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 November 13th, 2017, 07:06 PM #1 Senior Member   Joined: Jan 2017 From: Toronto Posts: 200 Thanks: 2 Stokes's theorem revisit Let F=. The plane of $\displaystyle z=2x+2y-1$ 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.. $\displaystyle \int_{0}^{2} \int_{- (\sqrt{1-(x-1)^2} + 1 ) }^{\sqrt{1-(x-1)^2} + 1} -3 ~dy~dx$ I would be much appreciated if someone could also show me the polar coordinate version.
November 14th, 2017, 05:04 AM   #2
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Quote:
 Originally Posted by zollen Let F=. The plane of $\displaystyle z=2x+2y-1$ and the paraboloid $\displaystyle z=x^2+y^2$ intersect in a closed curve.
What were you asked to do with these surfaces? Stoke's theorem gives the integral over a surface in terms of an integral over boundary of the surface. You have not stated the problem completely.

Quote:
 Answer: $\displaystyle -3 \pi$ Here is my attempt of Stokes theorem, but I am unable to get the above answer.. $\displaystyle \int_{0}^{2} \int_{- (\sqrt{1-(x-1)^2} + 1 ) }^{\sqrt{1-(x-1)^2} + 1} -3 ~dy~dx$
It looks like you are attempting to integrate over a surface. "Using Stokes Theorem" would mean integrating the curl of the function over the boundary instead.

Quote:
 I would be much appreciated if someone could also show me the polar coordinate version.
I see no reason to use polar coordinates. Just apply Stokes theorem by changing from a surface integral to an integral around the boundary.

'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
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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:

$\displaystyle \int_{0}^{2} \int_{1}^{ \sqrt { 1 - (x - 1)^2 } + 1} -3 ~dy~dx ~+~ \int_{0}^{2} \int_{ 1 - \sqrt { 1 - ( x - 1 )^2 } }^{1} -3 ~dy~dx = -3 \pi$
Attached Images
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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: 200 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: 3,261 Thanks: 894 You could use polar coordinate after changing the coordinate system: x'= x- 1 y'= y- 1. I'm not sure that helps here. Thanks from zollen
November 14th, 2017, 09:19 AM   #6
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Quote:
 Originally Posted by zollen 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.
this isn't correct.

$x = r \cos(\theta)+1$

$y = r \sin(\theta)+1$

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
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I could have... but I feel like cheating because the result of curl(f) * n is a constant and does not affect by x and y anyway.

Quote:
 Originally Posted by romsek 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.

 Tags revisit, stokes, theorem

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