# Surface area theory

• Oct 6th 2009, 08:22 PM
genlovesmusic09
Surface area theory
Find the area of the surface generated by revolving the arc
$x = {t^{\frac{2}
{3}}},y = \frac{{{t^2}}}
{2},0 \leqslant t \leqslant 2,
$

I'm not sure how to graph or where to begin
• Oct 6th 2009, 08:26 PM
tonio
Quote:

Originally Posted by genlovesmusic09
Find the area of the surface generated by revolving the arc
$x = {t^{\frac{2}
{3}}},y = \frac{{{t^2}}}
{2},0 \leqslant t \leqslant 2,
$

I'm not sure how to graph or where to begin

No need to graph: pay attention that y = t^2/2 = t^[(2/3)]^3/2 = x^3/2, so you need the integral INT{0,2} (x^3/2)^2 dx and then multiply this by Pi = 3.14159...and that's the area.

Tonio
• Oct 6th 2009, 08:34 PM
genlovesmusic09
Quote:

Originally Posted by tonio
No need to graph: pay attention that y = t^2/2 = t^[(2/3)]^3/2 = x^3/2, so you need the integral INT{0,2} (x^3/2)^2 dx and then multiply this by Pi = 3.14159...and that's the area.

Tonio

i don't see how
y = t^2/2 = t^[(2/3)]^3/2 = x^3/2
• Oct 6th 2009, 08:52 PM
tonio
Quote:

Originally Posted by genlovesmusic09
i don't see how
y = t^2/2 = t^[(2/3)]^3/2 = x^3/2

Well, x = t^(2/3) , right? So x^3 = [t^(2/3)]^3 = t^2...

Tonio
• Oct 6th 2009, 08:58 PM
genlovesmusic09
okay i see that now
but how does that equal y?
• Oct 6th 2009, 09:05 PM
tonio
Quote:

Originally Posted by genlovesmusic09
okay i see that now
but how does that equal y?

Uuh?? We're given y = t^2/2 , but we already know t^2 = x^3 ==>
y = x^3/2.
I think you should read more carefully the answers you're given and give a little more thinking time to them before rushing to ask back stuff.

Tonio
• Oct 6th 2009, 09:45 PM
genlovesmusic09
Quote:

Originally Posted by tonio
No need to graph: pay attention that y = t^2/2 = t^[(2/3)]^3/2 = x^3/2, so you need the integral INT{0,2} (x^3/2)^2 dx and then multiply this by Pi = 3.14159...and that's the area.

Tonio

so i keep trying this and i get 14.36 instead of the correct answer of 13.676

this is what i am doing:
$A = \pi \int_0^2 {{{\left( {\frac{{{x^3}}}
{2}} \right)}^2}dx = \pi } \int_0^2 {\frac{{{x^6}}}
{4}} dx
$

$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, 05:05 AM
tonio
Quote:

Originally Posted by genlovesmusic09
so i keep trying this and i get 14.36 instead of the correct answer of 13.676

this is what i am doing:
$A = \pi \int_0^2 {{{\left( {\frac{{{x^3}}}
{2}} \right)}^2}dx = \pi } \int_0^2 {\frac{{{x^6}}}
{4}} dx
$

$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
$

If x = t^(2/3) and 0 <= t <= 2, then 0 <= x <= 4^(1/3) and not 2 ==> the integral's upper limit is 4^(1/3). But any way you don't get what you say the answer is.

Tonio
• Oct 7th 2009, 06:13 AM
Calculus26
Y'all are using the wrong formula See attachment
• Oct 7th 2009, 08:59 AM
Calculus26
For the parametric solution see attachment