A tank in the shape of a sphere of radius 9 ft is full of water. Find a single integral whose value is the work done when the tank is emptied by pumping all of the water through a spigot located at the top of the tank. Use the fact that water weights 62.5 lb/ft3.
the circle of water at depth y has radius r=9sinθ where 0≤θ≤π
If the slice of water has thickness dy, then its weight is 62.5*πr^2 dy
Since work = force * distance, the work done to empty the tank is
∫[0,π] 62.5*π*81sin^2θ dy
Now, just express dy as a function of θ, and you can do the integral.
To find the work done when the tank is emptied, we need to calculate the force required to pump all of the water through the spigot, and then multiply it by the distance the water is lifted.
First, let's find the volume of water in the tank. Since the tank is in the shape of a sphere, the volume of water is equal to the volume of the sphere.
The volume of a sphere with radius r is given by the formula V = (4/3)πr³.
In this case, the radius of the sphere is 9 ft, so the volume of water in the tank is V = (4/3)π(9^3) = 3053.63 ft³.
Next, let's find the weight of the water in the tank. We are given that water weighs 62.5 lb/ft³, so the weight of the water in the tank is W = 3053.63 ft³ * 62.5 lb/ft³ = 190851.88 lb.
To lift the water out of the tank, we need to overcome the force of gravity acting on it, which is equal to the weight of the water. The force required to lift an object can be calculated using the formula F = m * g, where F is the force, m is the mass, and g is the acceleration due to gravity (approximately 32 ft/s²).
In this case, the mass of the water is given by m = W/g = 190851.88 lb / 32 ft/s² ≈ 5964.12 slugs.
Therefore, the force required to lift the water is F = 5964.12 slugs * 32 ft/s² = 190853.44 lb*ft/s².
Finally, let's calculate the work done to empty the tank. Work is equal to force times distance, so the work done is equal to W = F * d, where d is the distance the water is lifted.
Since the spigot is located at the top of the tank, the water is lifted to a height equal to the diameter of the sphere, which is 2 * 9 ft = 18 ft.
Therefore, the work done to empty the tank is W = 190853.44 lb*ft/s² * 18 ft = 3435361.92 lb*ft²/s².
To express the answer as a single integral, we can use the integral form of work, which is given by W = ∫ F * dx, where x represents the distance over which the force is applied.
In this case, the force F is constant, so the integral simplifies to W = F * ∫ dx.
Since we're lifting the water straight up, the distance x is equal to the height h of the water column, which ranges from 0 to 18 ft.
Therefore, the single integral representing the work done when the tank is emptied is W = ∫ 190853.44 lb*ft/s² * dx, integrated from 0 to 18.