Question

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

Answer 1

revolving implies a volume

Answer 2

.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 ?

Answer 3

@ valedrine85