A spherical tank of radius a feet is buried in the ground. The top of the tank lies b feet below the surface of the ground.(δ=64lbs/ft^3)Find a definite integral whose value equals total work done in emptying the tank through a pipe whose opening is at ground level. I have W=62.4π*int([-(a+a),0](a^2-y^2)(b-y)dy)
Not sure if the limits [-(a+a),0] are correct.
I'd integrate from -a to a
To find the definite integral that represents the total work done in emptying the tank, we need to calculate the work done to lift each thin slice of water from the tank to the ground level.
Let's break down the problem into smaller steps and then construct the integral.
Step 1: Determine the volume of each thin slice of water.
Since the tank is spherical, the volume of each slice can be represented as the cross-sectional area multiplied by the thickness.
Let's consider a slice lying at a height y from the bottom of the tank. The radius of this slice can be calculated using the equation of a spherical segment:
r = √(a^2 - y^2)
where r is the radius of the slice.
The cross-sectional area of this slice is A = πr^2.
The thickness of the slice can be approximated by considering the change in height of adjacent slices. Therefore, the thickness of the slice is dy.
The volume element of the slice is then dV = A * dy = π(√(a^2 - y^2))^2 * dy = π(a^2 - y^2) * dy.
Step 2: Determine the weight of each thin slice.
The weight of each slice can be calculated multiplying the volume of the slice by the density of water.
The weight of the slice is w = δ * dV = 64 * π(a^2 - y^2) * dy.
Step 3: Calculate the work done to lift each slice.
The work done to lift each slice is given by the product of the weight of the slice and the distance it is lifted. In this case, since the slices are lifted from the depth b to ground level (depth 0), the distance is (b - y).
Therefore, the work done to lift each slice is dw = w * (b - y).
Step 4: Construct the integral for the total work.
The integral to calculate the total work done in emptying the tank can be obtained by summing up the work done to lift each slice. This can be expressed as:
W = ∫[(a+a),0] dw.
Substituting the value of dw, we get:
W = ∫[(a+a),0] (w * (b - y)) dy
= ∫[(a+a),0] ((64 * π(a^2 - y^2) * dy) * (b - y)).
Simplifying and factoring out the constants:
W = 62.4π ∫[(a+a),0] ((a^2 - y^2) * (b - y)) dy.
So, the definite integral that represents the total work done in emptying the tank is:
W = 62.4π ∫[(a+a),0] ((a^2 - y^2) * (b - y)) dy.