# Problem 18 Solution

First, ignore the length of the tank until the end of the problem.

Next, define x as the shortest distance from the level of gas line segment to the center of the a circular cross section of the tank. By the pythagorean formula the length of the level of gas line is 2×(r2-x2)1/2.

Next measure the area of the triangle with the level of gas segment as a base and the center of the circle as the opposite angle, which is the the sum of two right triangles, or x×(r2-x2)1/2.

Next, determine the degree of the angle opposite of the base of gas line segment in the triange above. With a little trigonometry it is found that this angle is 2×cos-1(x/r).

Next, find the area of the slice of the circle made between the two equal sides of the triangle mentioned above, which is the product of the area of the circule and the ratio of the angle of the wedge to 2×pi: pi×r2 × (2×cos-1(x/r)/2×pi) = r2 × (cos-1(x/r)) .

From this wedge subtract the area of the triangle, leaving r2 × (cos-1(x/r)) - x×(r2-x2)1/2.

Next, multiply by the length of the tank: l × ((r2 × (cos-1(x/r)) - x×(r2-x2)1/2).

If g is the depth of the gas then g=r-x. The solution in terms of g is l × ((r2 × (cos-1((r-g)/r)) - (r-g)×(2rg-g2)1/2).

Thanks for Craig Katz for this solution.

Here is my old calculus solution:

First, ignore the length of the tank until the end of the problem.

Next, define x as the shortest distance from the level of gas line segment to the center of the a circular cross section of the tank. By the pythagorean formula the length of the level of gas line is 2×(r2-x2)1/2.

Next, take the integral from 0 to the depth of the gas, measured in distance from the center of the side. For example if the level of gas is y (where y is less than x) then the answer is the integral from r-y to r of 2×(r2-x2)1/2.

To solve this integral it is helpful to know that the integral of (r2-x2)1/2 is 1/2×[x×(r2-x2)1/2 + r2×sin-1(x/r).

I'm sure if you can get this far you can do the rest yourself.

Michael Shackleford, A.S.A.