# Calculate integral using Stokes Theorem

• MHB
• mathmari
In summary: Shouldn't that be $\Sigma (x,y)=\left (x, y, 1-x-y\right )$?Or alternatively $\Sigma(r,\theta)=(r\cos\theta, r\sin\theta, 1 - r\cos\theta - r\sin\theta)$?
mathmari
Gold Member
MHB
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)

mathmari said:
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)

mathmari said:
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.

I like Serena said:
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)

mathmari said:
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)

## What is Stokes Theorem?

Stokes Theorem is a mathematical theorem that relates the integral of a vector field over a surface to the line integral of the field around the boundary of the surface.

## How is Stokes Theorem used to calculate integrals?

Stokes Theorem allows us to convert a difficult surface integral into a more manageable line integral, making it easier to calculate the value of the integral.

## What are the prerequisites for using Stokes Theorem?

Before using Stokes Theorem, we need to have a good understanding of vector fields, line integrals, and surface integrals. It is also important to have knowledge of vector calculus and multivariable calculus.

## How do you determine the direction of the line integral in Stokes Theorem?

The direction of the line integral in Stokes Theorem is determined by the right-hand rule. The thumb of the right hand should point in the direction of the normal vector of the surface, and the fingers should curl in the direction of the line integral.

## Can Stokes Theorem be used for any surface?

Stokes Theorem can be used for any surface that is smooth and has a well-defined boundary. It is not applicable for surfaces with sharp edges or corners.

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