Find the area of the surface generated by revolving the arc
$\displaystyle x = {t^{\frac{2}
{3}}},y = \frac{{{t^2}}}
{2},0 \leqslant t \leqslant 2,
$
about the x-axis
I'm not sure how to graph or where to begin
so i keep trying this and i get 14.36 instead of the correct answer of 13.676
this is what i am doing:
$\displaystyle A = \pi \int_0^2 {{{\left( {\frac{{{x^3}}}
{2}} \right)}^2}dx = \pi } \int_0^2 {\frac{{{x^6}}}
{4}} dx
$
$\displaystyle A = \frac{\pi }
{4}\left[ {\frac{{{x^7}}}
{7}} \right]_0^2 = \frac{\pi }
{4}\left[ {\frac{{{2^7}}}
{7} - \frac{{{0^7}}}
{7}} \right]_0^2
$