and

This will give you the answer correct to the third decimal place (if you want more accuracy you can include more terms of the related Maclurin series). Therefore, in this case, the previous formulae will become:

and

So there is no trigonometry at all and I think the 9 graders can do the arithmetic operations.

]]>There is a circle theorem which will help for the surface area.

If chords AB and CD, for any circle, intersect at E then AE.EB = CE.ED You can prove this using similar triangles.

So if the width of the surface area is 2x (CD) then x^2 = 16 times 68

You can also use Pythag on triangle CEO. (x^2 = 42^2 - 26^2)

To get the volume you'd need the area of cross section . Let O be the centre of the circle.

You can work out the area of triangle CDO. Then you'd need the area of the sector CBD which requires the angle COD (obtuse). I cannot see a non trig way of getting that with 68 as a measurement. With another length it might be possible.

Bob

]]>If we let d and r to be the water height in the pipe and the radius of the base of the pipe, respectively, then we would have:

where h is the length of the arc of the circle covered by the rain water, and hence the surface area a of the pipe contacted with the water would be given by:

where l is the length of the water pipe. To get the volume of the water in the pipe, we first calculate the area A of the base covered by the water use the integration:

where the integration as been performed using trigonometric substitutions (if you would like to have the details I can provide it to you). Hence the volume of the water in the pipe is:

Note: We get the Volume of the cylinder (the water pipe) in the case that r-d=-r (that is when the pipe is full of rain water).

]]>Determine:

a. The surface area which gets contact with the water

b. The volume of the water (in liters)

So... How do I do? What is the simple way to determine the area of a... truncated circle? (dunno what the proper term is)

]]>