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.
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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.
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