Chapter 6.1, Problem 22E

(a)

To determine

The S of the water volume approximating the water with eight circumscribed solid cylinders.

Answer to Problem 22E

The volume of S is $372.27873{\text{m}}^3$ .

Explanation of Solution

Given information:

Volume of Water in a Reservoir A reservoir shaped like a hemispherical bowl of radius 8 m is filled with water to a depth of 4 m.

Formula used:

The volume of cylinder is given by

  $V=\pi r^{2}h$

Calculation:

Since hemisphere is approximated with cylinders.

So, volume of cylinder is given by

  $V=\pi r^{2}h$

Here,

  $\begin{array}{l}r=\sqrt{64-x^2}\\ h=\Delta x\end{array}$

So, use LRAM with

  $\begin{array}{l}f\left(x\right)=\pi {\left(\sqrt{64-x^2}\right)}^2\\ f\left(x\right)=\pi \left(64-x^2\right)\end{array}$

And interval [4, 8] which is partitioned into eight equal subintervals so, length of a subinterval is 0.5 so, substitutes the value of x with 0.5 differences in f(x) from 4, 4.5.... up to 7.5 and then add and multiply with 0.5 then it is equal to S

  $\begin{array}{l}S=0.5\pi \left[48+43.75+39+33.75+28+21.75+15+7.75\right]\\ S=372.27873{\text{m}}^3\end{array}$

Conclusion:

The volume of S is $372.27873{\text{m}}^3$

(b)

To determine

The percentage of V to the nearest percentage.

Answer to Problem 22E

The percentage error is 11%.

Explanation of Solution

Given information:

Volume of Water in a Reservoir A reservoir shaped like a hemispherical bowl of radius 8 m is filled with water to a depth of 4 m.

Formula used:

The Volume of hemisphere is $V=\frac{2}{3}\pi r^3$

Calculation:

Volume of hemisphere is calculated as

  $\begin{array}{l}V=\frac{320}{3}{\text{m}}^3\\ V=335.1032{\text{m}}^3\end{array}$

Now, for percentage error

  $\begin{array}{l}\frac{V-S}{V}\times 100=0.11\times 100\\ =11\%\end{array}$

Conclusion:

The percentage error is 11%.