Question

Find the volume of water in the swimming pool. The swimming pool is in circular shape of diameter\(40ft.\)

Short answer

The volume of water in the swimming pool is\(1800\pi {\rm{. }}\)

Step-by-step solution

The concept of the polar rectangle

If\(f\)is a polar rectangle\(R\)given by\(0 \le a \le r \le b,\alpha \le \theta \le \beta {\rm{,}}\)where\(0 \le \beta - \alpha \le 2\pi {\rm{,}}\)then,

…… (1)

If\(g(x)\)is the function of\(x\)and\(h(y)\)is the function of\(y\)then\(\int_a^b {\int_c^d g } (x)h(y)dydx = \int_a^b g (x)dx\int_c^d h (y)dy\) …… (2)

Use the concept of polar rectangle to find the volume of swimming pool

The swimming pool is in a circular shape of diameter\(40{\rm{ft}}.\)

Since the swimming pool is in the circular shape, \(r\)varies from \(0\) to \(20\) and \(\theta \)varies from \(0\) to\(2\pi .\)It is given that the depth of the pool increases linearly from South to North.

Thus, assume the function as\({\rm{h(y) = a y + b}}\) and from the given conditions it is observed that,\(\begin{aligned}{l}h( - 20) = 2\\h(20) = 7\end{aligned}\)

Substitute the values in the equation\({\rm{h(y) = a y + b}}\) and obtain the values of\({\rm{a}}\) and\({\rm{b}}{\rm{.}}\)

Case 1:

\(\begin{aligned}{l}{\rm{ }}h( - 20) = 2\\h(y) = ay + b\\h( - 20) = a( - 20) + b\\2 = - 20a + b\end{aligned}\) …… (3)

Case 2:

\(\begin{aligned}{l}h(20) = 7\\h(y) = ay + b\\h(20) = a(20) + b\\7 = 20a + b\end{aligned}\) …… (4)

Solve the equation (3) and (4). That is, add both the equation yields,

\(\begin{aligned}{l}9 = 2b\\b = \frac{9}{2}\end{aligned}\)

And, substitute the value of\(b\) in equation (3),

\(\begin{aligned}{l}7 = 20a + \frac{9}{2}\\7 - \frac{9}{2} = 20a\\\frac{5}{2} = 20a\\a = \frac{1}{8}\end{aligned}\)

Therefore, \(h(y) = \frac{1}{8}y + \frac{9}{2}\) and by equation (1) the volume of water in the swimming pool becomes,

Integrate the function with respect to\(r\) and\(\theta \) by using the equation (2).

\(\begin{aligned}{l}\int_0^{2\pi } {\int_0^{20} {\frac{1}{8}} } {r^2}\sin \theta drd\theta + \int_0^{2\pi } {\int_0^{20} {\frac{{9r}}{2}} } drd\theta = \frac{1}{8}\int_0^{2\pi } {\sin } \theta d\theta \int_0^{20} {{r^2}} dr + \frac{9}{2}\int_0^{2\pi } d \theta \int_0^{20} r dr\\ = \frac{1}{8}( - \cos \theta )_0^{2\pi }\left( {\frac{{{r^3}}}{3}} \right)_0^{20} + \frac{9}{2}(\theta )_0^{2\pi }\left( {\frac{{{r^2}}}{2}} \right)_0^{20}\end{aligned}\)

Apply the limit of value\(r\) and\(\theta \),

\(\begin{aligned}{l}\int_0^{2\pi } {\int_0^{20} {\frac{1}{8}} } {r^2}\sin \theta drd\theta + \int_0^{2\pi } {\int_0^{20} {\frac{{9r}}{2}} } drd\theta = \frac{1}{8}( - \cos (2\pi ) + \cos (0))\left( {\frac{{{{20}^3}}}{3} - \frac{{{0^3}}}{3}} \right) + \frac{9}{2}(2\pi - 0)\left( {\frac{{{{20}^2}}}{2} - \frac{{{0^2}}}{2}} \right)\\ = \frac{1}{8}( - (1) + 1)\left( {\frac{{8000}}{3}} \right) + \frac{9}{2}(2\pi )\left( {\frac{{400}}{2}} \right)\\ = \frac{1}{8}(0)\left( {\frac{{8000}}{3}} \right) + \frac{9}{2}(2\pi )\left( {\frac{{400}}{2}} \right)\\ = 0 + 1800\pi \\ = 1800\pi \end{aligned}\)

Thus, the volume of water in the swimming pool is\(1800\pi {\rm{. }}\)