volume question :( (1 Viewer)

[norelle]

A hole of diameter R is drilled through the centre of a solid sphere of diameter 2R. Show that the remaining solid has volume [(sqrt root 3)/2 pi R^3] by using:

1) cylindrical shells method
2) slicing


thanks guys

 

[jet]

Do you really need it done by slicing?

 

[jet]

 

[norelle]

thanksss


but btw, why cant i use the normal volume calculation method?
I calculate the top part first, limit from R to R/2, then times 2

2 x limit (R~R/2) pi x^2 dy
= 2 pi limit (R~R/2) R^2 - y^2 dy
= 2 pi [R^2 y - y^3 /3 ] (R ~ R/2)
= 2 pi [R^3 - R^3/3 - R^3 / 2 + R^3 / 24]
= 2pi [ 2/3 R^3 - 11/24 R^3]
= 2pi [ 15 R^3 / 24]
= 8piR^3 ?!!

 

[omniscience]

 

[jet]

thanksss


but btw, why cant i use the normal volume calculation method?
I calculate the top part first, limit from R to R/2, then times 2

2 x limit (R~R/2) pi x^2 dy
= 2 pi limit (R~R/2) R^2 - y^2 dy
= 2 pi [R^2 y - y^3 /3 ] (R ~ R/2)
= 2 pi [R^3 - R^3/3 - R^3 / 2 + R^3 / 24]
= 2pi [ 2/3 R^3 - 11/24 R^3]
= 2pi [ 15 R^3 / 24]
= 8piR^3 ?!!

it should equal 5R^3/24 inside the brackets.
And i think, the way you are treating the volume, you are already calculating the bottom bit, in a way, as you are still rotating it about the x-axis. Try doing the topic bit around the y-axis, and then doubling it. THat should work..

 

[norelle]

I dont get the limit part,
why is not from R to -R ?

 

[jet]

Sorry, that y in the inequality is meant to be an x.

 

[norelle]

noo i mean the limit part

 

[jet]

You mean how I got the limits?
I used the lines which were the boundaries for the hole in the sphere, and solved them with the equation for the circle. This gave me the x-values for the ends of the hole. If you want a diagram, just tell me.

 

[norelle]

You mean how I got the limits?
I used the lines which were the boundaries for the hole in the sphere, and solved them with the equation for the circle. This gave me the x-values for the ends of the hole. If you want a diagram, just tell me.

But why arent we use the limit from x=R to x= -R ?
if we use the point from (sqrt rt3 /2 , R/2 ) to (-sqrt rt3 /2 , R/2 )
then wht abt the volume between x=sqrt rt3 /2 to x=R ?

 

[jet]

But, that will be taken out by the hole....
Let me draw a diagram.

 

[jet]

YOu can see how the hole removes that bit of volume, so we need to solve the equations simultaneously to get the limits.

 

[norelle]

YOu can see how the hole removes that bit of volume, so we need to solve the equations simultaneously to get the limits.

OH..i get it now!!!!!!! thanks a lot!!!!!!