Answer to Question #346271 in Calculus for doni

Question #346271

Find the center of mass of the solid generated by the area bounded by x = 1, x = 3, y = 0

and y = x^2 by revolving about the x-axis.


1
Expert's answer
2022-05-30T23:24:01-0400

When a curve "y=f(x)" is rotated about the "x" - axis, the centre of mass of the solid generated will lie on the "x" – axis because of the symmetry of the curve.


"\\bar{x}=\\dfrac{\\displaystyle\\int_{1}^{3}xy^2dx}{\\displaystyle\\int_{1}^{3}y^2dx}=\\dfrac{\\displaystyle\\int_{1}^{3}x^5dx}{\\displaystyle\\int_{1}^{3}x^4dx}"

"=\\dfrac{[\\dfrac{x^6}{6}]\\begin{matrix}\n 3 \\\\\n 1\n\\end{matrix}}{[\\dfrac{x^5}{5}]\\begin{matrix}\n 3 \\\\\n 1\n\\end{matrix}}=\\dfrac{910}{363}"

"(\\dfrac{910}{363}, 0)"


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