The height, say h, left in the tank can be easily measured. However the volume will be proportional to the cross-sectional area - not the height. Suppose the cross section is elliptical. Your task...
The height, say h, left in the tank can be easily measured. However the volume will be proportional to the cross-sectional area - not the height. Suppose the cross section is elliptical. Your task is to find the area of ellipse, and so provide a link between area and volume.
You need to remember that the area of the circle is `A = pi*r^2` .
If the radius of the circle measures 1 unit and if you stretch the circle in x direction by A and in y direction by B, the area of the new shape is `A = pi*A*B*1^2` .
This new geometric shape represents an ellipse and A denotes one half of major axis length and B denotes one half of minor axis length.
You should get the area of ellipse integrating the equation of ellipse:
`(x^2)/(A^2) + (y^2)/(B^2) = 1`
`` You need to solve for y hence: `y^2 =B^2(1 - (x^2)/(A^2)) `
`y^2 = (B^2)/(A^2)*(A^2 - x^2) =gt y = +-(B/A)sqrt(A^2 - x^2)`
You need to evaluate definite integral of the equation of ellipse to determine the area such that:
Area = `int_(-A)^A (2B/A)sqrt(A^2 - x^2) dx`
Area `= (1/2)*(xsqrt(A^2 - x^2) + A^2*arcsin (x/A))|_(-A)^A`
Area `= A*B(pi/2 + pi/2) = pi*A*B`
The volume of tank is area of ellipse multiplied by the height, hence:
`V = pi*A*B*h`
Hence, the area of the ellipse is Area = `pi*A*B` and volume of the tank is V = `pi*A*B*h` .
The cross-sectional area of the ellipse = pi * a * b / 4,
where a and b are major and minor axis.
If h is the height of fluid left in tank then
Volume of the fluid in the tank = pi * a * b * h / 4
The length of major and minor axis can also be measured as one time activity.