Let "S" be a solid and suppose that the area of the cross section in the plane perpendicular to the
"x-"axis is "A(x)" for "a\\leq x\\leq b."
"V=\\displaystyle\\int_{0}^{1\/2}{2\\over \\sqrt{1-4x^2}}dx=\\displaystyle\\int_{0}^{1\/2}{1\\over \\sqrt{{1 \\over 4}-x^2}}dx"
Trigonometric substitution
"V=\\displaystyle\\int_{0}^{1\/2}{2\\over \\sqrt{1-4x^2}}dx=\\displaystyle\\int_{0}^{1\/2}{1\\over \\sqrt{{1 \\over 4}-x^2}}dx="
"=[\\arcsin(2x)]\\begin{matrix}\n 1\/2 \\\\\n 0\n\\end{matrix}={\\pi \\over 2}-0={\\pi \\over 2} (units^3)"
"V={\\pi \\over 2}\\ cubic\\ units"
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