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

\[xy\]

plane, and a diameter of the circle from \[x=-5\]

to \[x=5\]

is raised to a height 6 units above the \[xy\]

plane. Lines are drawn in the \[xy\]

plane perpendicular to the \[x\]

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\[\begin{equation} \begin{aligned} V &= 6 \int^5_0 \sqrt{25-x^2}dx \\ &= 6 [ \frac{1}{2} x \sqrt{25-x^2} + \frac{25}{2} sin^{-1} ( \frac{x}{5})]^5_0 \\ &= \frac{75 \pi}{2} \end{aligned} \end{equation}\]