Volumes question (1 Viewer)

[cutemouse]

Hello,

Got a problem this question. Could anyone please help me out?

I wrote the whole question so it makes sense. I'm happy if you just wanna do (a), and I could probably work out the rest from it. Thanks!

A solid is formed such that its cross-sections by the plane parallel to the base and at a height h above the base are squares of side s(h). In each solid, the dimension of the top square is half the dimension of the base square, find the volume of the solid if s(h) is given by:

a) 2-3h
b) sqrt(4-h)
c) 1/sqrt(4+h)

 

[Trebla]

Hello,

Got a problem this question. Could anyone please help me out?

I wrote the whole question so it makes sense. I'm happy if you just wanna do (a), and I could probably work out the rest from it. Thanks!

A solid is formed such that its cross-sections by the plane parallel to the base and at a height h above the base are squares of side s(h). In each solid, the dimension of the top square is half the dimension of the base square, find the volume of the solid if s(h) is given by:

a) 2-3h
b) sqrt(4-h)
c) 1/sqrt(4+h)

Might be helpful to draw a diagram to visualise it.

Since the side of each square is s, the area of the square is s²
δV = s²δh
We need s as a function of h, so in a) it's given as s = 2 - 3h
=> δV = (2 - 3h)²δh
We need the limits before we integrate this. Obviously the lower limit is 0.
Now the 'dimension' (I'm assuming this means one side) of the top square is half the dimension of the base square. At h = 0 (the lower limit), s = 2 given in a) which is the side of the base square, hence the side of the top square must be 1 and at s = 1, h = 1/3. So the limits of h are 0 and 1/3.
V = ∫₀^1/3(2 - 3h)²dh
= [- (2 - 3h)³ / 9]₀^1/3
= - 1 / 9 + 8 / 9
= 7 / 9

 

[cutemouse]

Hey Trebla thanks for that. I get it now.

But I have a problem with another question. It'd be great if anybody could help me out.

Question is:

The figure (attached) shows a cistern whose top and bottom rectangular faces measure 4cm by 3cm and 8cm by 5cm respectively. If these parrallel faces are 10cm apart, find the volume of the cistern.

Thanks.

 

[azureus88]

Hey Trebla thanks for that. I get it now.

4cm by 3cm and 8cm by 5cm respectively.
Thanks.

you sure it isnt 4cm by 3cm and 8cm by 6cm respectively ?

 

[cutemouse]

Nope, it's 4cm by 3cm and 8cm by 5cm respectively.

 

[jet]

OKay, here goes.
Takes slices of the solid parallel to the rectangular base with thickness ∂h and at a height h.

Let the width of each slice be x(h)=ah + b
When h=0, x(0)=8 therefore b=8
When h=10, x(10)=4 therefore 10a + 8 = 4, a=-(2/5)
Hence x(h) = 8 - (2h/5)

Let the length of each slice be y(h)= ch + d
h=0, y(0)=5 therefore d=5
h=10, y(10)=3 therefore 10c +5 = 3, c=-(1/5)

Now, the area of each slice A = x(h) x y(h)
= (8 - (2h/5))(5 - (h/5))
= 40 - (18h/5) + (2h^2)/25

Hence, by summing the slices the volume of the solid
∂V = lim(h to 0)∑(h=0 to 10)[40 - (18h/5) + (2h^2)/25]∂h
Therefore V = ∫(0 to 10) 40- (18h/5) + (2h^2)/25 dh
=[40h - (9h^2)/5 + (2h^3)/75](0 to 10)
=(740/3) cm ^3

Thats what i got. Tell me if im wrong

 

[cutemouse]

Yeah you got the answer correctly, but I'm not sure about one thing:

OKay, here goes.
Let the width of each slice be x(h)=ah + b
...
Let the length of each slice be y(h)= ch + d

Where did you get x(h)=ah + b and y(h)= ch + d from?

Thanks

 

[jet]

Well, its just an assumed fact that, because the sides are straight, the relationship between height and width/length will be linear, much in the same way that Area is a quadratic.

Another way to think of it is that the units (cm) are to the first power, whilst the units of area are to the second power and the units of volume are to the third power.