Solving Solid of Revolution HW: Find V(L) for 0<=L<=2R

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SUMMARY

The discussion focuses on calculating the volume of water V(L) in a spherical tank with radius R, where L represents the distance from the hole to the water surface. Participants explore the integration of a semi-circle rotated around the x-axis to derive the volume formula. A key point raised is the discrepancy between the user's solution and the solution manual, highlighting the importance of correct integration limits. Ultimately, the conversation reveals that the sum of the user's volume expression and the solution manual's expression yields the volume of a sphere, emphasizing the need for accurate calculations in this context.

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Homework Statement


Peter has a spherical shaped water tank with radius R. At the top of the tank there's a small hole. Peter wants to know how much water there is left in the tank by measuring the distance L from the hole to the water surface.
Find an explicit form for the water volume V(L), 0 <= L <= 2R

Homework Equations

The Attempt at a Solution


So,

I considered the semi-circle
gif.gif


which I then rotated around the x-axis,
i.e.
gif.gif


But this yields the wrong answer, in the solution manual they have the following solution:
gif.gif


Why is my solution wrong? I am trying to do the exact same thing as the solution proposes, just different integration limits.
 

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What is the sum of your V and that from the solution ?
 
Mine is:
gif.gif


Solution is:

gif.gif
 

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@Kqwert, you didn't answer the question that @BvU asked.
BvU said:
What is the sum of your V and that from the solution ?
 
Not sure if I understand the question? Do you mean the two expressions 'I've given in post #3 added together?
 
Yes, of course :smile:. Add them up and be surprised.
 
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uhm, it gives the volume of a sphere? How does that help me?
 
So together you integrate from top to bottom.
What is the volume in the tank when L = 0 ? (Your integral would yield 0 in that case)
 

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