2001 HSC - Q4(b) (1 Viewer)

[Stefano]

I don't get it...

i) How the **** does 'h' come into it ?

ii) How is R²-r²=b² ? I can integrate it after that, so no need to integrate it for me.

Thanks in advance

 

[KFunk]

I'll explain ii) first. It's tough to describe without an adequate graph but I'll try.

Let the centre of the sphere be O (as in the diagram). Where they have that point at the top indicating the distance 'r', let that point be R. Let the point where that arrow indicating distance 'r' touches the face of the sphere be Q.

You have a right angled triangle. RQ = r as indicated. OR = b since OR is half of the distance 2b. The tricky bit is realising that OQ = R. It's a sphere so the distance from the centre to any point on its surface is R.

OQ is the hypotenuse so R² = r² + b²

∴ R² - r² = b²

(The diagram is here if you need it : http://www.boardofstudies.nsw.edu.au/hsc_exams/hsc2001exams/pdf_doc/mathemat_ext2_01.pdf )

 

[KFunk]

I don't get it...

i) How the **** does 'h' come into it ?

This is something where I think people would use a variety of methods. What I did was consider a cross section parallel to the x-axis and perpendicular to the y-axis. I then drew a graph of part of this cross section: the quarter-circle x = √(R² - z²) (z ≥ 0).

In the diagram 'h' is just an arbitrary distance travelled up from O parallel to the z-axis (which is unnamed but goes up through the centre of the sphere).

Considering the quarter-circle diagram I mentioned you'll realise that when an arbitrary distance 'h' has been travelled that x = √(R² - h²). Making the connection between the diagram and the sphere you should notice that this value 'x' is the outer radius of the cross section taken at a height 'h' above the central one.

Yeah.. it's not a convenient one to explain given the 3-d nature of it. Alternatively you could try to work out something with similar triangles (which they like in these questions).

 

[Stefano]

KFunk Rules

I'll explain ii) first. It's tough to describe without an adequate graph but I'll try.

Let the centre of the sphere be O (as in the diagram). Where they have that point at the top indicating the distance 'r', let that point be R. Let the point where that arrow indicating distance 'r' touches the face of the sphere be Q.

You have a right angled triangle. RQ = r as indicated. OR = b since OR is half of the distance 2b. The tricky bit is realising that OQ = R. It's a sphere so the distance from the centre to any point on its surface is R.

OQ is the hypotenuse so R² = r² + b²

∴ R² - r² = b²

(The diagram is here if you need it : http://www.boardofstudies.nsw.edu.au/hsc_exams/hsc2001exams/pdf_doc/mathemat_ext2_01.pdf )

Well explained. Makes PERFECT sense in my mind. Thank you VERY much!

 

[Stefano]

This is something where I think people would use a variety of methods. What I did was consider a cross section parallel to the x-axis and perpendicular to the y-axis. I then drew a graph of part of this cross section: the quarter-circle x = √(R² - z²) (z ≥ 0).

In the diagram 'h' is just an arbitrary distance travelled up from O parallel to the z-axis (which is unnamed but goes up through the centre of the sphere).

Considering the quarter-circle diagram I mentioned you'll realise that when an arbitrary distance 'h' has been travelled that x = √(R² - h²). Making the connection between the diagram and the sphere you should notice that this value 'x' is the outer radius of the cross section taken at a height 'h' above the central one.

Yeah.. it's not a convenient one to explain given the 3-d nature of it. Alternatively you could try to work out something with similar triangles (which they like in these questions).

Genius... thanks dude.

 

[KFunk]

Glad I could help.

 

[Stefano]

Glad I could help.

Me too!