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Line Integral Over a Vector Field of a set of points where a sphere intersects $2z$ $+$ $x$ $=$ $0$

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1

$begingroup$

One of my practice problems asks me to compute $int_C zdx+xdy$ where $C$ is the set of points satisfying $$x^2+y^2+z^2=4 quadtextandquad 2z+x=0$$ where $C$ is oriented counterclockwise.

If I set $z=-frac12x,$ I get the ellipse $frac58x^2+frac14y^2=1.$ I'm not sure this the correct curve $C$ to parametrize. Supposing I am correct I can parametrize it to $$vecg(t) = left(frac2sqrt105cos t,;2sin tright)$$ for $0leq t leq 2pi.$ I have no idea how to proceed.

The crux of my issues lies in my not understanding what $zdx + xdy$ represents in the integrand.

integration multivariable-calculus line-integrals

share|cite|improve this question

edited Mar 12 at 18:42

Trevor Mason

asked Mar 12 at 18:17

Trevor MasonTrevor Mason

288

$endgroup$

add a comment | 

1

$begingroup$

One of my practice problems asks me to compute $int_C zdx+xdy$ where $C$ is the set of points satisfying $$x^2+y^2+z^2=4 quadtextandquad 2z+x=0$$ where $C$ is oriented counterclockwise.

If I set $z=-frac12x,$ I get the ellipse $frac58x^2+frac14y^2=1.$ I'm not sure this the correct curve $C$ to parametrize. Supposing I am correct I can parametrize it to $$vecg(t) = left(frac2sqrt105cos t,;2sin tright)$$ for $0leq t leq 2pi.$ I have no idea how to proceed.

The crux of my issues lies in my not understanding what $zdx + xdy$ represents in the integrand.

integration multivariable-calculus line-integrals

share|cite|improve this question

edited Mar 12 at 18:42

Trevor Mason

asked Mar 12 at 18:17

Trevor MasonTrevor Mason

288

$endgroup$

add a comment | 

$begingroup$

One of my practice problems asks me to compute $int_C zdx+xdy$ where $C$ is the set of points satisfying $$x^2+y^2+z^2=4 quadtextandquad 2z+x=0$$ where $C$ is oriented counterclockwise.

If I set $z=-frac12x,$ I get the ellipse $frac58x^2+frac14y^2=1.$ I'm not sure this the correct curve $C$ to parametrize. Supposing I am correct I can parametrize it to $$vecg(t) = left(frac2sqrt105cos t,;2sin tright)$$ for $0leq t leq 2pi.$ I have no idea how to proceed.

The crux of my issues lies in my not understanding what $zdx + xdy$ represents in the integrand.

integration multivariable-calculus line-integrals

share|cite|improve this question

edited Mar 12 at 18:42

Trevor Mason

asked Mar 12 at 18:17

Trevor MasonTrevor Mason

288

$endgroup$

share|cite|improve this question

edited Mar 12 at 18:42

Trevor Mason

asked Mar 12 at 18:17

Trevor MasonTrevor Mason

288

share|cite|improve this question

edited Mar 12 at 18:42

Trevor Mason

asked Mar 12 at 18:17

Trevor MasonTrevor Mason

288

asked Mar 12 at 18:17

Trevor MasonTrevor Mason

288

asked Mar 12 at 18:17

Trevor MasonTrevor Mason

288

1 Answer
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0

$begingroup$

The curve is a circle, so, its projection on the $x-y$ plane is an ellipse (better $frac516x^2+frac14y^2=1$ and $x(t)=dfrac4sqrt55cos t$, $y(t)=2sin t$), the one you got. Nevertheless, you almost have it: we need $vecg(t) = left(x(t),y(t),z(t)right)$, a vector with three components, but you did the parametrization for the $x$ and the $y$ ones. We can complete it because we know that $z=-x/2$, so is, $z= -dfrac12dfrac4sqrt55cos t$

$$vecg(t) = left(frac4sqrt55cos t,;2sin t,frac-2sqrt55cos tright)$$

Now, for the line integral, $mathbb dx=x'(t),mathbb dt=-dfrac4sqrt55sin t,mathbb dt$ and $mathbb dy=y'(t),mathbb dt=2cos t,mathbb dt$

$$int_C zdx+xdy=int_0^2pifrac-2sqrt55cos tdfrac-4sqrt55sin t,mathbb dt+int_0^2pifrac4sqrt55cos t,2cos tmathbb dt=$$

$$=int_0^2pileft(dfrac85cos tsin t+frac8sqrt55cos^2 tright)mathbb dt$$

Being $C$ anticlockwise as you parametrized its projection in that way.

share|cite|improve this answer

edited Mar 13 at 19:42

answered Mar 13 at 5:34

Rafa BudríaRafa Budría

5,9101825

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  We have z parametrized that way because its the one half of x correct? We also need to parametrize the three components because we need to account for the entire three-dimensional vector field?
  $endgroup$
  – Trevor Mason
  Mar 13 at 15:31
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes to both questions. I suggest you to use an online program for drawing surfaces and curves, they make some problems more intuitive.
  $endgroup$
  – Rafa Budría
  Mar 13 at 19:35
  

  
  
  
  

add a comment | 

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0

$begingroup$

The curve is a circle, so, its projection on the $x-y$ plane is an ellipse (better $frac516x^2+frac14y^2=1$ and $x(t)=dfrac4sqrt55cos t$, $y(t)=2sin t$), the one you got. Nevertheless, you almost have it: we need $vecg(t) = left(x(t),y(t),z(t)right)$, a vector with three components, but you did the parametrization for the $x$ and the $y$ ones. We can complete it because we know that $z=-x/2$, so is, $z= -dfrac12dfrac4sqrt55cos t$

$$vecg(t) = left(frac4sqrt55cos t,;2sin t,frac-2sqrt55cos tright)$$

Now, for the line integral, $mathbb dx=x'(t),mathbb dt=-dfrac4sqrt55sin t,mathbb dt$ and $mathbb dy=y'(t),mathbb dt=2cos t,mathbb dt$

$$int_C zdx+xdy=int_0^2pifrac-2sqrt55cos tdfrac-4sqrt55sin t,mathbb dt+int_0^2pifrac4sqrt55cos t,2cos tmathbb dt=$$

$$=int_0^2pileft(dfrac85cos tsin t+frac8sqrt55cos^2 tright)mathbb dt$$

Being $C$ anticlockwise as you parametrized its projection in that way.

share|cite|improve this answer

edited Mar 13 at 19:42

answered Mar 13 at 5:34

Rafa BudríaRafa Budría

5,9101825

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  We have z parametrized that way because its the one half of x correct? We also need to parametrize the three components because we need to account for the entire three-dimensional vector field?
  $endgroup$
  – Trevor Mason
  Mar 13 at 15:31
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes to both questions. I suggest you to use an online program for drawing surfaces and curves, they make some problems more intuitive.
  $endgroup$
  – Rafa Budría
  Mar 13 at 19:35
  

  
  
  
  

add a comment | 

0

$begingroup$

The curve is a circle, so, its projection on the $x-y$ plane is an ellipse (better $frac516x^2+frac14y^2=1$ and $x(t)=dfrac4sqrt55cos t$, $y(t)=2sin t$), the one you got. Nevertheless, you almost have it: we need $vecg(t) = left(x(t),y(t),z(t)right)$, a vector with three components, but you did the parametrization for the $x$ and the $y$ ones. We can complete it because we know that $z=-x/2$, so is, $z= -dfrac12dfrac4sqrt55cos t$

$$vecg(t) = left(frac4sqrt55cos t,;2sin t,frac-2sqrt55cos tright)$$

Now, for the line integral, $mathbb dx=x'(t),mathbb dt=-dfrac4sqrt55sin t,mathbb dt$ and $mathbb dy=y'(t),mathbb dt=2cos t,mathbb dt$

$$int_C zdx+xdy=int_0^2pifrac-2sqrt55cos tdfrac-4sqrt55sin t,mathbb dt+int_0^2pifrac4sqrt55cos t,2cos tmathbb dt=$$

$$=int_0^2pileft(dfrac85cos tsin t+frac8sqrt55cos^2 tright)mathbb dt$$

Being $C$ anticlockwise as you parametrized its projection in that way.

share|cite|improve this answer

edited Mar 13 at 19:42

answered Mar 13 at 5:34

Rafa BudríaRafa Budría

5,9101825

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  We have z parametrized that way because its the one half of x correct? We also need to parametrize the three components because we need to account for the entire three-dimensional vector field?
  $endgroup$
  – Trevor Mason
  Mar 13 at 15:31
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes to both questions. I suggest you to use an online program for drawing surfaces and curves, they make some problems more intuitive.
  $endgroup$
  – Rafa Budría
  Mar 13 at 19:35
  

  
  
  
  

add a comment | 

$begingroup$

The curve is a circle, so, its projection on the $x-y$ plane is an ellipse (better $frac516x^2+frac14y^2=1$ and $x(t)=dfrac4sqrt55cos t$, $y(t)=2sin t$), the one you got. Nevertheless, you almost have it: we need $vecg(t) = left(x(t),y(t),z(t)right)$, a vector with three components, but you did the parametrization for the $x$ and the $y$ ones. We can complete it because we know that $z=-x/2$, so is, $z= -dfrac12dfrac4sqrt55cos t$

$$vecg(t) = left(frac4sqrt55cos t,;2sin t,frac-2sqrt55cos tright)$$

Now, for the line integral, $mathbb dx=x'(t),mathbb dt=-dfrac4sqrt55sin t,mathbb dt$ and $mathbb dy=y'(t),mathbb dt=2cos t,mathbb dt$

$$int_C zdx+xdy=int_0^2pifrac-2sqrt55cos tdfrac-4sqrt55sin t,mathbb dt+int_0^2pifrac4sqrt55cos t,2cos tmathbb dt=$$

$$=int_0^2pileft(dfrac85cos tsin t+frac8sqrt55cos^2 tright)mathbb dt$$

Being $C$ anticlockwise as you parametrized its projection in that way.

share|cite|improve this answer

edited Mar 13 at 19:42

answered Mar 13 at 5:34

Rafa BudríaRafa Budría

5,9101825

$endgroup$

share|cite|improve this answer

edited Mar 13 at 19:42

answered Mar 13 at 5:34

Rafa BudríaRafa Budría

5,9101825

answered Mar 13 at 5:34

Rafa BudríaRafa Budría

5,9101825

answered Mar 13 at 5:34

Rafa BudríaRafa Budría

5,9101825