Section 8.3 in stewart calculus problem 8

1. May 24, 2014

TitoSmooth

So I'm working On hydrostatic pressure. The problem #8 in stewart

Is a trapezoid with the base above water.

4

\------------------------/ 1(above watee)
\ / 1(in watee)
\ /
---------------
2

So if the base was a number say
50. N the bottom was 15.

I COULD split the trapezoid in half.

15 +10 base

And bottom would be 15.

Similar triangles for the rectangle slab would give me. A/(height of object minus height in water-Y)=10/20

Then I'll solve for A.

The problem here is that the number 4 does not break up as nicely as this does. How would I relate the sides?

Thanks

2. May 24, 2014

Simon Bridge

I'm afraid you are not making sense.
You need to relate the numbers to dimensions in order to make sense.

A trapezoidal trough may have length L, bottom-width B, top-width W, and overall-height H.
It may be filled with water to height y.

We'd expect W>B, and L>W, so it's like a pig-trough. But it does not have to be that way.

w=B+(W-B)y/H would be the width of the trough at the water-height,
and the volume of water in the trough would be V=y(B+w)L/2.

Or it could be a trapezoidal boat, like a punt with a flat prow and stern.
In that case, w is the width of the punt at the water-line and V is the volume of water displaced.

Now... what are you talking about?