3D shapes (Volume, Surface Area)

This question is a quite a common problem about geometrical solids. We can break the shape into a few sections. There is a hemisphere cap, and a cylindrical body, but I will go further and break it into the SWEPT volume and the CLEARANCE volume.

The question explains that the clearance volume is 95% of the cylinder, so lets find that first:

So now we can work on finding the clearance volume, this is made up of two sections, the hemisphere above and a small 4cm cylinder:

Now lets work on the SWEPT volume, its a simple cylinder

to find "the clearance volume as a percentage of the swept volume" we will need to divide one by the other and then multiply by 100 to turn into a percentage:

First we will find the volume of the bearing. And then we will find the volume of the cone

We can think of this bearing, as a big external cylinder minus a small internall cylinder.

I have done the whole calculation in mm.

Next we will look at the volume of the cone

The only thing left to do is to find out how many cones we need, so we divide:

As the cube and the sphere have the same mass and are made from the same material, we can assume them to be of the same volume. Also, we have no dimensions of the cube, so we must start with the sphere. We can do the calculation in mm or in cm, this time I have done it in mm.

now that we know the volume required, let's find the length of the rod

we can use the formula for the volume of a cylinder. We KNOW the volume, and we can transpose to find for h

This is a simple question. They are usually hidden towards the end of 2 or 3 parters, so don't give up! We just apply a simple formula for the curved surface area:

Ok, so we need to calculate the surface area of the sphere and the cone, the measurements are given, so its all pretty straight forwards, just remember to convert those diameters into radii (radiuses!)

So you can see that the Cone is larger!

First let's draw a diagram and write down the formula for the volume of a cone

The question is asking for the volume in meters cubed. It's best we start that way, so let's convert these measurements to meters. We will also need the base, so we can use Pythagoras to calculate the base:

so now we have the height of the cone is 0.3m and the radius is 0.16m, we can calculate the volume in meters cubed directly.

We need to find the angle from the base. We have everything we need, and we can keep the measurements in the original mm to avoid typos:

Let's use SOH CAH TOA to solve for the missing angle