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Q32E

Expert-verifiedFound in: Page 370

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 \(74.7\% \).

**The equation of circular tank is** \({x^2} + {y^2} = r{}^2\)

**Pythagorean Theorem** \(A{B^2} + B{C^2} = A{C^2}\)

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).

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

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).

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

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 \(74.7\% \).

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