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valedrine85
find the surface area swept over by the portion of the parabola 9y^2=4x lying between the lines x=0 and x=1 when revolves through 2 right angles about the x-axis
revolving implies a volume
.the parabola 9y^2=4x y=sqrt(4x/9) \[\int\limits_{0}^{1}\pi[\sqrt{\frac{ 4x }{ 9 }}]^{2}dx\] =\[=\pi \int\limits_{0}^{1}\frac{ 4x }{ 9 }dx\] \[=\frac{ 4\pi }{ 9 } \int\limits_{0}^{1}xdx\] can you do the next step ?