Results 1 to 10 of 10

Math Help - Surface area theory

  1. #1
    Member
    Joined
    Sep 2009
    Posts
    124

    Surface area theory

    Find the area of the surface generated by revolving the arc
    x = {t^{\frac{2}<br />
{3}}},y = \frac{{{t^2}}}<br />
{2},0 \leqslant t \leqslant 2,<br />
    about the x-axis

    I'm not sure how to graph or where to begin
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Banned
    Joined
    Oct 2009
    Posts
    4,261
    Thanks
    2
    Quote Originally Posted by genlovesmusic09 View Post
    Find the area of the surface generated by revolving the arc
    x = {t^{\frac{2}<br />
{3}}},y = \frac{{{t^2}}}<br />
{2},0 \leqslant t \leqslant 2,<br />
    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
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Sep 2009
    Posts
    124
    Quote Originally Posted by tonio View Post
    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
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Banned
    Joined
    Oct 2009
    Posts
    4,261
    Thanks
    2
    Quote Originally Posted by genlovesmusic09 View Post
    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
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member
    Joined
    Sep 2009
    Posts
    124
    okay i see that now
    but how does that equal y?
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Banned
    Joined
    Oct 2009
    Posts
    4,261
    Thanks
    2
    Quote Originally Posted by genlovesmusic09 View Post
    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
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Member
    Joined
    Sep 2009
    Posts
    124
    Quote Originally Posted by tonio View Post
    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}}}<br />
{2}} \right)}^2}dx = \pi } \int_0^2 {\frac{{{x^6}}}<br />
{4}} dx<br />
    A = \frac{\pi }<br />
{4}\left[ {\frac{{{x^7}}}<br />
{7}} \right]_0^2 = \frac{\pi }<br />
{4}\left[ {\frac{{{2^7}}}<br />
{7} - \frac{{{0^7}}}<br />
{7}} \right]_0^2<br />
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Banned
    Joined
    Oct 2009
    Posts
    4,261
    Thanks
    2
    Quote Originally Posted by genlovesmusic09 View Post
    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}}}<br />
{2}} \right)}^2}dx = \pi } \int_0^2 {\frac{{{x^6}}}<br />
{4}} dx<br />
    A = \frac{\pi }<br />
{4}\left[ {\frac{{{x^7}}}<br />
{7}} \right]_0^2 = \frac{\pi }<br />
{4}\left[ {\frac{{{2^7}}}<br />
{7} - \frac{{{0^7}}}<br />
{7}} \right]_0^2<br />

    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
    Follow Math Help Forum on Facebook and Google+

  9. #9
    MHF Contributor Calculus26's Avatar
    Joined
    Mar 2009
    From
    Florida
    Posts
    1,271
    Y'all are using the wrong formula See attachment
    Attached Thumbnails Attached Thumbnails Surface area theory-surfacearea.jpg  
    Follow Math Help Forum on Facebook and Google+

  10. #10
    MHF Contributor Calculus26's Avatar
    Joined
    Mar 2009
    From
    Florida
    Posts
    1,271
    For the parametric solution see attachment
    Attached Thumbnails Attached Thumbnails Surface area theory-surfacearea.jpg  
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Surface Area of Surface
    Posted in the Calculus Forum
    Replies: 1
    Last Post: November 1st 2010, 10:53 AM
  2. Calculate the surface area of the surface
    Posted in the Calculus Forum
    Replies: 2
    Last Post: June 26th 2009, 05:03 AM
  3. Lateral Area and Total Surface Area
    Posted in the Geometry Forum
    Replies: 1
    Last Post: May 25th 2009, 05:28 PM
  4. Help finding surface area of a surface
    Posted in the Calculus Forum
    Replies: 1
    Last Post: October 3rd 2008, 05:11 PM
  5. Volume, Surface Area, and Lateral Surface Area
    Posted in the Geometry Forum
    Replies: 1
    Last Post: April 15th 2008, 12:40 AM

Search Tags


/mathhelpforum @mathhelpforum