MHB Calculate integral using Stokes Theorem

[mathmari]

Hey!

I want to calculate $\int_{\sigma}\left (-y^3dx+x^3dy-^3dz\right )$ using the fomula of Stokes, when $\sigma$ is the curve that is defined by the relations $x^2+y^2=1$ and $x+y+z=1$.

Is the curve not closed? Because we have an integral of the form $\int_{\sigma}$ and not of the form $\oint_{\sigma}$ ? (Wondering)

We have that $$\int_{\sigma}\left (-y^3dx+x^3dy-^3dz\right )=\int_{\sigma}\left (-y^3, x^3, -z^3\right )\cdot \left (dx, dy, dz\right )=\int_{\sigma}f\cdot d\sigma$$ with $f(x,y,z)=\left (-y^3, x^3, -z^3\right )$, right?

From the formula of Stokes we have that $$\int_{\sigma} f\cdot d\sigma=\iint_{\Sigma}\left (\nabla \times f\right )\cdot N\ dA$$ Which is the surface $\Sigma$ ? Do we take the intersection of the given relations $x^2+y^2=1$ and $x+y+z=1$ ? (Wondering)

 

[I like Serena]

Hey!

I want to calculate $\int_{\sigma}\left (-y^3dx+x^3dy-^3dz\right )$ using the fomula of Stokes, when $\sigma$ is the curve that is defined by the relations $x^2+y^2=1$ and $x+y+z=1$.

Is the curve not closed? Because we have an integral of the form $\int_{\sigma}$ and not of the form $\oint_{\sigma}$ ?

Hey mathmari!

The intersection of $x^2+y^2=1$ and $x+y+z=1$ is a closed curve.
So yes, it's closed.
The fact that the symbol $\int$ is used instead of $\oint$ doesn't mean it's not closed. The circle in the integral symbol is just an indication. Leaving it out doesn't mean it's not closed. (Thinking)

We have that $$\int_{\sigma}\left (-y^3dx+x^3dy-^3dz\right )=\int_{\sigma}\left (-y^3, x^3, -z^3\right )\cdot \left (dx, dy, dz\right )=\int_{\sigma}f\cdot d\sigma$$ with $f(x,y,z)=\left (-y^3, x^3, -z^3\right )$, right?

From the formula of Stokes we have that $$\int_{\sigma} f\cdot d\sigma=\iint_{\Sigma}\left (\nabla \times f\right )\cdot N\ dA$$ Which is the surface $\Sigma$ ? Do we take the intersection of the given relations $x^2+y^2=1$ and $x+y+z=1$ ?

Yes. (Nod)
Actually, any surface with the same bounding curve will do, but the plane $x+y+z=1$ bounded by the cylinder $x^2+y^2=1$ seems to be a good choice.

 

[mathmari]

Hey mathmari!

The intersection of $x^2+y^2=1$ and $x+y+z=1$ is a closed curve.
So yes, it's closed.
The fact that the symbol $\int$ is used instead of $\oint$ doesn't mean it's not closed. The circle in the integral symbol is just an indication. Leaving it out doesn't mean it's not closed. (Thinking)
Yes. (Nod)
Actually, any surface with the same bounding curve will do, but the plane $x+y+z=1$ bounded by the cylinder $x^2+y^2=1$ seems to be a good choice.

We have that $x^2+y^2=1 \Rightarrow y^2=1-x^2 \Rightarrow y=\pm \sqrt{1-x^2}$ and $x+y+z=1 \Rightarrow z=1-x-y$.

So, do we consider $\Sigma (x,y)=\left (x, \pm \sqrt{1-x^2}, 1-x-y\right )$ ? (Wondering)

 

[I like Serena]

We have that $x^2+y^2=1 \Rightarrow y^2=1-x^2 \Rightarrow y=\pm \sqrt{1-x^2}$ and $x+y+z=1 \Rightarrow z=1-x-y$.

So, do we consider $\Sigma (x,y)=\left (x, \pm \sqrt{1-x^2}, 1-x-y\right )$ ?

Shouldn't that be $\Sigma (x,y)=\left (x, y, 1-x-y\right )$?

Or alternatively $\Sigma(r,\theta)=(r\cos\theta, r\sin\theta, 1 - r\cos\theta - r\sin\theta)$? (Wondering)