# Thread: Application of definite integrals

1. ## Application of definite integrals

1.)A cable hangs between two poles of equal height
and 20 feet apart. Set up a coordinate system where the poles are placed at x =
-10 and x = 10, where x is measured in feet. The height (in feet) of the cable
at position x is h(x) = 12 \cosh(x/12),where \cosh(x) = (e^x + e^{-x})/2 is
the hyperbolic cosine.
How long is the cable in feet?

2.)A bucket of water of mass 20 kg is pulled at constant velocity up to a platform 40 meters
above the ground. This takes 16 minutes, during which time 7 kg of water drips
out at a steady rate through a hole in the bottom. Find the work needed to raise
the bucket to the platform. (Use g = 9.8 {m/s}^2.)

3.)A gas station stores its gasoline in a tank under the ground. The tank is a cylinder lying
horizontally on its side. (In other words, the tank is not standing vertically
on one of its flat ends.) If the radius of the cylinder is 0.5 meters, its
length is 2 meters, and its top is 1 meter under the ground, find the total
amount of work needed to pump the gasoline out of the tank. (The density of
gasoline is 673 kilograms per cubic meter; use g=9.8{ m/s}^2.)

4.)A large tank is designed with ends in the shape of the region between the curves
y=x^2/2 and y=10, measured in feet. Find the hydrostatic force (in lb) on one
end of the tank if it is filled to a depth of 8 ft with gasoline. Assume the
gasoline's weight density is 42.0 lb/ft^3. (In British units, "weight density"
is the equivalent of g in SI units.)

5.)Calculate the fluid force on the shaded side of the triangular plate submerged in fluid with mass density =
850{kg/m}^3 shown in the figure. (Length is in meters. Note also that the base
triangle shown in the figure is parallel to the surface of the fluid, and the
angle 60 degree is the angle between the shaded triangle and based triangle.)

2. ## Re: Application of definite integrals

1) y= h(x)
The distance ds of a tiny piece of the cable is given by
$\displaystyle ds= \sqrt{(dx)^2+(dy)^2}$
Get dy in terms of dx and put it into that equation so that you have ds in terms of dx only. Then integrate ds between x=-10 and x=10.

2)Let the rate at which mass leaves the bucket be k where k is in kilograms per meter
Since the mass leaves at a constant rate the mass as a function of distance is
M(s)= 20-ks
You can find k with the information you are given.

You know that work= force*distance, then
dW= Fds

The force is equal to M(s)a= 9.8M(s)

Integrate
dW= 9.8 M(s)ds

The velocity isn't needed, they probably expected you to get the rate which mass leaves the bucket in kilograms per second and then get that in terms of distance using the velocity but this step isn't required.