Thread: work done in pumping something

A cylindrical tank 4 feet in diameter and 5 feet long is carried on the back of a truck and is used to fuel tractors. The axis of the tank is horizontal. Write the integral representing the following:

a. The work done in pumping the entire contents of the fuel tank into a tractor if the opening on the tractor is 3 feet above the tank in the truck. (assume the tank is full and the gasoline weighs 42 lbs per cubic foot.)

b. The fluid force on a circular end of the tank, assuming the tank is full and the gasoline weighs 42 lbs per cubic foot.

for the first one i got 420 (integral -2 to 2) (5-y)(sqroot(2^2-y^2))dy
and for b i got 42 (integral -2 to 2) 2(sqroot(2^2-y^2)(2-y))dy

this was on a previous exam im looking over while studying for the final. i was given full points for this question, but a tutor id seen is telling me that i got it wrong. i dont remember exactly how i got those answers. i dont remember the formula to the equation. and the constant "420" i think is what the tutor said i got wrong...

ok....tutor showed me
volume of the stripe is 5 x sqrt(2^2 - y^2) x dy
distance is (5-y)
integral from -2 to 2
weight of gasoline is 42

so 42(5)(integ -2 to 2) (5-y)(sqrt(2^2 - y^2))dy
=210(integ -2 to 2) (5-y)(sqrt(2^2 - y^2))dy

i got 420(integ -2 to 2) (5-y)(sqrt(2^2 - y^2))dy
so the only thing i had different was is that my constant in front of the integral is double his.

i think what i got was 42(5)(2). the 2 coming from (2)sqrt(2^2 - y^2) because its the radius, so you need to double it in order to get the length/diameter of the strip.

i dont know who is right though...i wish i had written down all my steps when i did this on the exam....