Question #195686

1.Find the centroid of the region bounded by the curve π¦ = π₯3 and π¦ = 4π₯ in the first quadrant.Β

2. Find the centroid of the solid of revolution generated by revolving the region bounded by π¦ = 4π₯ β π₯2 and the x-axis about the y-axis.

Expert's answer

1

"Mx=\\frac{1}{2}\\int_a^b(f(x)^2-g(x)^2)dx=\\frac{1}{2}\\int_0^1[(4x)^2-(x^3)]"

"=2.59523"

"My=\\int_a^bx[f(x)-g(x)]dx=\\int_0^1[x[(4x)^2-(x^3)dx]"

"=\\:3.85714"

"M=\\int_a^b[f(x)-g(x)]dx=\\int_0^1[4x-x^3]"

"=1.75"

"\\bar{x},\\bar{y}=(\\frac{My}{M},\\frac{Mx}{M})=(2.20408,1.48289)"

2

"y=4x-x^2"

"4x-x^2=x"

"x=0,3"

we are interested in the point in the first quadrant, the transition point is

"x_t,y_t=3,3"

"\\bar{y}=\\int\\frac{ydv}{v}" where v = volume

"V=\\int dV=\\int \\pi x^2dy"

"V=\\int_0^{y_t}\\pi^2dy+\\int_{y_t}^4 \\pi(\\sqrt{4-y)}dy"

"V=\\pi[(\\frac{1}{3}y^3)^{y_t}_0+(4y-\\frac{1}{2}y^2)]^4_{y_t}"

"V=\\frac{\\pi}{12}(121-17\\sqrt{17)}"

"\\int ydV=\\int y(\\pi x^2 dy)"

"\\int ydV=\\int_0^{y_t}\\pi^3dy+\\int_{y_t}^4\\pi y(4-y)dy"

"\\int ydV=\\pi [(\\frac{1}{4}y^4)_0^{y_t}+(2y^2-\\frac{1}{3}y^3)_{y_t}^4]"

"\\int ydV=\\frac{x}{24}(135+17\\sqrt{17)}"

finally we find the the centroid:

"\\bar{y}=\\frac{\\frac{x}{24}(135+17\\sqrt{17)}}{\\frac{x}{12}(121-17\\sqrt{17)}}"

"\\bar{y}=\\frac{83+17\\sqrt{17}}{76}\\approx 2.014379"

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