Results 1 to 6 of 6

Math Help - Stokes' Theorem

  1. #1
    Junior Member
    Joined
    Aug 2008
    Posts
    44

    Stokes' Theorem

    Let C be the intersection of the sphere x^2 +y^2 +z^2 = 2 and the plane z=1, oriented counter-clockwise as viewed from high above the z-axis. Use Stokes' Theorem to evaluate.

    \int_{c} (-y^3 +z) dx + (x^3 +2y) dy + (y - x) dz

    Is my solution right?

    Set \vec F=(-y^3 +z) \vec i + (x^3 +2y) \vec j + (y - x) \vec k
    curl(\vec F)= \vec i + \vec 2j + (3x^2 + 3y^2) \vec k

    By Stokes' Theorem, the line integral is equal to the surface integral
    \int \int_{\Sigma} (\nabla \times F) dS

    \int \int_{\Sigma} (\nabla \times F) dS = \int \int_{\Sigma}' (\nabla \times F) \cdot \vec n ds<br />
\int \int_{\Sigma}' (\nabla \times F) \cdot \vec k ds

    Sigma': { (x,y,z): x^2 +y^2 \leq 1, z=1} flat disk
    so (x, y, z) \rightarrow (cos \theta, sin \theta, 1)

    so
    \int \int_{\Sigma} ' 3x^2 +3y^2 ds

    3 \int_{0}^{2 \pi} \int_{0}^{\sqrt{2}} r dr d\theta  = 6\pi
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Apr 2005
    Posts
    15,572
    Thanks
    1412
    Quote Originally Posted by yzc717 View Post
    Let C be the intersection of the sphere x^2 +y^2 +z^2 = 2 and the plane z=1, oriented counter-clockwise as viewed from high above the z-axis. Use Stokes' Theorem to evaluate.

    \int_{c} (-y^3 +z) dx + (x^3 +2y) dy + (y - x) dz

    Is my solution right?

    Set \vec F=(-y^3 +z) \vec i + (x^3 +2y) \vec j + (y - x) \vec k
    curl(\vec F)= \vec i + \vec 2j + (3x^2 + 3y^2) \vec k

    By Stokes' Theorem, the line integral is equal to the surface integral
    \int \int_{\Sigma} (\nabla \times F) dS

    \int \int_{\Sigma} (\nabla \times F) dS = \int \int_{\Sigma}' (\nabla \times F) \cdot \vec n ds<br />
\int \int_{\Sigma}' (\nabla \times F) \cdot \vec k ds

    Sigma': { (x,y,z): x^2 +y^2 \leq 1, z=1} flat disk
    so (x, y, z) \rightarrow (cos \theta, sin \theta, 1)

    so
    \int \int_{\Sigma} ' 3x^2 +3y^2 ds
    No. To integrate on the surface (disk), x= rcos(\theta), y= r sin(\theta) and you must integrate over both r and \theta, with r going from 0 to 1 and \theta going from 0 to 2\pi.

    3\int_{r=0}^1\int{\theta= 0}^{2\pi} (r^2cos^2(\theta)+ r^2sin^2(\theta)) rd\theta dr
    = 3\int{r= 0}^1\int{\theta= 0}^{2\pi} r^3 d\theta dr
    = 6\pi \int_{r= 0}^1 r^3 d\theta= \frac{3}{2}\pi

    3 \int_{0}^{2 \pi} \int_{0}^{\sqrt{2}} r dr d\theta  = 6\pi
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Senior Member
    Joined
    Dec 2008
    Posts
    319
    The integral to be found is

    \oint_C(-y^3+z)\,dx+(x^3+2y)\,dy+(y-x)\,dz=\int_{\Sigma}\mbox{curl}\,\mathbf{F}\cdot\m  athbf{n}\,d\sigma.

    In our case,

    \begin{aligned}<br />
z&=\sqrt{2-x^2-y^2}\;\;\;\;\;\;\;\;\;\;\mbox{(Surface)}\\<br />
\mathbf{F}&=\langle-y^3+z,x^3+2y,y-x\rangle \\<br />
\mathbf{n}&=\left<-\frac{\partial z}{\partial x},-\frac{\partial z}{\partial y},1\right>\cdot\left(\sqrt{\left(\frac{\partial z}{\partial x}\right)^2+\left(\frac{\partial z}{\partial y}\right)^2+1}\right)^{-1}<br />
\end{aligned}<br />

    Since

    \int_{\Sigma}\mbox{curl}\,\mathbf{F}\cdot\mathbf{n  }\,d\sigma=\int_D\mbox{curl}\,\mathbf{F}\cdot\math  bf{n}\,\sec\gamma\,dA,

    where \gamma is the angle made by the upper unit normal of the surface with the z-axis, and

    \sec\gamma=\sqrt{\left(\frac{\partial z}{\partial x}\right)^2+\left(\frac{\partial z}{\partial y}\right)^2+1},

    our integral becomes

    \int_D\mbox{curl}\,\mathbf{F}\cdot\left<-\frac{\partial z}{\partial x},-\frac{\partial z}{\partial y},1\right>\,dA.

    We can now evaluate it with the mnemonic rule

    \mbox{curl}\,\mathbf{F}=\nabla\times\mathbf{F}=<br />
\left|\begin{array}{ccc}<br />
\mathbf{i} & \mathbf{j} & \mathbf{k} \\<br />
{\partial}/{\partial x} & {\partial}/{\partial y} & {\partial}/{\partial z} \\<br />
-y^3+z & x^3+2y & y-x<br />
\end{array}\right|.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Junior Member
    Joined
    Aug 2008
    Posts
    44
    Quote Originally Posted by Scott H View Post
    Is below way also right?


    Quote Originally Posted by HallsofIvy View Post
    No. To integrate on the surface (disk), x= rcos(\theta), y= r sin(\theta) and you must integrate over both r and \theta, with r going from 0 to 1 and \theta going from 0 to 2\pi.

    3\int_{r=0}^1\int{\theta= 0}^{2\pi} (r^2cos^2(\theta)+ r^2sin^2(\theta)) rd\theta dr
    = 3\int{r= 0}^1\int{\theta= 0}^{2\pi} r^3 d\theta dr
    = 6\pi \int_{r= 0}^1 r^3 d\theta= \frac{3}{2}\pi
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Junior Member
    Joined
    Aug 2008
    Posts
    44
    Quote Originally Posted by Scott H View Post
    our integral becomes

    \int_D\mbox{curl}\,\mathbf{F}\cdot\left<-\frac{\partial z}{\partial x},-\frac{\partial z}{\partial y},1\right>\,dA.
    if I were to do this in your way, what is the D refer to? the unit disk? then go to polar coordinates?
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Senior Member
    Joined
    Dec 2008
    Posts
    319
    Yes.

    When z=1, the equation for the surface x^2+y^2+z^2=2 becomes

    \begin{aligned}<br />
x^2+y^2+1&=2\\<br />
x^2+y^2&=1,<br />
\end{aligned}<br />

    defining a circle of radius 1. Our limits for \mbox{curl}\,\mathbf{F}\cdot\left<-\frac{\partial z}{\partial x},-\frac{\partial z}{\partial y},1\right> in polar coordinates will therefore be

    \int_0^{2\pi}\int_0^1.

    I believe that HallsofIvy was just giving the integral

    \int_D 3x^2+3y^2\,dA,

    which was not what I calculated for the surface integral in question. In fact, what you have tried to use is called Stoke's theorem in the plane, and while you have applied it correctly, it only works for planar surfaces. Here, we must use the general form of Stoke's theorem as the surface z=\sqrt{2-x^2-y^2} is curved. In this case, there is a useful theorem that

    \int_{\Sigma} \mbox{curl}\,\mathbf{F}\cdot\mathbf{n}\,d\sigma=\i  nt_R \mbox{curl}\,\mathbf{F}\cdot\left<-\frac{\partial z}{\partial x},-\frac{\partial z}{\partial y},1\right>\,dA,

    which we just proved by multiplying the unit normal \mathbf{n} by \sec\gamma in calculating the surface integral.
    Last edited by Scott H; May 19th 2009 at 03:59 AM.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 3
    Last Post: May 14th 2010, 10:04 PM
  2. Replies: 2
    Last Post: April 3rd 2010, 04:41 PM
  3. Stokes Theorem
    Posted in the Calculus Forum
    Replies: 1
    Last Post: June 24th 2009, 12:09 AM
  4. Stokes' Theorem
    Posted in the Calculus Forum
    Replies: 3
    Last Post: April 20th 2009, 02:22 PM
  5. Stokes Theorem help
    Posted in the Calculus Forum
    Replies: 0
    Last Post: November 25th 2007, 11:39 PM

Search Tags


/mathhelpforum @mathhelpforum