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## anonymous 3 years ago 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

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1. anonymous

revolving implies a volume

2. anonymous

.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 ?

3. anonymous

@ valedrine85

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