Results 1 to 4 of 4

Math Help - Stoke's Theorem

  1. #1
    Junior Member
    Joined
    Oct 2009
    Posts
    54

    Stoke's Theorem

    Hi all,
    I've been given a question based on Stoke's Theorem, and the question is as follows:
    Given the surface, S=x^2+y^2+2(z-1)^2=6, z\geq0 and the vector field \underline{F}=(xz-y^3cos(z))\underline{i}+(x^3e^z)\underline{j}+(xyz  e^{x^2+y^2+z^2})\underline{k}. Use stokes formula to evaluate the integral \int\int_{S}curl\underline{F}.\underline{S}.

    My answer:
    Using Stokes Theorem: \int\int_{S}curl\underline{F}.d\underline{S}=\int_  {C}\underline{F}.d\underline{r}.
    So, I need to find d\underline{r}. I can find a vector equation for S:
    \underline{r}(t)=2cos(t)\underline{i}+2sin(t)\unde  rline{j}+2\underline{z}.

    Then, I can calculate  \underline{r}'(t) & \underline{f}(\underline{r}(t)). However, the latter leaves me with something that looks hideously ugly. I was just wondering if any could go over this, see whether I've done the right thing.

    Thank you in advance!
    Last edited by Silverflow; May 9th 2010 at 12:12 AM. Reason: wrong script
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Jester's Avatar
    Joined
    Dec 2008
    From
    Conway AR
    Posts
    2,373
    Thanks
    48
    Quote Originally Posted by Silverflow View Post
    Hi all,
    I've been given a question based on Stoke's Theorem, and the question is as follows:
    Given the surface, S=x^2+y^2+2(z-1)^2=6, z\geq0 and the vector field \underline{F}=(xz-y^3cos(z))\underline{i}+(x^3e^z)\underline{j}+(xyz  e^{x^2+y^2+z^2})\underline{k}. Use stokes formula to evaluate the integral \int\int_{S}curl\underline{F}.\underline{S}.

    My answer:
    Using Stokes Theorem: \int\int_{S}curl\underline{F}.d\underline{S}=\int_  {C}\underline{F}.d\underline{r}.
    So, I need to find d\underline{r}. I can find a vector equation for S:
    \underline{r}(t)=2cos(t)\underline{i}+2sin(t)\unde  rline{j}+2\underline{z}.

    Then, I can calculate  \underline{r}'(t) & \underline{f}(\underline{r}(t)). However, the latter leaves me with something that looks hideously ugly. I was just wondering if any could go over this, see whether I've done the right thing.

    Thank you in advance!
    In your vector function for the curve of intersection of your surface with z = 0, why are you setting z = 2? i.e.

    \vec{r}(t)=2cos(t)\vec{i}+2sin(t)\vec{j}+2\vec{z}.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Senior Member AllanCuz's Avatar
    Joined
    Apr 2010
    From
    Canada
    Posts
    384
    Thanks
    4
    Quote Originally Posted by Silverflow View Post
    Hi all,
    I've been given a question based on Stoke's Theorem, and the question is as follows:
    Given the surface, S=x^2+y^2+2(z-1)^2=6, z\geq0 and the vector field \underline{F}=(xz-y^3cos(z))\underline{i}+(x^3e^z)\underline{j}+(xyz  e^{x^2+y^2+z^2})\underline{k}. Use stokes formula to evaluate the integral \int\int_{S}curl\underline{F}.\underline{S}.

    My answer:
    Using Stokes Theorem: \int\int_{S}curl\underline{F}.d\underline{S}=\int_  {C}\underline{F}.d\underline{r}.
    So, I need to find d\underline{r}. I can find a vector equation for S:
    \underline{r}(t)=2cos(t)\underline{i}+2sin(t)\unde  rline{j}+2\underline{z}.

    Then, I can calculate  \underline{r}'(t) & \underline{f}(\underline{r}(t)). However, the latter leaves me with something that looks hideously ugly. I was just wondering if any could go over this, see whether I've done the right thing.

    Thank you in advance!
    Are you using "Calculus: A complete Course" by Robert A Adams? This is a question directly from the text.

    Our surface  x^2 + y^2 + 2(z-1)^2 = 6 is an ellipsoid of revolution about the z-axis, centred at (0,0,1). Setting z=0 we find our XY domain and it's the circle  x^2 + y^2 = 4 but since our condition was  z \ge 0 our domain becomes  x^2 + y^2 \le 4

    Noting that,

     \iint_S Curl F \cdot \hat N dS = \int_C F \cdot dr = \iint_D Curl F \cdot k dA

    We get,

     curl F \cdot k = [ \frac{ \partial }{ \partial x } x^3e^z - \frac{ \partial  }{ \partial y} (xz-y^3 cosz) ]_{|z=0} = 3(x^2 + y^2)

    We are evaluating at z=0 because we transformed our stokes equation into 2D which is only in the XY plain!

    Thus,

    \iint_D Curl F \cdot k dA = \iint_{x^2+ y^2 \le 4} 3(x^2 + y^2) dxdy

    Transforming to polar co-ordinates

    \iint_{x^2 + y^2 \le 4} 3(x^2 + y^2) dxdy = \int_0^{2 \pi } d \pi \int_0^2 3 r^2 r dr = 24 \pi
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Junior Member
    Joined
    Oct 2009
    Posts
    54
    Thanks guys!
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Stoke's theorem
    Posted in the Calculus Forum
    Replies: 0
    Last Post: October 20th 2010, 04:22 AM
  2. Stoke's Theorem
    Posted in the Calculus Forum
    Replies: 2
    Last Post: April 11th 2010, 06:19 AM
  3. Stoke Theorem
    Posted in the Calculus Forum
    Replies: 0
    Last Post: November 30th 2009, 07:41 PM
  4. Stoke's Theorem Help
    Posted in the Calculus Forum
    Replies: 0
    Last Post: November 30th 2009, 07:39 PM
  5. Stoke's Theorem
    Posted in the Calculus Forum
    Replies: 3
    Last Post: April 12th 2008, 12:37 AM

Search Tags


/mathhelpforum @mathhelpforum