Let us find volume of solid generated by revolving the region bounded by "y = \\sqrt x" and the lines "y = 2" and "x = 0" about the "x" -axis.
Let us sketch the graph of the region bounded by "y = \\sqrt x" and the lines "y = 2" and "x = 0":
"V=\\pi\\int\\limits_0^4(2^2-(\\sqrt x)^2)dx=\\pi\\int\\limits_0^4(4- x)dx=\\pi(4x-\\frac{x^2}{2})|_0^4=\\pi(16-\\frac{16}{2})=8\\pi"
Comments
Leave a comment