Very Hard Question from SBHS EX1 Paper. (1 Viewer)

[Sleight_of_Hand]

As the title states, solve:

A cylindrical hole of radius c is bored symettrically though a sphere of radius a. Show that the remaining volume is 4/3 pi (a^2 - c^2)^(2/3)

 

[random-1006]

As the title states, solve:

A cylindrical hole of radius c is bored symettrically though a sphere of radius a. Show that the remaining volume is 4/3 pi (a^2 - c^2)^(2/3)

doesnt look that bad, looks like it will be something like V sphere- V cylinder
(4/3)pi a^3 - pi ( c) (a^2)


i think i got an idea, give me few minutes

it will involve the theorm "radius is perpendicular to the tangent of the circle i think", that gets a right angle, and the diff of sqaures most likely came from phythag

got it i think,writing up now

 

[pOtentiaLeaver]

i think question is wrong, i got 4/3 (a^2-c^2)^(3/2)

 

[random-1006]

Volume = Vol Sphere- Vol Cylinder

Draw picture, the height for the cylinder is not the radius, it is the third side of the right angled triangle formed between the radius of the sphere and the radius of the cylinder),

sorry i cant put up picture, so im doing a good description as possible

For cylinder: height= 2 sqrt(a^2 -c^2) [ half the height is sqrt (a^2-c^2) ]using phythag ( as the radius is perpendicular to tangt)




Vol= 4/3 pi a^3 - pi (c^2)(2sqrt(a^2-c^2))
= (4/3)pi ( a^3 - 3/2 c^2 sqrt(a^2 -c^2))
but height cylinder= a= sqrt(a^2-c^2) the radius of a sphere.
= (4/3) pi ( a^2 sqrt(a^2 -c^2) -3/2 c^2 sqrt(a^2 -c^2) )
= (4/3) pi (sqrt(a^2-c^2)) [ a^2 -3/2 c^2]
i cant see how you could get the second bracket in (a^2 -c^2) form for simplication)


damm so close, but a bloody good shot.

check my arithmetic, its probably ( i hope lol, a small mistake somewhere)

 

[pOtentiaLeaver]

actually no noobs...

you need to take away the top and bottom bits...

 

[pOtentiaLeaver]

with x=this shiit below

juggle them a bit u will get the answer with 3/2 and not 2/3

 

[random-1006]

with x=this shiit below

juggle them a bit u will get the answer with 3/2 and not 2/3

yeh, that makes sense, the two parts at the end are made by rotating around x axis and it is a circle x^2 +y^2 = r^2, with radius of revolution [ a - sqrt(a^2 -c^2) ], but c is along the x axis, so instead it is c^2 +y^2 = r^2.


wait a minute, wouldnt that mean you would only need to minus one x, because the parts at the end of the cylinder are half spheres, not whole spheres [ the integral above will give volume of a whole revolution] , try it with only subtracting one x

they make them tough lol

 

[pOtentiaLeaver]

dude, are u kidding me? it doesn't matter what x and y is because it's a circle... and even if it matters, the answer is just a number that you sub in....

also i edited my latex code, don't quote my wrong one haha

 

[pOtentiaLeaver]

no, i don't get what you're saying but just imagine when you inscribe a cylinder in a sphere, there is a top and bottom hemispheres(not really a hemisphere because they're not half sphere, but you get the point) which you need to take away to get the volume

 

[random-1006]

no, i don't get what you're saying but just imagine when you inscribe a cylinder in a sphere, there is a top and bottom hemispheres(not really a hemisphere because they're not half sphere, but you get the point) which you need to take away to get the volume

yes, i understand, but what the intergal you gave is doing is going from the end of the cylinder to the edge of the sphere and rotating that whole area ( 360 degress) about the x axis ( we only want half this, ie 180 degress rotation) ( this will create a sphere, not the hemisphere we want) , hence you should only substract away one x, not two

 

[pOtentiaLeaver]

sorry, still don't get what you're saying
my x is refering to this volumn:

my integral could be wrong (forgot about this crap already) but i still "THINK" im right

 

[pOtentiaLeaver]

why the fk am i double posting...? i swear i click the post button once... this site is really laggy lol

 

[random-1006]

sorry, still don't get what you're saying
my x is refering to this volumn:

my integral could be wrong (forgot about this crap already) but i still "THINK" im right

you probably are right, i had the picture round the other way, ie the extra volume on the top and bottom so that when they were rotated ( not about x axis i should say, about the edge of the cylinder) it would create a sphere

your method looks correct

im going to sleep now lol

 

[pOtentiaLeaver]

i had my cylinder standing straight too...

i drew this new diagram to help find the extra bit which is the same size no matter where you view it at

 

[Carrotsticks]

To those thinking that it's just sphere - cylinder, you're wrong.

 

[Carrotsticks]

I'm taking the top half of the circle. My limit will then be from c to a, so then it ignores the middle bit. I then rotate that around the x axis.

 

[nat_doc]

how would u integrate that with out a trig sub? u ca'nt use trig since u are given a and c, unless ut going to use sin^-1(a)?

 

[random-1006]

how would u integrate that with out a trig sub? u ca'nt use trig since u are given a and c, unless ut going to use sin^-1(a)?

why cant you use trig substitution, a and c are constants (from the question), its exactly like int sqrt(1-x^2) dx

harder 3 unit, you should be able to select your own substitutions , they arent always given ( however, most of the time the substitution is obvious), theres a reason why its harder sydney boys question!

 

[nat_doc]

why cant you use trig substitution, a and c are constants (from the question), its exactly like int sqrt(1-x^2) dx

harder 3 unit, you should be able to select your own substitutions , they arent always given ( however, most of the time the substitution is obvious), theres a reason why its harder sydney boys question!

well for a trig sub, let x=acos@... then u have to find what theta is in terms of a and c, since it is a definite integral ......so ur gonna get theta to equal the sin inverse of some fraction with a sqrt in it....... ill put up my solution up now... gimme time to play with latex

 

[nat_doc]

(since even)










XD