ago
Best ResponseYou've already chosen the best response.0yes this y is confusing to us.

Shayaan_Mustafa
 4 years ago
Best ResponseYou've already chosen the best response.0@shaan_iitk. I got the same ans. :(

Shayaan_Mustafa
 4 years ago
Best ResponseYou've already chosen the best response.0is this question complete? may be he forget to give description about y.

Mertsj
 4 years ago
Best ResponseYou've already chosen the best response.0Perhaps y is the radius of the small cylinder and it has height 2x since the problem specifies that one end is on the end of the first cylinder.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0That the question I got nothing else specified. Can it be the second cylinder is the same size

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0well then the question is incomplete ..isn't it.. there has to be some meaning of y

Shayaan_Mustafa
 4 years ago
Best ResponseYou've already chosen the best response.0really i am an electronics engineer, solid state physicist, semi conductor physicist, interest in cosmology. but not a good mathematician.

Mertsj
 4 years ago
Best ResponseYou've already chosen the best response.0Could that be what is meant?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0It stated that the other cylinder only touches one side so it should be outside

Mertsj
 4 years ago
Best ResponseYou've already chosen the best response.0And it also says that is it inscribed.

Shayaan_Mustafa
 4 years ago
Best ResponseYou've already chosen the best response.0@Mertsj Read line. "Between one end ot this cylinder & sphere", another cylinder is inscribed cylinder and sphere.

Shayaan_Mustafa
 4 years ago
Best ResponseYou've already chosen the best response.0we need another help. until we don't figure it. we can't move ahead.

Shayaan_Mustafa
 4 years ago
Best ResponseYou've already chosen the best response.0not this too. i am sure.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I think this is the right diagram

Mertsj
 4 years ago
Best ResponseYou've already chosen the best response.0So do we agree that the volume of the large cylinder is 2pi(x)(a^2x^2)?

Shayaan_Mustafa
 4 years ago
Best ResponseYou've already chosen the best response.0kindly review again this line. Between one end ot this cylinder & sphere, "another cylinder is inscribed" another cylinder means the whole other cylinder. can't see it? huh.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Let try. First take vol of sphere vol of fist cyl. We got 4 equal outer region that does not touch the cyl

Shayaan_Mustafa
 4 years ago
Best ResponseYou've already chosen the best response.0ok try. i will just see.

Shayaan_Mustafa
 4 years ago
Best ResponseYou've already chosen the best response.0will any one try to solve it?

Mertsj
 4 years ago
Best ResponseYou've already chosen the best response.0\[V=2\pi x(a ^{2}x ^{2})\] Volume of first cylinder

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0It'd be lovely to have a definitive answer of what y is still... :)

Shayaan_Mustafa
 4 years ago
Best ResponseYou've already chosen the best response.0hmm... may be you are right Mertsj. this seems to be the volume of the first cylinder.

Shayaan_Mustafa
 4 years ago
Best ResponseYou've already chosen the best response.0yes this y creating confusion. still.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0What I get is v = 4 Pi. ( a x )^2

Mertsj
 4 years ago
Best ResponseYou've already chosen the best response.0Let y be the distance that the small cylinder extends below the large cylinder. Then the height of the small cylinder is y+2x and the radius of the small cylinder is a^2(y+x)^2 and it's volume is pi(y+2x)[(a^2(y+x)^2]^2

Mertsj
 4 years ago
Best ResponseYou've already chosen the best response.0And the total volume of the two cylinders is \[2\pi x(a ^{2}x ^{2})+\pi (y+2x)[a ^{2}(y+x)^{2}]^{2}\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Where did this problem come from? I would be willing to nearly bet my life that there is some indication of what y is supposed to be in the problem, variables are never implicitly defined like that.... additionally, could we get a clarification on the correct answer? Because dimensionally what's typed in the box can't be right, so was there a typo, or what?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0That was the question given to me

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0In that case, I would ask you to clarify these questions with the source.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0It says the 2 cylinder are collinear. Thus, I think both the drawings are incorrect?

Shayaan_Mustafa
 4 years ago
Best ResponseYou've already chosen the best response.0yes. i agree with accessibm. That is what i am also saying.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Volume of first cylinder =2πX(a^2−x^2) this looks fine. if we confirm the vol of 1st cylinder is as above. we can substract from the sum of vol. which gives us the vol of the 2nd cylinder. perform some reverse engineering and hopefully get some ans? so far cannot conclude any ans. maybe there is typo error in the sum of the vol. stated in the question. Do confirm.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Hi Mertsj, shayaan, accessibm, Jemurray already check with the source, there is some mistake in the question. The 2nd cylinder is of length 2y is inscribed. The equation is V= 2(Pi) (x+y) (a^2  x^2 4y ) Thanks

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Then the vol of the sm cyl will be 2πY(a^2−(X+2Y)^2). add both vol. Still dont not add up to the eqn. Do confirm if there is typo error? I believe the approach is correct. The diagram should look like a urn in a sphere from 2D front view.
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