November 13th, 2017, 07:06 PM  #1 
Senior Member Joined: Jan 2017 From: Toronto Posts: 163 Thanks: 2  Stokes's theorem revisit
Let F=<z,x,y>. The plane of $\displaystyle z=2x+2y1$ and the paraboloid $\displaystyle z=x^2+y^2$ intersect in a closed curve. Answer: $\displaystyle 3 \pi $ Here is my attempt of Stokes theorem, but I am unable to get the above answer.. $\displaystyle \int_{0}^{2} \int_{ (\sqrt{1(x1)^2} + 1 ) }^{\sqrt{1(x1)^2} + 1} 3 ~dy~dx $ I would be much appreciated if someone could also show me the polar coordinate version. 
November 14th, 2017, 05:04 AM  #2  
Math Team Joined: Jan 2015 From: Alabama Posts: 2,922 Thanks: 785  Quote:
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'Stokes Theorem' says $\displaystyle \int \vec{F}\cdot d\vec{r}= \int\int \nabla\times\vec{F}\cdot d\vec{S}$ where line integral is over the curve bounding surface S. Here the surface is of two parts, the paraboloid $\displaystyle z= x^2+ y^2$ and the plane $\displaystyle z= x+ y+ 1$. The vector function to be integrated is $\displaystyle \vec{F}= z\vec{i}+ x\vec{j}+ y\vec{k}$ so that $\displaystyle \nabla\times\vec{F}= \vec{i}+ \vec{j}+ \vec{k}$. The curve where the two surfaces $\displaystyle z= 2x+ 2y 1$ and the paraboloid $\displaystyle z= x^2+ y^2$ intersect is given by $\displaystyle x^2+ y^2= 2x+ 2y 1$ so $\displaystyle x^2 2x+ y^2 2y= 1$. Completing the squares, $\displaystyle x^2 2x+ 1+ y^2 2y+ 1= (x 1)^2+ (y 1)^2= 1$. That is a circle with center at (1, 1) and radius 1. We can parameterize that by taking $\displaystyle x= cos(t)+ 1$, $\displaystyle y= sin(t)+ 1$ and $\displaystyle z= x^2+ y^2= cos^2(t)+ 2cos(t)+ 1+ sin^2(t)+ 2sin(t)+ 1= 2(cos(t)+ sin(t)+ 1)$. $\displaystyle \vec{r}= cos(t)\vec{i}+ sin(t)\vec{j}+ 2(cos(t)+ sin(t)+ 1)\vec{k}$. $\displaystyle d\vec{r}= (sin(t)\vec{i}+ cos(t)\vec{j}+ 2(cos(t) sin(t))\vec{k})dt$. $\displaystyle \nabla\times\vec{F}\cdot d\vec{r}= (sin(t)+ cos(t)+ 2cos(t) 2sin(t))dt= 3(cos(t) sin(t)dt$. So the integral reduces to $\displaystyle 3\int_0^{2\pi} (cos(t) sin(t))dt$  
November 14th, 2017, 07:23 AM  #3 
Senior Member Joined: Jan 2017 From: Toronto Posts: 163 Thanks: 2 
Yes. I was attempting to integrate the intersected surface between $\displaystyle z = 2x + 2y  1$ and $\displaystyle z = x^2 + y^2$. curl(f) = <1, 1, 1> N = <2, 2, 1> curl(f) * N = 3 I think I got my initial solution wrong. Here is the correct double integral: $\displaystyle \int_{0}^{2} \int_{1}^{ \sqrt { 1  (x  1)^2 } + 1} 3 ~dy~dx ~+~ \int_{0}^{2} \int_{ 1  \sqrt { 1  ( x  1 )^2 } }^{1} 3 ~dy~dx = 3 \pi $ Last edited by zollen; November 14th, 2017 at 07:31 AM. 
November 14th, 2017, 07:25 AM  #4 
Senior Member Joined: Jan 2017 From: Toronto Posts: 163 Thanks: 2 
Correct me if I am wrong. It is not possible to use polar coordinate for this problem because the intersected 'circle' is away from the origin.

November 14th, 2017, 08:38 AM  #5 
Math Team Joined: Jan 2015 From: Alabama Posts: 2,922 Thanks: 785 
You could use polar coordinate after changing the coordinate system: x'= x 1 y'= y 1. I'm not sure that helps here.

November 14th, 2017, 09:19 AM  #6  
Senior Member Joined: Sep 2015 From: USA Posts: 1,696 Thanks: 861  Quote:
$x = r \cos(\theta)+1$ $y = r \sin(\theta)+1$ for $r \in [0,1], ~\theta \in [0, 2\pi]$ $z$ turns out to be $z = 2r(\cos(\theta)+\sin(\theta)  1$ or as CountryBoy suggested translate the entire problem to $(0,0)$ then $f = (z, x+1, y+1)$ The curl is unaffected by this translation.  
November 14th, 2017, 12:28 PM  #7 
Senior Member Joined: Jan 2017 From: Toronto Posts: 163 Thanks: 2  

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