Volume of Solid Formed by Circle in xy P;ane and Lines To A Raised Diameter

A circle 5 units is drawn in the  
  plane, and a diameter of the circle from  
  is raised to a height 6 units above the  
  plane. Lines are drawn in the  
  plane perpendicular to the  
  axis from a point on the circle at  
\[(x, - \sqrt{25-x^2} )\]
  to the point on the circle at  
\[(x, \sqrt{25-x^2} )\]
, and an vertical isosceles triangle drawn between these two points and a point on the raised diameter. In this way a solid is formed. What is the diameter of this solid?
Each isosceles triangle formed as above has a volume  
\[dV= \frac{1}{2} base \times height = \frac{1}{2} \times 2 \sqrt{25-x^2}dx \times 6= 6 \sqrt{25-x^2}dx\]
  and the volume of the solid is
\[\begin{equation} \begin{aligned} V &= 6 \int^5_0 \sqrt{25-x^2}dx \\ &= 6 [ \frac{1}{2} x \sqrt{25-x^2} + \frac{25}{2} sin^{-1} ( \frac{x}{5})]^5_0 \\ &= \frac{75 \pi}{2} \end{aligned} \end{equation}\]

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