Question 1178403
.
a closed cylindrical tank is 8 feet long and 3 feet in diameter. 
when lying in a horizontal position, the water is 2 feet deep. 
if the tank is the vertical position, the depth of water in the tank is?
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        What is written in the post by @mananth,  is  FATALLY  and  TOTALLY  wrong.


        First,  he  MISSED  the radius and diameter values in his calculations,  using  MISTAKENLY 

        value of  36 inches  (or 3 ft)  as the radius  R,  while the true value of the radius is 1.5 ft.


        Second, the basic calculation formula in his post is written  INCORRECTLY.

        A reader can see that even the dimension of the second term in the formula 

        is not the dimension of area.


        Third, he suddenly uses gallons in the mid of his calculations, although it is totally different

        unit from cubic feet or cubic inches that are used in the input.


        Fourth, the final answer absents in the post by @mananth.


        From these my notes,  it is clear  that the solution by @mananth should be re-done from scratch.



<pre>
The formula for the volume of water in horizontal cylindrical tank is 

    V = {{{L*(r^2*arccos((r-h)/r) - (r-h)*sqrt(2rh-h^2)))}}},    (1)

where 'r' is the radius of the cylindrical tank and 'h' is the depth of water; L is the length of the cylinder.

This formula represents the product of the length of the container by the area of the cross-section
of the tank, occupied by water.


Notice that in this problem the depth 'h' of 2 feet is greater than the radius of the tank, which is 1.5 ft.

So, the horizontal axis of the container is BELOW the water level.

Nevertheless, the formula works in this case too, without change.


Indeed, when h > r, the first term represents the area of the major sector of the circle, 
while the second term represents the area of the triangle, which complement the sector to the major segment.


So, we are ready to calculate. Insert the numbers instead of symbols

    V = {{{8*(1.5^2*arccos((1.5-2)/1.5) - (1.5-2)*sqrt(2*1.5*2-2^2))}}}.    (2)


We have  {{{arccos((1.5-2)/1.5)}}} = {{{arccos(-1/3)}}} = 1.910633 radians,

         {{{(1.5-2)*sqrt(2*1.5*2-2^2)}}} = {{{-0.5*sqrt(2)}}} = -0.707107.


so we can continue formula (2) this way

    V = {{{8*(1.5^2*1.910633 - (-0.707107))}}} = 40.04825 ft^3.


Now, to get the height of the water in vertical cylinder, we should divide this volume
by the area of the base  {{{pi*r^2}}} = {{{3.14159*1.5^2}}} = 7.0685775.

Thus we find

    the height of the water in vertical container = {{{40.04825/7.0685775}}} = 5.665673185

or about 5.666 ft.


<U>ANSWER</U>.  The height of the water in vertical container is about 5.666 ft.
</pre>

At this point, the problem is solved completely.


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After completing my solution, &nbsp;I found a solution to this problem by tutor @math_helper at this forum,
which provides the &nbsp;SAME &nbsp;answer


https://www.algebra.com/algebra/homework/Bodies-in-space/Bodies-in-space.faq.question.1198022.html