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Expert-verified Found in: Page 370 ### Essential Calculus: Early Transcendentals

Book edition 2nd
Author(s) James Stewart
Pages 830 pages
ISBN 9781133112280 # A water storage tank has the shape of a cylinder with diameter $$10{\rm{ft}}$$. It is mounted so that the circular cross-sections are vertical. If the depth of the water is $$7{\rm{ft}}$$, what percentage of the total capacity is being used?

The percentage of usage capacity of water storage tank is $$74.7\%$$.

See the step by step solution

## Equation of circle and Pythagorean Theorem

The equation of circular tank is $${x^2} + {y^2} = r{}^2$$

Pythagorean Theorem $$A{B^2} + B{C^2} = A{C^2}$$

## Find area of cross section of tank

Consider the cross section of the tank as circular shape.

Draw the cross section of storage tank as in figure 1 The equation of circular tank is as follows:

$${x^2} + {y^2} = 25{\rm{ }}$$ $${\rm{(1) }}$$

Rearrange Equation (1)

\begin{aligned}{l}{x^2} = 25 - {y^2}\\x = \sqrt {25 - {y^2}} \end{aligned}

Calculate the area of cross section of circle tank:

$$A = \int_a^b x dy$$ $$(2)$$

Substitute $$( - 5)$$ for a, 2 for $$b$$, and $$\left( {\sqrt {25 - {y^2}} } \right)$$ for $$x$$ in Equation (2).

$$A = \int_{ - 5}^2 {\sqrt {25 - {y^2}} } dy$$ $$(3)$$

Consider $$y = 5\sin \theta$$ (4)

Differentiate both sides of the Equation (4).

$$dy = 5\cos \theta d\theta$$

Substitute $$(5\sin \theta )$$ for $$y$$ and $$(5\cos \theta d\theta )$$ for in Equation (3).

\begin{aligned}A&= \int_{ - 5}^2 {\sqrt {25 - {{(5\sin \theta )}^2}} } (5\cos \theta d\theta )\\&= 5\int_{ - 5}^2 {\sqrt {25 - 25{{\sin }^2}\theta } } \cos \theta d\theta \\&= 5\int_{ - 5}^2 {\sqrt {25\left( {1 - {{\sin }^2}\theta } \right)} } \cos \theta d\theta \\&= 5\int_{ - 5}^2 5 \sqrt {{{\cos }^2}\theta } \cos \theta d\theta \\&= 5 \times 5\int_{ - 5}^2 {\cos } \theta \cos \theta d\theta \\&= 25\int_{ - 5}^2 {{{\cos }^2}} \theta d\theta \\&= 25\int_{ - 5}^2 {\left( {1 + {{\sin }^2}\theta } \right)} d\theta \,\end{aligned}

Integrate Equation

\begin{aligned}A&= 25\left[ {\theta + \frac{1}{2}\sin 2\theta } \right]_{ - 5}^2\\&= 25\left[ {\theta + \frac{1}{2}(2\sin \theta \cos \theta )} \right]_{ - 5}^2\end{aligned}

$$= (25\theta + 25\sin \theta \cos \theta )_{ - 5}^2$$

Rearrange Equation (4).

\begin{aligned}\sin \theta&= \frac{y}{5}\\\theta&= {\sin ^{ - 1}}\left( {\frac{y}{5}} \right)\end{aligned}

## Use Pythagorean Theorem and solve

Represent Equation (7) with Pythagorean Theorem as shown in Figure 2.

$$A{B^2} + B{C^2} = A{C^2}$$ Refer to Figure 2.

\begin{aligned}A{B^2} + {y^2}&= {5^2}\\A{B^2} + {y^2}&= 25\\A{B^2}&= 25 - {y^2}\\AB&= \sqrt {25 - {y^2}} \end{aligned}

Calculate the value of $$\cos \theta$$ from Figure 2.

\begin{aligned}\cos \theta &= \frac{{AB}}{{AC}}\\ &= \frac{{\sqrt {25 - {y^2}} }}{5}\end{aligned}

Substitute $${\sin ^{ - 1}}\left( {\frac{y}{5}} \right)$$ for $$\theta ,\frac{y}{5}$$ for $$\sin \theta$$, and $$\frac{{\sqrt {25 - {y^2}} }}{5}$$ for $$\cos \theta$$ in Equation (6).

\begin{aligned}A&= \left[ {25{{\sin }^{ - 1}}\left( {\frac{y}{5}} \right) + 25 \times \frac{y}{5}\frac{{\sqrt {25 - {y^2}} }}{5}} \right]_{ - 5}^2\\&= \left[ {25{{\sin }^{ - 1}}\left( {\frac{y}{5}} \right) + \frac{{25y\sqrt {25 - {y^2}} }}{{25}}} \right]_{ - 5}^2\\&= \left[ {25{{\sin }^{ - 1}}\left( {\frac{2}{5}} \right) + 2\sqrt {25 - {2^2}} } \right] - \left[ {25{{\sin }^{ - 1}}\left( {\frac{{ - 5}}{5}} \right) + ( - 5)\sqrt {25 - {{( - 5)}^2}} } \right]\\&= (10.29 + 9.16) - ( - 39.25 - 0)\\&= 58.7{\rm{f}}{{\rm{t}}^2}\end{aligned}

## The percentage of usage capacity of water storage tank.

Calculate the area of cross section of circle tank:

$$a = \pi {r^2}{\rm{ (8) }}$$

Substitute 5 for $$r$$ in Equation (8).

\begin{aligned}a = \pi {(5)^2}\\ = 25\pi {\rm{f}}{{\rm{t}}^2}\end{aligned}

Calculate the percentage of the total capacity in used:

$$Pt = \frac{A}{a} \times 100{\rm{ (9) }}$$

Substitute $$58.7{\rm{f}}{{\rm{t}}^2}$$ for $$A$$ and $$25\pi {\rm{f}}{{\rm{t}}^2}$$ for $$a$$ in Equation (8).

$$Pt{\rm{ }} = \frac{{58.7}}{{25\pi }} \times 100 = 74.7\%$$

Hence, the percentage of usage capacity of water storage tank is $$74.7\%$$. ### Want to see more solutions like these? 