## valedrine85 Group Title 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 one year ago one year ago

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1. Posideon Group Title

revolving implies a volume

2. mark_o. Group Title

.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. mark_o. Group Title

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