Let us find the volume of the solid generated by revolving about the x-axis bounded by the curve, "y^2 = 9x" and the line "y = 3x." Firstly, let us find the points of intersection of the curve and the line: "(3x)^2=9x" implies "9x^2=9x," and hence "x_1=0" and "x_2=1." Let us use the formula "V=\\pi\\int_{x_1}^{x_2}(y_1^2(x)-y^2_2(x))dx."
"V=\\pi\\int_0^1(9x-(3x)^2)dx=9\\pi\\int_0^1(x-x^2)dx=9\\pi(\\frac{x^2}{2}-\\frac{x^3}{3})|_0^1=\n9\\pi(\\frac{1}{2}-\\frac{1}{3})=9\\pi\\cdot\\frac{1}{6}=\\frac{3}{2}\\pi."
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