SOLUTION: I have cylinder on its side holding a liquid. the volume of the cylinder is no problem to determine but i need help as the volume in the tank drops. Cylinder measures 11 ft x 4

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Question 47737: I have cylinder on its side holding a liquid. the volume of the cylinder is no problem to determine but i need help as the volume in the tank drops. Cylinder measures 11 ft x 4ft
and current tank level is 1 ft or three ft down. What is the volume in tank at 1 ft? What is the calculation to determine volume at different tank levels?
Answer by venugopalramana(3286) (Show Source):
You can put this solution on YOUR website!
I have cylinder on its side holding a liquid. the volume of the cylinder is no problem to determine but i need help as the volume in the tank drops. Cylinder measures 11 ft x 4ft
and current tank level is 1 ft or three ft down. What is the volume in tank at 1 ft? What is the calculation to determine volume at different tank levels?
GOOD QUESTION.THE METHOD IS TO FIND CROSS SECTION AREA OF THE CYLINDER AT ANY TIME.WHEN FULL IT IS CIRCLE WITH RADIUS = 2'..BUT WHEN LEVEL DROPS BY 3',IT IS A SECTOR .SINCE A SECTOR IS A PART OF THE CIRCLE ,ITS AREA IS DETERMINED AS A FRACTION OF THE TOTAL ARE OF CIRCLE
TOTAL AREA OF CIRCLE =PI*R^2...FOR A 360 DEGREE ANGLE AT THE CENTRE.
AREA OF CIRCLE FORMING X DEGRES AT CENTRE =(X/360)*PI*R^2

IF YOU KNOW TRIGNOMETRY THIS ANGLE IS FOUND AS FOLLOWS.
LET CENTRE BE O AND WATER LEVEL BE AB .LET D BE THE MIDPOINT OF AB.
ODA IS A RIGHT ANGLED TRIANGLE WITH ANGLE ADO=90
OA=2
OD=2-1=1
HENCE AD = SQRT(2^2-1^2)=SQRT(3)
AD/OA=SQRT(3)/2=SIN(ANGLE AOD)
ANGLE AOD =60 DEG.
ANGLE AOB =2 *60=120
HENCE AREA OF CROSS SECTION OF WATER IN THE TANK IS (120/360)*PI*2^2=4PI/3
VOLUME = LENGTH*AREA OF CROSS SECTION = 11*4PI/3=44PI/3