Question 1210593
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Help me for ths question please.

a. The region bounded by the circle x2+y2=a2 is the base of a solid. Cross sections taken perpendicular to the base and parallel to the y-axis are equilateral triangles.

Find the volume of the solid.

b.  A cylindrical hole of constant radius r and height h is bored through the centre of a sphere with radius R.

Find the volume of the solid in terms of h.

c.  The region bounded by the curve y=−x2+4x−3 and the x-axis is rotated about the line x=3 to form a solid.
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                In this my post, I provide the solution for question (a), ONLY.



        To solve this problem, we should take the integral of the area of equilateral triangles in

        vertical sections and simply integrate this expression for the equilateral triangle areas 

        from the y=0 section to the y=a section.



<pre>
So, we consider the section of the solid y=b by the plane perpendicular to the base z=0
and parallel to z-axis.


This plane makes a chord in the circle  x^2+y^2 = a^2  at the base.

The length of this chord is  {{{2*sqrt(a^2-b^2)}}}.   (1)

This chord is the side of the equilateral triangle - so the area of this triangle is

    area(b) = {{{(sqrt(3)/4)*c^2}}} = {{{(sqrt(3)/4)*4*(a^2-b^2)}}} = {{{sqrt(3)*(a^2-b^2)}}}.    (2)


So, now we should integrate this expression over  'b' from b=0 to b=a.


It is a simple table integral. In order for do not strain my mind, I asked Artificial Intelligence
to calculate this integral. The AI successfully performed this job and produced the answer

    integral of the area expression {{{sqrt(3)*(a^2-b^2)}}} over 'b' from b=0 to b=a  is {{{2*(sqrt(3)/3)*a^3}}}.     (3)


The whole volume consists of two symmetrical parts, one in the space 0 <= b <= a  
and the other in the space  -a <= b <= 0.


Therefore, the volume of the whole solid is doubled expression (3).


<U>ANSWER</U>.  The volume of the solid under the problem's question (a) is  {{{4*(sqrt(3)/3)*a^3}}}.
</pre>

Solved.