a1.2m internal diameter pipe is running almost full?

Question

what is the area of free space above the water level if the height is 1.05m

Answer 1

If the pipe were full, the area would be:

A = πr² = π(0.6 m)² = 0.36π m²

But it's short by 1.2 m - 1.05 m = 0.15 m.

If you draw a picture of the situation, then draw radii to where the water meets the pipe, you'll see that you have an isosceles triangle with two equal sides of length 0.6 m (the radius), and height 0.6 m - 0.15 m = 0.45 m.

You can split that triangle into two right triangles, where 0.45 m is the height and 0.6 m is the hypotenuse.

So to find the center angle of the big triangle, you can find 1/2 of it using one of the two right triangles:

cos θ/2 = 0.45 m/0.6 m = 0.75
θ/2 = arccos 0.75 = 41.41°
θ = 82.82°

So from this, you can calculate the area of the "piece of the pie" using:

Ap = 0.36π m² (82.82°/360°) = 0.0828π m²

Now you can calculate the area of the triangle of water and subtract it from the area of the piece of pie to get your answer.

You already have the height (0.45 m), and you can calculate 1/2 the base using the Pythagorean theorem with the radius as the hypotenuse:

(0.45 m)² + b² = (0.6 m)²
b² = 0.1575 m²
b = 0.3969 m

So the length of the base of the triangle is 2(0.3969 m) = 0.7937 m

The area is 1/2(0.7937 m)(0.45 m) = 0.7586 m²

So the area of the of the free space should be:

0.0828π m² - 0.7586 m² = 0.0816 m²

Hopefully I finally got it right.

Answer 2

Height (h) about the water - (1∙2 -1∙05) = 0∙15m
Diameter - 1∙2m
Radius - 0∙6m
Volume - Vol.

I think the question should be, the Volume of free space about the water.
Vol. = π r² h
Vol. = π (0∙6)² (0∙15)
Vol. = 0∙169 646 003 m^3
Vol. ≈ 0∙17 m^3