another volumes question (1 Viewer)

[freaking_out]

A hole of radius r is bored through a sphere of radius R. Find the remaining volume if:

i) R=8, r=2
ii) R/r=2/1

 

[underthesun]

My teacher used slicing method for example by subtraction of the cylinder "hole", and calculating the sphere volume with the poles "chipped off".. if you get what I mean.

I forgot, but I tried to use shell method. I forgot if it worked or not. Why don't you try it?

 

[freaking_out]

hey, this question is under the slicing disks and washers section. we still not meant to know cylindrical shells yet....but anyway i tried it but got it wrong!!!! i atleast wanna know what answer you got...

 

[wogboy]

Firstly draw up a nice big diagram of the sphere, with a cylindric hole bored thru it (don't be too artistic with the 3D toning & stuff, this just wastes time and can confuse you more). If you can visualise it in your mind, that's good enough.

Split this huge volume into smaller pieces. You have a sphere, minus a hole in the shape of a cylinder, minus two caps at each end of the hole.

First find the volume of the entire sphere including the hole, which is V = (4/3)*pi*R^3 (note the capital V)

Then find the volume of the cylindrical hole. To do this you have to find out where the bored hole starts & ends. You know the sphere has radius R, so the equation of the circle, which represents the sphere is: x^2 + y^2 = R^2
Also the cylindric hole has radius r, which means at y=r, the hole ends.

x^2 + r^2 = R^2
x = sqrt(R^2 - r^2) <-- This is where the cylindric hole ends. (ignore the -ve value for now, which is the beginning of the hole)

x = sqrt(R^2 - r^2)
so the volume of the cylinder is (note the lower case v):
v = pi*(r^2)*h
= 2*pi*(r^2)*sqrt(R^2 - r^2)

Now for the hardest part, to find the volumes of the caps. Let the volume of each cap be C. The cross section of volume C (the cap) is circular, and it's bounded by the large sphere, so you can find it's volume by slicing:

r = y
x^2 + y^2 = R^2
hence r^2 = R^2 - x^2

so dC = pi*(r^2)*dx
dC = pi*(R^2 - x^2)*dx
dC/dx = pi*(R^2 - x^2)

to find C, integrate with the limits sqrt(R^2 - r^2) to R (where the cap starts & ends)

C = pi * I{sqrt(R^2 - r^2) --> R} R^2 - x^2
= pi * (- (R^3)/3 + (Rr)^2 + {(R^2 - r^2)^(3/2)}/3)

Now the volume (U) of the final product (answer) is:

U = V - v - 2C
= pi * [({4R^3}/3 - 2*pi*(r^2)*sqrt(R^2 - r^2) + 2*((R^3)/3 - (Rr)^2 - {(R^2 - r^2)^(3/2)}/3)]

Now you have U in terms of R & r (that horrible mess up there )

Sub in R=8, r=2 to get the answer for i), and sub r = R/2 to get the answer for ii) (most probably in terms of R)

To make things easier, maybe you can sub R & r to find V, v, and C and then find U.

 

[Affinity]

theres are better ways to do this:

The volume is just rotating the area bounded by
y=sqrt(R^2 -x^2)
y=r

about the x axis.

you can either do this by method of cylindrical shells or annular slices. some necessary working are omitted in the following.

shells:

dV = 2*Pi*(2*sqrt(R^2 -y^2))*y dy

V = integral {r->R} 2*Pi*(2*sqrt(R^2 -y^2))*y dy

V = [-(4Pi)*(1/3)(R^2-y^2)^(3/2)] (R, r)

V = [(4Pi/3)(R^2 -r^2)^(3/2)]

annular slices:

the x values for intersections are:

a=-sqrt(R^2 -r^2) and b=sqrt(R^2 - r^2)

dV = Pi*(y^2 - r^2) dx

V = Integral {a->b} Pi*(y^2 - r^2) dx

V = Integral {a->b} Pi*(R^2- x^2 - r^2) dx

V = 2* Integral {0->b} Pi*(R^2- x^2 - r^2) dx

V = 2*Pi*(xR^2- (1/3)x^3 - xr^2) [0,b]

V = 2Pi*(sqrt(R^2 - r^2) *(R^2 -r^2) -(1/3)(R^2 - r^2)^(3/2))

V = (4Pi/3)*(R^2 - r^2)(3/2)

by the way, it has been assumed that the hole and the sphere share the same center in the textbooks and all solutions posted. as a challenge, try the same question but with the hole off center.

 

[freaking_out]

try the same question but with the hole off center

now that is confusing...... how would you go about doing it???

 

[Affinity]

try it if you have time, I will post the solution when I figure it out (I haven't tried it myself yet)

 

[Affinity]

should be do-able with slices
at the ends, the slices are not annular, but 'cresent' shaped so might mess things up a little.

 

[freaking_out]

Originally posted by Affinity
should be do-able with slices
at the ends, the slices are not annular, but 'cresent' shaped so might mess things up a little.

hey does these sorta questions appear in the exam????? if so hurry up with the solutions! otherwise.....take your time coz i can't seem to do it.:mad1:

 

[Affinity]

Oh.. they wouldn't be that hard in the exams since this will involve many calculations, but you should be able to do it .

I am in the process of figuring out what my chemistry teachers mean by their questions(their questions do not corrospond to what they want you to write) so.. I might try it after the trials