A cylindrical gasoline tank is placed so that the axis of the cylinder is horizontal. Find the fluid force on a circular end of the tank if the tank is full, assuming that the diameter is Feet and the gasoline weighs pounds per cubic foot. (Evaluate one integral by a geometric formula And the other by observing that the integrand is an odd function.) Round your answer to two decimal
Places.
A) $135.09 \mathrm { lb }$
B) $128.81 \mathrm { lb }$
C) $64.40 \mathrm { lb }$
D) $1,030.44 \mathrm { lb }$
E) $122.52 \mathrm { lb }$

Question

A cylindrical gasoline tank is placed so that the axis of the cylinder is horizontal. Find the fluid force on a circular end of the tank if the tank is full, assuming that the diameter is Feet and the gasoline weighs pounds per cubic foot. (Evaluate one integral by a geometric formula And the other by observing that the integrand is an odd function.) Round your answer to two decimal
Places. 
A) $135.09 \mathrm { lb }$
B) $128.81 \mathrm { lb }$
C) $64.40 \mathrm { lb }$
D) $1,030.44 \mathrm { lb }$
E) $122.52 \mathrm { lb }$

Quizplus · Accepted Answer

Let the diameter of the tank be $d$, then the radius is $r=\frac{d}{2}$. Let the height of the tank be $h$. The volume of the gasoline in the tank is given by $\pi r^2 h$. Since the tank is full, $h=2r$.

The fluid force on a circular end of the tank can be found by integrating the pressure over the area of the circular end. The pressure at a depth $y$ below the surface of the gasoline is given by $p(y)=ho g (h-y)$, where $ho$ is the density of gasoline, $g$ is the acceleration due to gravity, and $h$ is the total height of the tank. Note that $p(y)$ is an odd function with respect to $y$. Thus, we can find the fluid force on one circular end of the tank and double it to get the total fluid force.

Using geometric formulas, the area of one circular end is $\frac{\pi}{4}d^2$. Thus, the fluid force on one circular end of the tank is given by

\begin{align*}
F &= \int_0^r p(y) \cdot \frac{\pi}{4}d^2\ dy\
&= \int_0^r ho g (h-y) \cdot \frac{\pi}{4}d^2\ dy\
&= ho g \cdot \frac{\pi}{4}d^2 \cdot \int_0^r (h-y) \ dy\
&= ho g \cdot \frac{\pi}{4}d^2 \cdot \left[hy - \frac{y^2}{2}ight]_0^r\
&= ho g \cdot \frac{\pi}{4}d^2 \cdot \left(hr - \frac{r^2}{2}ight)\
&= \frac{1}{2}ho g \pi r^2 h\
&= \frac{1}{2}ho g \pi (\frac{d}{2})^2 (2r)\
&= \frac{ho g\pi d^2}{16}r\
&= \frac{1}{8}ho g \pi d^3
\end{align*}

Substituting in $ho=42$ pounds per cubic foot, $g=32.2$ feet per second squared, and $d=10$ feet, we get

$$F = \frac{1}{8} \cdot 42 \cdot 32.2 \cdot \pi \cdot 10^3 \approx 128.81 	ext{lb}$$

Double this to get the total fluid force, which is $\boxed{	extbf{(B)}}$.

A Cylindrical Gasoline Tank Is Placed So That the Axis 135.09lb135.09 \mathrm { lb }135.09lb

A Cylindrical Gasoline Tank Is Placed So That the Axis $135.09 \mathrm { lb }$