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

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- Oct 6th 2009, 07:22 PMgenlovesmusic09Surface area theory
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 - Oct 6th 2009, 07:26 PMtonio
- Oct 6th 2009, 07:34 PMgenlovesmusic09
- Oct 6th 2009, 07:52 PMtonio
- Oct 6th 2009, 07:58 PMgenlovesmusic09
okay i see that now

but how does that equal y? - Oct 6th 2009, 08:05 PMtonio
- Oct 6th 2009, 08:45 PMgenlovesmusic09
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

$ - Oct 7th 2009, 04:05 AMtonio
- Oct 7th 2009, 05:13 AMCalculus26
Y'all are using the wrong formula See attachment

- Oct 7th 2009, 07:59 AMCalculus26
For the parametric solution see attachment