A tank is is half full of oil that has a density of 900 kg/m3. Find the work W required to pump the oil out of the spout. (Use 9.8 m/s2 for g. Assume r = 3 m and h = 1 m.)

Question

aum · Answer

Find the volume of oil.  Multiply by density to find its mass.  Multiply by g to find its weight.

Have you done calculus yet?  If not we have to use some another method to find the work done.

anonymous · Answer

Work = Force x Distance, right? I did found the force to equal 158760pi, but I have no idea how to find the distance in this case.

aum · Answer

Have you been taught calculus yet?

anonymous · Answer

Yes.

aum · Answer

Is h = 1m the height of the oil tank or height of the oil in the tank?

anonymous · Answer

http://www.webassign.net/scalcet7/6-4-022-alt.gif This is the picture of the tank.

aum · Answer

|dw:1412833286283:dw|

aum · Answer

$$
x^2 + y^2 = 3^2 = 9\
x^2 = 9 - y^2 \
dV = \pi x^2 dy \
dF =  \pi x^2 dy * 900 * 9.8 ~N = 27,708x^2dy = 27,708(9-y^2)dy\
dW = F * d = 27,708(9-y^2)dy * (3-y+1) = 27,708(9-y^2)(4-y)dy\
$$

aum · Answer

Work done = 
$$\large
\int_{-3}^027,708(9-y^2)(4-y)dy
$$

anonymous · Answer

Can you explain to me why it's -3 to 0 as opposed to 0 to 3?

aum · Answer

The tank is half full.  The origin is at the center of the sphere with positive x axis to the right and positive y-axis up.

anonymous · Answer

That makes sense! Thank you for your help!

aum · Answer

You are welcome.

aum · Answer

Do you have the answer to this?

anonymous · Answer

Yes! 2556063 J

aum · Answer

Yay!! http://www.wolframalpha.com/input/?i=integrate+-3+to+0+27%2C708%289-y^2%29%284-y%29dy