Question 1189058
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A {{{cross(hallow)}}} <U>HOLLOW</U> metal sphere 45cm in diameter is as a float. If the metal sphere sinks of the depth of 15cm, 
what is the area  of the wetted surface?
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<pre>
Use the formula for the surface area of a spherical cap

    A = {{{2*pi*R*h}}},


where pi = 3.14159...,  R is the sphere radius (45 cm), h is the depth of the spherical cap (15 cm).


According to the formula, the wetted area of the sphere is

    A = 2*3.14159*45*15 = 4241.1465 cm^2    <U>ANSWER</U>
</pre>

Solved.


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For the area of a spherical segment,  &nbsp;see these Internet sources


https://en.wikipedia.org/wiki/Spherical_segment


https://mathworld.wolfram.com/Zone.html


https://www.math10.com/en/geometry/sphere.html