September 8th, 2015, 10:43 AM  #1 
Senior Member Joined: Oct 2014 From: Complex Field Posts: 119 Thanks: 4  Line Integral? Stokes? What is it?
Hello, I need to solve the next integral: The integral of z^3dx+3y^2dyx^3dz On the curve C, when C is the circle: {(x,y,z)x^2+z^2=1,y=1}, and the direction is from the point (0,1,1) through (1,1,0) back to (0,1,1) What am I supposed to do here? Thanks! Last edited by greg1313; September 8th, 2015 at 01:20 PM. 
September 8th, 2015, 01:22 PM  #2 
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 7,739 Thanks: 1000 Math Focus: Elementary mathematics and beyond 
It's good to know that you've solved your problem but please do not delete the original content. This is unfair to others who may have an interest. I've restored the OP to its original content. 
September 8th, 2015, 01:27 PM  #3 
Senior Member Joined: Oct 2014 From: Complex Field Posts: 119 Thanks: 4 
Thanks do you have an interest ? solve it I challenge you 
September 9th, 2015, 07:17 AM  #4  
Math Team Joined: Jan 2015 From: Alabama Posts: 2,953 Thanks: 800  Quote:
direct integration or by using Stokes theorem. Did you try either of those? Fairly standard parametric equations for a circle, with center at (0, 0) and radius 1 is $\displaystyle x= sin(\theta)$, $\displaystyle z= cos(\theta)$ (I have replaced "y" with "z" since the circle is in the y= 1 plane). What do you get when you put those into the integral? "Stokes theorem" (actually "Green's theorem" since the path is in a plane): $\displaystyle \int(Ldx+ Mdz)= \int\int (\partial M/\partial x \partial L/\partial z) dxdz$ $\displaystyle L= z^3$ and $\displaystyle M= x^3$ (Since y is a constant, the term "$\displaystyle 3y^2 dy$" is 0). The partial derivatives are $\displaystyle \partial L/\partial z= 3z^2$ and [math]\partial M/\partial x= 3x^2[math]. The integral becomes $\displaystyle 3\int \int (x^2 z^2)dxdz$ over the disk. Both of those methods seem reasonable. So why did you waste time worrying about which method to use rather than trying one or the other? Last edited by Country Boy; September 9th, 2015 at 07:21 AM.  
September 9th, 2015, 10:56 AM  #5 
Senior Member Joined: Oct 2014 From: Complex Field Posts: 119 Thanks: 4 
I was not sure about a few things: If I am allowed to say that z=cos(theta) (Because I was thinking maybe it's only for y) How you use Green, when green is only for Pdx+Qdy (2 variables), and here I have Pdx+Qdy+Rdz. Only then I realized that z^3,3y^2,x^3 is the field, and I should find the curl of the field and use Stokes to solve the integral on the disk. You seem to have a mistake (I can't find it), because my integrand is 3x^2+3z^2 (Yours is 3x^2+z^2 if we put the 3 back inside  just for comparison) What I did: I found the Curl which is (0,3x^2+3z^2,0), the normal of the disk plane is (0,1,0), and then you calculate the scalar product of the two: (0,3x^2+3z^2)o(0,1,0)=3x^2+3z^2, and after parameterization you get that 3x^2+3z^2 = 3r^2*r (jacobian) = 3r^3, so the integrand is 3r^3, and you do the integral on the disk (0<r<1 ; 0<theta<2pi) Last edited by noobinmath; September 9th, 2015 at 11:10 AM. 

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integral, line, stokes 
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