Question 1173952
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A spheroid (or oblate spheroid)  is a surface obtained by rotating an ellipse around its minor axis 
the ball in figure 1.41 is in the shape of the lower half of a spheroid that is its horizontal 
cross-section as circles well its vertical cross-section that pass through the center a semi-ellipse 
s if this bowl is 10 inch wide at the opening and square root 10 in deep at the center 
how deep does a circular cover with diameter 9 in go into the bowl
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        The solution to the problem in the post by @CPhill, giving the answer of 0.02 inches 

        for the depth of the cover is  INCORRECT  conceptually,  since it uses wrong ideas.


        I came to bring a correct solution.



<pre>
To solve the problem, it is enough to consider vertical section through the vertical axis of rotation.


In vertical section, we have the lover half of the ellipse with the horizontal major semi-axis 
of a = 10/2 = 5 inches long and vertical minor semi-axis of b = {{{sqrt(10)}}} inches long.


The equation of the ellipse is

    {{{x^2/5^2}}} + {{{y^2/(sqrt(10))^2}}} = 1,    (1)

or

    {{{x^2/25}}} + {{{y^2/10}}} = 1.    (2)


We are given x = 4.5 inches for the edge of the cover, and we want to find y.


From equation (2)

    y = +/- {{{sqrt(10)*sqrt(1 - x^2/25)}}} = +/- {{{sqrt(10)*sqrt(1-4.5^2/25)}}} = 1.378405 inches.


We round it reasonably and get for the depth of the cover


<U>ANSWER</U>.  1.4 inches, approximately.
</pre>

Solved correctly.