help with this integral! if its possible.

[jeca86]

a solid metal object is to be hoisted to the top of a building using a crane. the object is composed of material which weighs 327.36kg per cubic meter. Determine the approximate weight of the object.

Ok so there is a diagram that goes along with it, but i think it would be hard to post up so i will describe it as much as possible. the diagram looks like two funnel shapes were put together, without the stick part of each funnel. So in other words, one funnel is the top with the big opening at the top and the other funnel is the bottom part with the big opening at the bottom. the diameter of the large openings is 120cm. the place where they join has a diameter of 60 cm. the height of the diagram is 136cm. Hopefully someone understands this. I need help on setting this up because i just dont get it. sorry this is so long.

 

[soroban]

Hello, jeca86!

We need more information . . .

A solid metal object is to be hoisted to the top of a building using a crane.
The object is composed of material which weighs 327.36kg per cubic meter.
Determine the approximate weight of the object.

The diagram looks like two funnel shapes were put together, without the stick part of each funnel.
So in other words, one funnel is the top with the big opening at the top
and the other funnel is the bottom part with the big opening at the bottom.
The diameter of the large openings is 120cm. the place where they join has a diameter of 60 cm.
The height of the diagram is 136cm. Hopefully someone understands this.

Can I assume that the slanted sides are straight?

              :  60   :
    - *-------+-------*
    :  \      :      /
    :   \     :     /
   136   *----+----*
    :   /     : 30  \
    :  /      :      \
    - *-------+-------*
                 60

There are two congruent frustums of a cone.

The volume of a frustum is: \(\displaystyle \,V\;=\;\frac{\pi}{3}h\left(R_1^2\,+\,R_1R_2\,+\,R_2^2\right)\)

We have: \(\displaystyle h\,=\,68,\;R_1\,=\,30,\;R_2\,=\,60\)
\(\displaystyle \;\;\)Hence: \(\displaystyle V\:=\:\frac{\pi}{3}(68)\left(30^2\,+\,30\cdot60\,+\,60^2\right)\:=\:142,800\pi\) cm³\(\displaystyle \;=\;0.1428\pi\) m³.

The total volume is: \(\displaystyle V\;=\;2\,\times\,0.1428\;=\;0.2856\pi\) m³.

At 327.36 kg/m³, the total weight is: \(\displaystyle \,327.36\,\times\,0.2856\pi\:\approx\;293.7\) kg.

 

[jeca86]

thanks so much for all your help.