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A tank has the shape of an inverted circular cone with height 10 m and base radius 4 m. It is filled with water to a height of 8 m. Find the work required to empty the tank by pumping all of the water to the top of the tank. (The density of water is 1000 kg/m^3 .)

I was following this formula $W=\int_a^bpA(y)D(y)dy$ which works fine for a tank with full water but I'm stuck on this case.

Using the base radius of 4, I found the equation of the line which was $y=\frac {5x} {2}$ and by solving for x gives us $x= \frac{2y} {5}$

The area of circle is $\pi{r^2}$

Since the water is not full I decided to integrate from 2 to 10 instead of 0 to 10

$$9.8*1000\int_2^{10} (\pi(\frac{2y} 5)^2)(10-y)dy$$

The correct answer is $3.4\times10^6$ but mine is around $3.9\times 10^6$

I think the part that is wrong is the distance $10-y$ and maybe the bounds. Can someone help?

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Here you are wrong. Limits should be 0 to 8.

So it becomes,

$$9.8*1000\int_0^{8} (\pi(\frac{2y} 5)^2)(10-y)dy$$

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  • $\begingroup$ Oh, thanks. The bounds were wrong. $\endgroup$ Commented Mar 13, 2017 at 2:53

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