MHB How to Find the Curve σ from Intersection of Surfaces?

[mathmari]

Hey!

I want t calculate $\int_{\sigma}(ydx+zdy+xdz)$ when $\sigma$ is the curve that traces once the intersection of the surfaces with equations $x+y=2$ and $x^2+y^2+z^2=2(x+y)$ with positive direction while we look the traces from the point $(0,0,0)$. We have that $$\int_{\sigma}(ydx+zdy+xdz)=\int_{\sigma}(y, z, x)\cdot (dx,dy,dz)=\int_{\sigma}f(x,y,z)\cdot d\sigma$$ with $f(x,y,z)= (y, z, x)$.

The curve $\sigma$ is the intersection of the surfaces with equations $x+y=2$ and $x^2+y^2+z^2=2(x+y)$.
So, we have that $x^2+y^2+z^2=2(x+y) \Rightarrow x^2+y^2+z^2=2\cdot 2 \Rightarrow x^2+y^2+z^2=4$.

How can we get the curve $\sigma$ ? Could you give me a hint? (Wondering)

 

[HallsofIvy]

Hey!

I want t calculate $\int_{\sigma}(ydx+zdy+xdz)$ when $\sigma$ is the curve that traces once the intersection of the surfaces with equations $x+y=2$ and $x^2+y^2+z^2=2(x+y)$ with positive direction while we look the traces from the point $(0,0,0)$. We have that $$\int_{\sigma}(ydx+zdy+xdz)=\int_{\sigma}(y, z, x)\cdot (dx,dy,dz)=\int_{\sigma}f(x,y,z)\cdot d\sigma$$ with $f(x,y,z)= (y, z, x)$.

The curve $\sigma$ is the intersection of the surfaces with equations $x+y=2$ and $x^2+y^2+z^2=2(x+y)$.
So, we have that $x^2+y^2+z^2=2(x+y) \Rightarrow x^2+y^2+z^2=2\cdot 2 \Rightarrow x^2+y^2+z^2=4$.

How can we get the curve $\sigma$ ? Could you give me a hint? (Wondering)

Since x+ y= 2, y= 2- x. Then $x^2+ y^2+ z^2= x^2+ x^2- 4x+ 4+ z^2= 2(x^2- 2x+ 2)+ z^2= 4$. Completing the square, $2(x^2- 2x+ 1)+ 1+ z^2= 2(x+ 1)^2+ z^2+ 1= 4$. $2(x+ 1)^2+ z^2= 3$, an ellipse. To parameterize it, let $x= \frac{\sqrt{6}}{2} cos(\theta)- 1$, $y= 2- x= 3- \frac{\sqrt{6}}{2}cos(\theta)$, and $z= \sqrt{3} sin(\theta)$.

 

[mathmari]

Since x+ y= 2, y= 2- x. Then $x^2+ y^2+ z^2= x^2+ x^2- 4x+ 4+ z^2= 2(x^2- 2x+ 2)+ z^2= 4$. Completing the square, $2(x^2- 2x+ 1)+ 1+ z^2= 2(x+ 1)^2+ z^2+ 1= 4$. $2(x+ 1)^2+ z^2= 3$, an ellipse. To parameterize it, let $x= \frac{\sqrt{6}}{2} cos(\theta)- 1$, $y= 2- x= 3- \frac{\sqrt{6}}{2}cos(\theta)$, and $z= \sqrt{3} sin(\theta)$.

Ah ok!

So, we have the following:

\begin{align*}x^2+y^2+z^2=4 & \Rightarrow x^2+(2-x)^2+z^2=4 \\ & \Rightarrow x^2+4-4x+x^2+z^2=4 \\ & \Rightarrow 2x^2-4x+z^2=0 \\ & \Rightarrow 2(x^2-2x)+z^2 =0 \\ & \Rightarrow 2(x^2-2x+1-1)+z^2 =0 \\ & \Rightarrow 2(x-1)^2-2+z^2=0 \\ & \Rightarrow 2(x-1)^2+z^2=2 \\ & \Rightarrow (x-1)^2+\frac{z^2}{2}=1 \\ & \Rightarrow (x-1)^2+\left (\frac{z}{\sqrt{2}}\right )^2=1\end{align*}
So, we set $x-1=\cos t\Rightarrow x=\cos t+1$ and $\frac{z}{\sqrt{2}}=\sin t\Rightarrow z=\sqrt{2}\sin t$. Then $y=2-x =1-\cos t$.

Therefore, we get the curve $\sigma (t)=(\cos t +1, 1-\cos t, \sqrt{2}\sin t)$, right? (Wondering)

with positive direction while we look the traces from the point $(0,0,0)$.

Do we use this information to determine the interval of $t$ ? (Wondering)

 

[I like Serena]

Yep. If we fill in the result in the original equations, everything fits! (Happy)

I don't how we're supposed to find the orientation, but the interval is 0 to 2pi, since that's the interval for an ellipse. (Thinking)

 

[mathmari]

Yep. If we fill in the result in the original equations, everything fits! (Happy)

I don't how we're supposed to find the orientation, but the interval is 0 to 2pi, since that's the interval for an ellipse. (Thinking)

Thank you! (Smile)