Hello, geeko!

A swimming pool is 3 ft. deep at the shallow end, 8 ft. deep at the deep end,

40 ft. long, 30 ft. wide, and the bottom is an incline plane.

Express the volume $\displaystyle V$ of water in the pool as a function of the height $\displaystyle h$

of the water at the deep end. (Hint: $\displaystyle V$ will be a piecewise-defined function) Code:

40
- * - - - - - - - - - - - - - - - - - *
: | |
3 | | 3
: | 40 |
- * - - - - - - - - - - - - - - - - - *
: | x *
: + - - - - - - - - - - - *
6 |:::::::::::::::::*
: h|:::::::::::*
: |:::::*
- *

For the first six feet, the volume of water is a triangular prism.

. . Its volume is: .$\displaystyle \text{Area of triangle} \times \text{30-foot width}$

From the similar right triangles: .$\displaystyle \frac{x}{h} = \frac{40}{6} \quad\Rightarrow\quad x \:=\:\frac{20}{3}h$ .[1]

The area of the triangle is: .$\displaystyle A \:=\:\frac{1}{2}hx $

Substitute [1]: .$\displaystyle A \:=\:\frac{1}{2}h\left(\frac{20}{3}h\right) \:=\;\frac{10}{3}h^2$

Then the volume is: .$\displaystyle V \:=\:\frac{10}{3}h^2 \times 30 \:=\:100h^2$

. . $\displaystyle \boxed{\text{For }0 \leq h \leq 6\!:\;\;V \;=\;100h^2\text{ ft}^3}$

For the last three feet of water, the bottom (triangular) portion is already filled.

. . It contains: .$\displaystyle 100(6^2) \:=\:3600$ ft³ of water.

The upper portion is a rectangular slab of water:

. . length 40, width 30, and height $\displaystyle h-6$

Its volume is: .$\displaystyle 40 \times 30 \times (h-6) \:=\:1200(h-6)$ ft³.

The total volume is: .$\displaystyle V \:=\:1200(h-6) + 3600$

. . $\displaystyle \boxed{\text{For }6 \leq h \leq 9\!:\;\;V \;=\;1200(h-3)\text{ ft}^3} $

Therefore: .$\displaystyle V \;=\;\bigg\{ \begin{array}{ccc}100h^2 & & 0 \leq h \leq 6 \\ 1200(h-3) && 6 \leq h \leq 9 \end{array} $