Question

Hi my geometry question is- A new part needs to be designed for a machine. The cube center of the part has a side length of 2 in. Each cylinder off the sides of the cube has a diameter and height that matches the sides of the cube with a 1 inch hole drilled out 1 inch deep. There are 4 cylinders. How much metal is needed to make one part?

Answer 1

Answer on Question #77820 – Math – Geometry

Question

Hi my geometry question is- A new part needs to be designed for a machine. The cube center of the part has a side length of 2 in. Each cylinder off the sides of the cube has a diameter and height that matches the sides of the cube with a 1 inch hole drilled out 1 inch deep. There are 4 cylinders. How much metal is needed to make one part?

Solution

As each of four cylinders is 1 in deep (the height of each cylinder is 1 in) we can say that every two cylinders form one cylinder if height 2 in, i.e. the same value that the side of a cube is. A solid formed by intersection of these two cylinders is the one that is drilled off the initial cube. The first hole is centered along the x axis, the second hole is centered along the y axis. The solid common to two (or three) right circular cylinders of equal radii intersecting at right angles is called the Steinmetz solid. Two cylinders intersecting at right angles are called a bicylinder.

The volume of the initial cube is:

$$V _ {\text {c u b e}} = a ^ {3} = 2 ^ {3} \text {i n} ^ {3} = 8 \text {i n} ^ {3}$$

To find common volume of two intersecting cylinders we should subtract volume of bicylinder from the volume of the two cylinders added together.

The volume of bicylinder is:

$$V _ {b i c y l} = \frac {1 6}{3} r ^ {3}$$

where $r$ is the radius of a cylinder.

As diameter of a cylinder is 1 in then radius is

$$r = \frac{d}{2} = \frac{1}{2} \quad \text{(in)}.$$ $$V_{bicyl} = \frac{16}{3} \times \left(\frac{1}{2}\right)^3 = \frac{16}{3 \times 8} = \frac{2}{3} \quad \text{(in}^3\text{)}$$

The volume of one cylinder is:

$$V_{cyl} = \pi r^2 h$$ $$V_{cyl} = \pi \times \left(\frac{1}{2}\right)^2 \times 2 = \pi \times \frac{1}{2} \quad \text{(in}^3\text{)}$$

The volume of the two cylinders added together is:

$$2 V_{cyl} = 2 \times \pi \times \frac{1}{2} = \pi \quad \text{(in}^3\text{)}$$

The volume of a solid obtained is:

$$V = V_{cube} - \left(2 V_{cyl} - V_{bicyl}\right) = 8 - \left(\pi - \frac{2}{3}\right) = 8 + \frac{2}{3} - \pi = \frac{26}{3} - \pi \cong 5.53 \quad \text{(in}^3\text{)}$$

The volume of a metal required is $5.53 \text{ in}^3$.

Answer: $5.53 \text{ in}^3$.

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