1

Find flux of field $(z,x,y)$ on curve $x^2+y^2=1, z=1$ oriented positive looking from positive part of z axis using Stokes formula. So this should come to calculating $3\iint_S dS$ or $3\int\limits_{0}^{2\pi}\int\limits_{0}^{1}drd\phi$. Is this correct? I think we should use plus in front of integral because it is oriented positively.

Trevor
  • 533
  • you only have a curve. you do not know the surface so the question is expecting you to do line integral over the curve. – Math Lover Feb 01 '21 at 12:16
  • so parametrize your curve as a function of $\phi$. Positive orientation here would mean anticlockwise which means you go from $0$ to $2\pi$. – Math Lover Feb 01 '21 at 12:17
  • Yes... And one more question, what is $z=1$ telling me? – Trevor Feb 01 '21 at 13:46
  • 1
    $z=1$ will be the value of $z$ component in parametrization which you will plug in vector field. – Math Lover Feb 01 '21 at 14:40
  • And how would I do this without(or with)Stokes formula, because I need on both ways. So using Stokes and without it. – Trevor Feb 01 '21 at 16:07
  • You seem to be mixing up between flux of a vector field through a surface vs. curl of a vector field. How did you get to $3\iint_S dS$? May be that will help me understand better how to answer. – Math Lover Feb 01 '21 at 16:30
  • I got that using Stokes formula. After calculating determinant, we are geting surface integral of second kind – Trevor Feb 01 '21 at 16:38
  • You mean you did $\nabla \times \vec{F}$? But that will give you curl which is a vector field. I am not clear. – Math Lover Feb 01 '21 at 16:43
  • Yes, I did that. I am probably missing when to use Stokes formula – Trevor Feb 01 '21 at 16:44
  • OK understood but that should not have given you a scalar. Here is how you need to see it. Stokes theorem as we generally call it, is applied when you have to find surface integral of a curl of a vector field. This is equal to the line integral on the curve that makes the boundary of the surface. When I say boundary, it can be a disc, a paraboloid cut by a plane, a hemisphere etc. A full sphere for example has no boundary. Instead of doing double integral of the curl, you can do line integral on the boundary curve. Depends on which is easier and what the questions asks. – Math Lover Feb 01 '21 at 16:50
  • 1
    The flux of a vector field is different than the curl. For flux through a surface, you do the flux integral - like the examples you had yesterday. Now that double integral is equivalent to triple integral of the divergence of the vector field. So you can either do the double integral or the triple integral. Again depending on what the question asks or what is easier. – Math Lover Feb 01 '21 at 16:54
  • OK, so when I have curves in plane, or intersection of two surfaces, I am working with line integral right? And that is equivalent with Stokes formula? – Trevor Feb 01 '21 at 16:55
  • Yes, either apply divergence theorem, or do it directly with dot product of field and normal vector. – Trevor Feb 01 '21 at 16:56
  • 1
    Yes intersection of surfaces can be a surface too. So it depends on the question. In this case as you can see, the question calls out the curve. So it is clear or if you see surface integral of the curl then again it is clear. – Math Lover Feb 01 '21 at 16:59
  • And parametrization is $x=\cos t, y=\sin t$? – Trevor Feb 01 '21 at 17:13

1 Answers1

1

The Stokes' Theorem states that $\displaystyle \int_C \vec{F} \cdot d\vec{r} = \iint_{S}(\nabla \times \vec{F}) \cdot d\vec S$

The surface integral of the curl of a vector field is equal to the line integral of the vector field over the smooth closed boundary curve $C$ which is the bound of the surface.

In this case, boundary curve is $x^2+y^2 = 1$ at $z=1$. We can parametrize the curve as $\vec{r}(t) = (\cos t, \sin t, 1), 0 \leq t \leq 2\pi$.
$\vec{r'}(t) = (-\sin t, \cos t, 0)$
$\vec {F} = (z,x,y) = (1, \cos t, \sin t)$
and then the line integral.

Now if you are not applying Stokes' Theorem and are doing a double integral, you need to find a surface whose boundary curve is $x^2+y^2 = 1$ at $z=1$. It can be any simple surface, let's take a disc itself that is given by $x^2+y^2 = 1$ in plane $z=1$. The normal surface to the plane is $(0,0,1)$ based on the orientation.

$curl(\vec{F}) = \nabla \times \vec{F} = (1,1,1)$

$(\nabla \times \vec{F}) \cdot \hat{n} = 1$ which is a constant so double integral equals to nothing but the surface area of the disc which is $\pi$. Check the result to see both sides (integrals) are same.

Math Lover
  • 51,819
  • Thanks a lot. Just have one question. For the second part(without Stokes formula) could we use that $x=\sqrt{1-y^2}$ and to everything with respect to $x$? – Trevor Feb 01 '21 at 17:47
  • Yes double integral is bounds for the area of the circle, what you said in cartesian or $\iint r \ dr d\theta$ in polar. But do you have to even do an integral? You can say it is area of the unit circle which is $\pi$. – Math Lover Feb 01 '21 at 18:12
  • Yes, get it. Thanks a lot again!! – Trevor Feb 01 '21 at 18:25