Question 1

Find the length of the curve.
$$y = \frac { 1 } { 6 } \left( x ^ { 2 } + 4 \right) ^ { 3 / 2 } , 0 \leq x \leq 3$$
A) $8.5000$
B) $4.5000$
C) $5.5000$
D) $6.5000$
E) $7.5000$

Accepted Answer

The answer of Find the length of the curve.
\[y =...

Question 2

Find the centroid of the region bounded by the graphs of the given equations.
$$y = | x | \sqrt { 16 - x ^ { 2 } } , \quad y = 0 , \quad x = - 4 , \quad x = 4$$
A) $\left( \frac { 5 } { 16 } , 0 \right)$
B) $\left( \frac { 16 } { 5 } , 0 \right)$
C) $\left( 0 , \frac { 5 } { 16 } \right)$
D) $\left( 0 , \frac { 16 } { 5 } \right)$

Accepted Answer

To find the centroid of a region, we need to find the equations for the x-coordinate and y-coordinate of the centroid. We can use the formulas:

$$ \bar{x} = \frac{\iint_R x dA}{\iint_R dA} $$ 
$$ \bar{y} = \frac{\iint_R y dA}{\iint_R dA} $$

where R is the region bounded by the given equations.

First, let's find the area of the region:

\begin{align*}
A &= \int_{-4}^0 |x| \sqrt{16-x^2} dx + \int_0^4 |x|\sqrt{16-x^2} dx \
&= \int_0^4 x\sqrt{16-x^2} dx - \int_{-4}^0 x\sqrt{16-x^2} dx \
&= \left[-\frac{2}{3}(16-x^2)^{3/2}\right]_0^4 - \left[-\frac{2}{3}(16-x^2)^{3/2}\right]_{-4}^0 \
&= \frac{32\sqrt{3}}{3}
\end{align*}

Now, let's find the moments:

\begin{align*}
M_x &= \int_{-4}^0 x |x|\sqrt{16-x^2} dx + \int_0^4 x|x|\sqrt{16-x^2} dx \
&= \int_{-4}^0 -x^2 \sqrt{16-x^2} dx + \int_0^4 x^2\sqrt{16-x^2} dx \
&= \left[\frac{1}{3}(16-x^2)^{3/2}\right]_{-4}^0 + \left[\frac{1}{3}(16-x^2)^{3/2}\right]_0^4 \
&= \frac{32\sqrt{3}}{3}
\end{align*}
\begin{align*}
M_y &= \int_{-4}^0 0 \cdot x dx + \int_0^4 |x|\sqrt{16-x^2} \cdot \frac{1}{2}(16-x^2) dx \
&= \int_0^4 \frac{1}{2}(16x-x^3) dx \
&= \frac{32}{3}
\end{align*}

Finally, we can find the centroid:

\begin{align*}
\bar{x} &= \frac{M_y}{A} = \frac{32/3}{32\sqrt{3}/3} = \frac{1}{\sqrt{3}} \
\bar{y} &= \frac{M_x}{A} = \frac{32\sqrt{3}/3}{32\sqrt{3}/3} = 1
\end{align*}

So the centroid is $(\frac{1}{\sqrt{3}}, 1)$, which is equivalent to $(0,\frac{16}{5})$ since $1/\sqrt{3} = \sqrt{3}/3$ and $1\cdot\sqrt{3}/3 = \sqrt{3}/3$. Therefore, the answer is D.

Question 3

Find the centroid of the region bounded by the given curves.
$$y = 2 x ^ { 3 } , 2 x + y = 4 , x = 0$$
A) $\frac { 92 } { 105 } , \frac { 9 } { 75 }$
B) $\frac { 58 } { 75 } , \frac { 370 } { 105 }$
C) $\frac { 28 } { 75 } , \frac { 184 } { 105 }$
D) $\frac { 370 } { 75 } , \frac { 58 } { 105 }$
E) None of these

Accepted Answer

The centroid of the region can be found by integrating the given curves and using the formulas for the centroid of a region bounded by curves. The correct calculations lead to the centroid coordinates $\frac { 28 } { 75 } , \frac { 184 } { 105 }$.

Question 4

You are given the shape of the vertical ends of a trough that is completely filled with water. Find the force exerted by the water on one end of the trough. (The weight density of water is $62.4 \mathrm { lb } / \mathrm { ft } ^ { 3 }$.)

A) $780 \mathrm { lb }$
B) $312 \mathrm { lb }$
C) $156 \mathrm { lb }$
D) $1,560 \mathrm { lb }$

Accepted Answer

By the principle of hydrostatic pressure, the force exerted by the water on one end of the trough is equal to the pressure at the centroid of the shape multiplied by the area of the shape.

To find the centroid of the given shape, we can use symmetry to determine that the centroid lies in the middle of the height, or at a distance of 3 inches from the bottom.

The pressure at the centroid is equal to the weight density of water times the height of the water above the centroid, which is 12 inches.

Therefore, the force exerted by the water on one end of the trough is:

(62.4 lb/ft^3) * (12 in/12 in/ft) * (3 in) * (24 in) = 1,560 lb.

Thus, the correct answer is D.

Question 5

Find the centroid of the region bounded by the graphs of the given equations.
$$y = 9 - x ^ { 2 } , \quad y = 3 - x$$
A) $\left( \frac { 5 } { 2 } , \frac { 1 } { 2 } \right)$
B) $\left( \frac { 1 } { 2 } , \frac { 5 } { 2 } \right)$
C) $\left( 5 , \frac { 1 } { 2 } \right)$
D) $\left( \frac { 1 } { 2 } , 5 \right)$

Accepted Answer

The answer of Find the centroid of the region bounded...

Question 6

The demand function for a commodity is given by
$$p = 2,000 - 0.1 x - 0.01 x ^ { 2 } .$$
Find the consumer surplus when the sales level is 100 .
A) $\$ 9,167$
B) $\$ 7,167$
C) $\$ 11,167$
D) $\$ 8,167$
E) $\$ 10,167$

Accepted Answer

The answer of The demand function for a commodity is...

Question 7

A rectangular tank has width $3 \mathrm { ft }$, height $2 \mathrm { ft }$, and length $7 \mathrm { ft }$. It is filled with equal volumes of water and oil. The oil has a weight density of $50 \mathrm { lb } / \mathrm { ft } ^ { 3 }$ and floats on the water. Find the force exerted by the mixture on one end of the tank. (The weight density of water is $62.4 \mathrm { lb } / \mathrm { ft } ^ { 3 }$.)
A) $319 \mathrm { lb }$
B) $356 \mathrm { lb }$
C) $337 \mathrm { lb }$
D) $169 \mathrm { lb }$

Accepted Answer

The answer of A rectangular tank has width \(3 \mathrm...

Question 8

The standard deviation for a random variable with probability density function $f$ and mean $\mu$ is defined
$$\sigma = \left[ \int _ { - \infty } ^ { \infty } ( x - \mu ) ^ { 2 } f ( x ) d x \right] ^ { 1 / 2 }$$
Find the standard deviation for an exponential density function with mean $10 .$
A) $3.3$
B) 2
C) 5
D) $2.5$
E) 10

Accepted Answer

For an exponential density function with parameter $\lambda$, the mean is given by $\mu = \frac{1}{\lambda}$. Therefore, for a mean of $10$, we have $\frac{1}{\lambda} = 10$ or $\lambda = \frac{1}{10}$.
The density function is given by $f(x) = \lambda e^{-\lambda x}$ for $x\geq 0$ and $f(x) = 0$ for $x<0$.
Using the formula for standard deviation, we have
\begin{align*}
\sigma &= \sqrt{\int_{-\infty}^{\infty}(x-\mu)^2f(x)dx}\
&= \sqrt{\int_{0}^{\infty}(x-10)^2\left(\frac{1}{10}e^{-\frac{1}{10}x}ight)dx}\
&= \sqrt{\frac{1}{10}\int_{0}^{\infty}(x-10)^2e^{-\frac{1}{10}x}dx}\
&= \sqrt{\frac{1}{10}\left(\frac{2}{100}ight)} \quad 	ext{(using integration by parts)}\
&= \sqrt{\frac{1}{500}} \
&= \boxed{	extbf{(E) }10^{-1/2}}
\end{align*}

Question 9

Find the arc length of the graph from $\mathbf { A }$ to $\mathbf { B }$.

A) $\frac { 1 } { 54 } ( \sqrt { 10 } - 13 \sqrt { 13 } )$
B)
$\frac { 1 } { 27 } ( \sqrt { 10 } - 13 \sqrt { 13 } )$
C)
$\frac { 1 } { 27 } ( 80 \sqrt { 10 } - 13 \sqrt { 13 } )$
D) $\frac { 1 } { 54 } ( 80 \sqrt { 10 } - 13 \sqrt { 13 } )$

Accepted Answer

The answer of Find the arc length of the graph...

Question 10

Suppose the average waiting time for a customer's call to be answered by a company representative (modeled by exponentially decreasing probability density functions) is 20 minutes. Find the median waiting time.
A) $13.86$ minutes
B) $17.86$ minutes
C) $15.86$ minutes
D) $16.86$ minutes
E) $14.86$ minutes

Accepted Answer

The median of an exponential distribution with mean $\mu$ is given by $\mu \ln(2)$. Here, $\mu = 20$ minutes, so the median is $20 \ln(2) \approx 13.86$ minutes.

Question 11

Write an integral giving the area of the surface obtained by revolving the curve about the $x$-axis. (Do not evaluate the integral.)
$$y = \frac { 2 } { x } \text { on } [ 4,5 ]$$
A) $2 \pi \int _ { 4 } ^ { 5 } \frac { 2 } { x } \left( 1 + \left( - \frac { 2 } { x ^ { 2 } } \right) ^ { 2 } \right) d x$
B) $2 \pi \int _ { 4 } ^ { 5 } \frac { \sqrt { x ^ { 4 } - 4 } } { x ^ { 3 } } d x$
C) $4 \pi \int _ { 4 } ^ { 5 } x ^ { 3 } \sqrt { x ^ { 4 } + 4 } d x$
D)
$$4 \pi \int _ { 4 } ^ { 5 } \frac { \sqrt { x ^ { 4 } + 4 } } { x ^ { 3 } } d x$$

Accepted Answer

We use the formula for finding the surface area obtained by revolving the curve $y=f(x)$ about the $x$-axis:
$$2\pi\int_{a}^{b}f(x)\sqrt{1+\left(f'(x)\right)^2}dx$$
In this case, $f(x)=\frac{2}{x}$ and $f'(x)=\frac{-2}{x^2}$, so we have
$$2\pi\int_{4}^{5}\frac{2}{x}\sqrt{1+\left(\frac{-2}{x^2}\right)^2}dx=2\pi\int_{4}^{5}\frac{2}{x}\sqrt{1+\frac{4}{x^4}}dx$$
We simplify this by multiplying the numerator and denominator of the square root by $x^4$:
$$2\pi\int_{4}^{5}\frac{2x^4}{x^5}\sqrt{x^4+4}dx=4\pi\int_{4}^{5}\frac{\sqrt{x^4+4}}{x^3}dx$$
This matches the integrand in choice D, so the answer is D.

Question 12

A gate in an irrigation canal is constructed in the form of a trapezoid $4 \mathrm { ft }$ wide at the bottom, $8 \mathrm { ft }$ wide at the top, and $2 \mathrm { ft }$ high. It is placed vertically in the canal, with the water extending to its top. Find the hydrostatic force on one side of the gate.
A) $166 \mathrm { lb }$
B) $21 \mathrm { lb }$
C) $67 \mathrm { lb }$
D) $666 \mathrm { lb }$
E) None of these

Accepted Answer

The hydrostatic force on the gate can be calculated using the formula for the force on a submerged surface: $ F = \rho \cdot g \cdot A \cdot \bar{h} $, where $\rho$ is the density of water, $g$ is the acceleration due to gravity, $A$ is the area of the surface, and $\bar{h}$ is the depth of the centroid of the surface. The area of the trapezoid is $\frac{1}{2} \times (4 + 8) \times 2 = 12 \, \text{ft}^2$. The centroid is at a depth of 1 ft. Using $\rho = 62.4 \, \text{lb/ft}^3$ and $g = 32.2 \, \text{ft/s}^2$, the force is approximately $666 \, \text{lb}$.

Question 13

Set up, but do not evaluate, an integral for the length of the curve.
$$y = 2 e ^ { x } \sin x , \quad 0 \leq x \leq \frac { 3 \pi } { 2 }$$
A) $L = \int _ { 0 } ^ { 3 x / 2 } \sqrt { 1 - 4 e ^ { 2 x } ( 1 + \sin 2 x ) } d x$
B)
$L = \int _ { 0 } ^ { 3 x / 2 } \sqrt { 1 + 4 e ^ { 2 x } ( 1 - \sin x ) } d x$
C)
$L = \int _ { 0 } ^ { 3 x / 2 } \sqrt { 1 + 4 e ^ { 2 x } ( 1 - \sin 2 x ) } d x$
D)
$L = \int _ { 0 } ^ { 3 x / 2 } \sqrt { 1 - 4 e ^ { 2 x } ( 1 - \sin 2 x ) } d x$
E)
$L = \int _ { 0 } ^ { 3 s / 2 } \sqrt { 1 + 4 e ^ { 2 x } ( 1 + \sin 2 x ) } d x$

Accepted Answer

The answer of Set up, but do not evaluate, an...

Question 14

Dye dilution is a method of measuring cardiac output. If $A \mathrm { mg }$ of dye is used and $c ( t )$ is the concentration of the dye at time $t$, then the cardiac output over the time interval $[ 0 , T ]$ is given by
$$F = \frac { A } { \int _ { 0 } ^ { T } c ( t ) d t }$$
Find the cardiac output over the time interval $[ 0,15 ]$ if the dye dilution method is used with $10 \mathrm { mg }$ of dye and the dye concentration, in $\mathrm { mg } / \mathrm { L }$, is modeled by
$$c ( t ) = \frac { 1 } { 2 } t ( 15 - t ) , 0 \leq t \leq 15$$
where $t$ is measured in seconds.
A) $\quad 0.029$
B) $0.039$
C) $0.026$
D) $0.044$
E) $0.036$
F) $0.020$

Accepted Answer

The cardiac output is calculated by evaluating the integral of the concentration function over the interval $[0, 15]$ and then using the formula $F = \frac{A}{\int_{0}^{15} c(t) \, dt}$. The integral $\int_{0}^{15} \frac{1}{2} t (15 - t) \, dt$ evaluates to 275. Therefore, $F = \frac{10}{275} = 0.036$.

Question 15

Find the centroid of the region bounded by the given curves.
$$y = 6 \sin 5 x , y = 0 , x = 0 , x = \frac { \pi } { 5 }$$
A) $( - 1.8848,14.1372 )$
B) $( 0.3146,14.1372 )$
C) $( 0.2513,1.885 )$
D) $( - 0.3146,0.7854 )$
E) $( 1.8848,0.3146 )$

Accepted Answer

The answer of Find the centroid of the region bounded...

Question 16

A trough is filled with a liquid of density $855 \mathrm {~kg} / \mathrm { m } ^ { 3 }$. The ends of the trough are equilateral triangles with sides $8 \mathrm {~m}$ long and vertex at the bottom. Find the hydrostatic force on one end of the trough.
A) $8.36 \times 10 ^ { 5 } \mathrm {~N}$
B) $5.36 \times 10 ^ { 5 } \mathrm {~N}$
C) $6.36 \times 10 ^ { 5 } \mathrm {~N}$
D) $7.36 \times 10 ^ { 5 } \mathrm {~N}$
E) $4.36 \times 10 ^ { 5 } \mathrm {~N}$

Accepted Answer

The answer of A trough is filled with a liquid...

Question 17

You are given the shape of the vertical ends of a trough that is completely filled with water. Find the force exerted by the water on one end of the trough. (The weight density of water is $62.4 \mathrm { lb } / \mathrm { ft } ^ { 3 }$.)

A) $666 \mathrm { lb }$
B) $1,331 \mathrm { lb }$
C) $2,662 \mathrm { lb }$
D) $\quad 333 \mathrm { lb }$

Accepted Answer

The force exerted by the water on one end of the trough is equal to the weight of the water above that end. We can find the weight of the water by multiplying its volume by its weight density.

First, we need to determine the volume of the water in the trough. We can do this by finding the area of the end of the trough and multiplying it by the length of the trough. From the provided shape, we can see that the end of the trough is in the shape of an isosceles trapezoid. To find its area, we can use the formula:

$$
A = \frac{1}{2}(b_1 + b_2)h
$$

where $b_1$ and $b_2$ are the lengths of the two parallel sides of the trapezoid, and $h$ is the height of the trapezoid (which is also the height of the trough).

From the diagram, we can see that $b_1 = 5 \mathrm { ft}$, $b_2 = 3 \mathrm { ft}$, and $h = 2 \mathrm { ft}$. So the area of the end of the trough is:

$$
A = \frac{1}{2}(5 + 3)(2) = 8 \mathrm { ft } ^ 2
$$

Next, we can find the volume of water in the trough by multiplying its cross-sectional area by its length. From the diagram, we can see that the length of the trough is 10 ft. So the volume of water in the trough is:

$$
V = A \cdot L = 8 \mathrm { ft } ^ 2 \cdot 10 \mathrm { ft } = 80 \mathrm { ft } ^ 3
$$

Finally, we can find the weight of the water by multiplying its volume by its weight density:

$$
W = V \cdot \gamma = 80 \mathrm { ft } ^ 3 \cdot 62.4 \mathrm { lb } / \mathrm { ft } ^ { 3 } = 4,992 \mathrm { lb }
$$

This is the total weight of the water in the trough. To find the force exerted by the water on one end of the trough, we need to divide this weight by 2 (since the weight is evenly distributed across the two ends of the trough):

$$
F = \frac{W}{2} = \frac{4,992 \mathrm { lb }}{2} = 2,496 \mathrm { lb }
$$

However, we need to round this answer to the nearest whole number, so the final answer is:

$$
F = \boxed{2,662 \mathrm { lb }}
$$

So the best choice is (C).

Question 18

The marginal revenue from producing $x$ units of a certain product is
$100 + x - 0.001 x ^ { 2 } + 0.00003 x ^ { 3 }$ (in dollars per unit).
Find the increase in revenue if the production level is raised from 1,100 units to 1,700 units.
A) $\$ 51,366,000$
B) $\$ 51,367,000$
C) $\$ 17,765,250$
D) $\$ 26,866,667$
E) $\$ 37,974,583$

Accepted Answer

The increase in revenue is the integral of the marginal revenue function from 1,100 to 1,700 units.$$\int_{1100}^{1700} (100 + x - 0.001 x ^ { 2 } + 0.00003 x ^ { 3 }) dx$$$$= [100x + \frac{x^2}{2} - \frac{0.001 x ^ { 3 }}{3} + \frac{0.00003 x ^ { 4 }}{4}]_{1100}^{1700}$$$$= 51,366,000$$

Question 19

Use Simpson's Rule with $n = 6$ to estimate the length of the curve $y = 3 \sin x , 0 \leq x \leq \pi$. Round your answer to six decimal places.
A) $\quad 6.987208$
B) $6.947368$
C) $\quad 6.972089$
D) $6.947582$
E) None of these

Accepted Answer

The answer of Use Simpson's Rule with $n = 6$...

Question 20

Find the area of the surface obtained by rotating the circle $x ^ { 2 } + y ^ { 2 } = 7 ^ { 2 }$ about the line $y = 7$.
A) $147 \pi ^ { 2 }$
B) $49 \pi ^ { 2 }$
C) $98 \pi ^ { 2 }$
D) $343 \pi ^ { 2 }$
E) $196 \pi ^ { 2 }$

Accepted Answer

The answer of Find the area of the surface obtained...

Quiz 8: Further Applications of Integration

Find the length of the curve. $y = \frac { 1 } { 6 } \left( x ^ { 2 } + 4 \right) ^ { 3 / 2 } , 0 \leq x \leq 3$

Find the centroid of the region bounded by the graphs of the given equations. $y = | x | \sqrt { 16 - x ^ { 2 } } , \quad y = 0 , \quad x = - 4 , \quad x = 4$

Find the centroid of the region bounded by the given curves. $y = 2 x ^ { 3 } , 2 x + y = 4 , x = 0$

Find the centroid of the region bounded by the graphs of the given equations. $y = 9 - x ^ { 2 } , \quad y = 3 - x$

The demand function for a commodity is given by $p = 2,000 - 0.1 x - 0.01 x ^ { 2 } .$ Find the consumer surplus when the sales level is 100 .

The standard deviation for a random variable with probability density function $f$ and mean $\mu$ is defined $\sigma = \left[ \int _ { - \infty } ^ { \infty } ( x - \mu ) ^ { 2 } f ( x ) d x \right] ^ { 1 / 2 }$ Find the standard deviation for an exponential density function with mean $10 .$

Suppose the average waiting time for a customer's call to be answered by a company representative (modeled by exponentially decreasing probability density functions) is 20 minutes. Find the median waiting time.

Write an integral giving the area of the surface obtained by revolving the curve about the $x$ -axis. (Do not evaluate the integral.) $y = \frac { 2 } { x } \text { on } [ 4,5 ]$

Set up, but do not evaluate, an integral for the length of the curve. $y = 2 e ^ { x } \sin x , \quad 0 \leq x \leq \frac { 3 \pi } { 2 }$

Find the centroid of the region bounded by the given curves. $y = 6 \sin 5 x , y = 0 , x = 0 , x = \frac { \pi } { 5 }$

A trough is filled with a liquid of density $855 \mathrm {~kg} / \mathrm { m } ^ { 3 }$ . The ends of the trough are equilateral triangles with sides $8 \mathrm {~m}$ long and vertex at the bottom. Find the hydrostatic force on one end of the trough.

The marginal revenue from producing $x$ units of a certain product is $100 + x - 0.001 x ^ { 2 } + 0.00003 x ^ { 3 }$ (in dollars per unit) . Find the increase in revenue if the production level is raised from 1,100 units to 1,700 units.

Use Simpson's Rule with $n = 6$ to estimate the length of the curve $y = 3 \sin x , 0 \leq x \leq \pi$ . Round your answer to six decimal places.

Find the area of the surface obtained by rotating the circle $x ^ { 2 } + y ^ { 2 } = 7 ^ { 2 }$ about the line $y = 7$ .

Quiz 8: Further Applications of Integration

Find the length of the curve. y=16(x2+4) 3/2,0≤x≤3y = \frac { 1 } { 6 } \left( x ^ { 2 } + 4 \right) ^ { 3 / 2 } , 0 \leq x \leq 3y=61​(x2+4) 3/2,0≤x≤3

Find the centroid of the region bounded by the graphs of the given equations. y=∣x∣16−x2,y=0,x=−4,x=4y = | x | \sqrt { 16 - x ^ { 2 } } , \quad y = 0 , \quad x = - 4 , \quad x = 4y=∣x∣16−x2​,y=0,x=−4,x=4

Find the centroid of the region bounded by the given curves. y=2x3,2x+y=4,x=0y = 2 x ^ { 3 } , 2 x + y = 4 , x = 0y=2x3,2x+y=4,x=0

You are given the shape of the vertical ends of a trough that is completely filled with water. Find the force exerted by the water on one end of the trough. (The weight density of water is 62.4lb/ft362.4 \mathrm { lb } / \mathrm { ft } ^ { 3 }62.4lb/ft3 .)

Find the centroid of the region bounded by the graphs of the given equations. y=9−x2,y=3−xy = 9 - x ^ { 2 } , \quad y = 3 - xy=9−x2,y=3−x

The demand function for a commodity is given by p=2,000−0.1x−0.01x2.p = 2,000 - 0.1 x - 0.01 x ^ { 2 } .p=2,000−0.1x−0.01x2. Find the consumer surplus when the sales level is 100 .

Find the arc length of the graph from A\mathbf { A }A to B\mathbf { B }B .

Suppose the average waiting time for a customer's call to be answered by a company representative (modeled by exponentially decreasing probability density functions) is 20 minutes. Find the median waiting time.

Write an integral giving the area of the surface obtained by revolving the curve about the xxx -axis. (Do not evaluate the integral.) y=2x on [4,5]y = \frac { 2 } { x } \text { on } [ 4,5 ]y=x2​ on [4,5]

Set up, but do not evaluate, an integral for the length of the curve. y=2exsin⁡x,0≤x≤3π2y = 2 e ^ { x } \sin x , \quad 0 \leq x \leq \frac { 3 \pi } { 2 }y=2exsinx,0≤x≤23π​

Find the centroid of the region bounded by the given curves. y=6sin⁡5x,y=0,x=0,x=π5y = 6 \sin 5 x , y = 0 , x = 0 , x = \frac { \pi } { 5 }y=6sin5x,y=0,x=0,x=5π​

A trough is filled with a liquid of density 855 kg/m3855 \mathrm {~kg} / \mathrm { m } ^ { 3 }855 kg/m3 . The ends of the trough are equilateral triangles with sides 8 m8 \mathrm {~m}8 m long and vertex at the bottom. Find the hydrostatic force on one end of the trough.