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I need help setting up the integral so that I can calculate the work done. I've tried it many times and have referred to Youtube, slader, the textbook, and also this site, but I still don't get how to solve it. I came up with a few solutions but they were all wrong. Please help, I've been at this for 2 hours.

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Your $P(x)$ is wrong because you took the base to be the top face, presumably because it is larger than the bottom face. The base is the bottom face. Instead of $5-(2/3)x$ it should be $3+(2/3)x$. Good on you for using parentheses. I didn't do the second part.

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  • $\begingroup$ thank you so much, I figured out the second part. If I needed to calculate the work done, would my integral be bound between $[0,4]$? I tried this bound, but it is always incorrect and I do not understand why, because it's the distance the water must be lifted $\endgroup$
    – sam
    Commented Jan 30, 2021 at 4:08
  • $\begingroup$ You don't show your integral, so I cannot be sure, but I would think the integral would be over $[0,3]$ because that is the water in the tank. You need to account for the extra meter in the work done for each layer, but there are only $3$ meters of layers. $\endgroup$
    – Ross Millikan
    Commented Jan 30, 2021 at 4:11
  • $\begingroup$ the integral is not required, but I just want to know how I could solve it for future reference if it were a question being tested. For my integral, I wrote $9800*pi*((3+(2/3)x)^2)(4-x)$ to be integrated (it's my answer from second part). $\endgroup$
    – sam
    Commented Jan 30, 2021 at 4:14
  • $\begingroup$ So do I need to account for that 1 meter where the spout sticks out? I understand that there are only 3 meters of layers, but what happens to the 1m? Is it not necessary to calculate? $\endgroup$
    – sam
    Commented Jan 30, 2021 at 4:15
  • $\begingroup$ You have that $1$ meter in your $4-x$ factor. That is the distance you lift the layer. If the spout were at the top of the reservoir it would be $3-x$. You are off a factor $4$ because $3+(2/3)x$ is the diameter, not the radius. $\endgroup$
    – Ross Millikan
    Commented Jan 30, 2021 at 4:20

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