A right circular conical tank, point down, with top radius of R and height H is fully filled with a liquid whose weight-density, d, depends on its depth from its surface as d(z) = az + b, where a and b are constant. What is the weight of the liquid in the tank.
To find the weight of the liquid in the tank, we need to calculate the integral of the weight-density function over the volume of the tank.
First, let's set up the coordinate system for the tank. We'll use a cylindrical system with the origin at the vertex of the cone. The z-axis will point downwards, so the top surface of the liquid will be at z = 0, and the bottom surface will be at z = -H.
The weight-density of the liquid is given by d(z) = az + b, where a and b are constants. We can rewrite the weight-density function as d(z) = bz + cz^2, where c = a/2.
Next, let's find the volume of the conical tank. The volume of a right circular cone can be calculated using the formula V = (1/3)πr^2h, where r is the radius and h is the height. In this case, the radius R is constant, but the height h changes with z. We can express h in terms of z using similar triangles:
h/R = (H - z)/H
Simplifying, we get h = R - Rz/H.
Now, the radius of the cone at a given height z can be calculated using similar triangles as well:
r = (Rz/H)
The volume of the conical tank then becomes:
V(z) = (1/3)π(r^2)(h) = (1/3)π((Rz/H)^2)(R - Rz/H) = (1/3)π(R^3/H^2)(z^3 - 3z^2 + 2z)
To find the weight of the liquid in the tank, we calculate the integral of the weight-density function over the range z = -H to z = 0:
Weight = ∫[d(z) * V(z)] dz
= ∫[(bz + cz^2) * ((1/3)π(R^3/H^2)(z^3 - 3z^2 + 2z))] dz
= ∫[bπ(R^3/H^2)(z^4 - 3z^3 + 2z^2) + cπ(R^3/H^2)(z^5 - 3z^4 + 2z^3)] dz
Evaluating this integral will give us the weight of the liquid in the tank.