Volume Questions (1 Viewer)

[matt_17]

1. Find the volume common to two right-circular cylinders, each having a base radius a, if the axes of the cylinders intersect at right angles.

2. The roof of a building is in the form of a frustum of a pyramid with a square base of side 8m and whose flat top is a square of side 2m. All the sloping sides are pitched at the same angle. The perpendicular distance between the top and the bottom planes is 4m. Calculate the volume enclosed by the roof.

I am having trouble doing the above two questions. Can anyone help?

 

[Jase]

I couldn't be bothered thinking about what the hell shape the first one is.. so i skipped it.
Is the answer to the second one 325 1/3 units^3?

 

[spaz1810]

i assumed that a frustrum was just a pyramid with its top cut off? like a 3d isosceles trapezium

i got 280 units^3 for the second one

side length (l) is proportional to the height of pyramid (h)
when h=0 l=8
when h=4 l=2
as they are pitched at equal angles and constant, it is always a square area of side length l = 8-6h/4

:. A = (8-6h/4)^2
dV = A.dh
V = lim (dh -> 0) sigma (0 to 4) dV
V = integral (0 to 4) (8-6h/4)^2 dh
V = 1/4(256h-6h^2+3h^3) (0 to 4)
V = 1/4(1024-96+192)
V = 280 units^3

 

[ngai]

:. A = (8-6h/4)^2
dV = A.dh
V = lim (dh -> 0) sigma (0 to 4) dV
V = integral (0 to 4) (8-6h/4)^2 dh
V = 1/4(256h-6h^2+3h^3) (0 to 4)
V = 1/4(1024-96+192)
V = 280 units^3

nice work
wats wrong with simplifying 6/4 = 3/2?
and the 6h^2......thats horrible

 

[Xayma]

I get 112m³ for the second one, by inspection.
By deriving the volume of a frustrum where all sides are at the same angle

 

[Jase]

my side length was l = 1/4h + 8..


EDIT : Woops you're right its 8 - 6/4h.
missed some numbers.

 

[mojako]

in HSC exams will they tell you the meaning of some words which are not very commonly used throughout the course?
(like frustum for example)
EDIT: I realise that an understanding of the word frustum is not important in this question.

 

[jumb]

I doubt it. Theyre real bitches sometimes

 

[spaz1810]

my answer earlier is wrong (simple algebra mistake) a much simpler way, from the equation l=8-6h/4 the height must be 16/3 (when l=0) and then by subtracting one pyramid from the other:

:. V = 1/3*(8*8*16/3-2*2*4/3) = 112 units^3

the same as xayma

however even with the correct algebra i still don't get 112 with calculus :S

 

[Xayma]

Or you could go since its on a common angle.
Let the height from the 2m square base to the apex of the pyramid created from the base

tan θ=tan θ
(x+4)/4=x/1
3x=4
x=4/3

∴ total height =16/3.

 

[wogboy]

1. Find the volume common to two right-circular cylinders, each having a base radius a, if the axes of the cylinders intersect at right angles.

I remember having answered this one some time back:

http://www.boredofstudies.org/community/showthread.php?t=35637

 

[Jase]

ah, i just took Spaz's answer for granted.. since it was the right eqaution.

Well if you use calculus you get 112 as well.. So check your working again Spaz ^^.
Problem solved.

Thanks for the link wogboy, although im too groggy to understand any of that right now.

 

[matt_17]

my teacher showed me how to do these questions today, but thanks for your help anyway. The answer to the second one is 112m3 and the answer to the second one as shown by wogboy is 16/3*a^3. I found it hard to derive the side length for this.

 

[Veck]

hmm by inspection I got the first one to be just 2.pi.a^3

any comments..?
it seemed really easy

 

[maths > english]

Good luck to all 4u students in their exam on Monday
Heres my solution to (1) (Attached)