Surface area theory

**genlovesmusic09** · October 6th 2009, 07:22 PM

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, $
about the x-axis

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

**tonio** · October 6th 2009, 07:26 PM

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, $
about the x-axis

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

**genlovesmusic09** · October 6th 2009, 07:34 PM

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

**tonio** · October 6th 2009, 07:52 PM

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

**genlovesmusic09** · October 6th 2009, 07:58 PM

okay i see that now
but how does that equal y?

**tonio** · October 6th 2009, 08:05 PM

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

**genlovesmusic09** · October 6th 2009, 08:45 PM

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 $

**tonio** · October 7th 2009, 04:05 AM

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

**Calculus26** · October 7th 2009, 05:13 AM

Y'all are using the wrong formula See attachment

**Calculus26** · October 7th 2009, 07:59 AM

For the parametric solution see attachment

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