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I want to calculate the integral $\mathop{\iint}_{\phi}xdx\land dy+ydy\land dz+z dz\land dx$ considering $\left\{ \phi\left(u,v\right):=\left(u+v,uv,u^{2}-v^{2}\right):u,v\in\left[0,1\right]\right\} $

And the Stokes Theorem says that integral of a differential form ω over the boundary of some orientable manifold Ω is equal to the integral of its exterior derivative dω over the whole of Ω.

I know that $d\left(Fdx+Gdy+Hdz\right)=\left(G_{x}-F_{y}\right)dx\land dy+\left(H_{y}-G_{z}\right)dy\land dz+\left(F_{z}-H_{x}\right)dz\land dx$

How should I do that? I only find examples of vector fields and I don't know if this should be converted to something similar to a vector field.

Jack Talion
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  • Are you trying to find a 1-form $\omega$ such that $d\omega$ is equal to your integrand? If so, you can use the method described here. You can find another example with a 2-form here. – amd Apr 17 '19 at 19:25
  • Do you need to use Stoke's Theorem? I think you could probably just do the integral directly, and it wouldn't be too hard. It just comes out to be a polynomial in $u$ and $v$ integrated over the rectangle $[0,1] \times [0,1]$. – Nick Apr 17 '19 at 23:01
  • @Nick I don't know the method of solving this integrals directly :S could you be more explicit? – Jack Talion Apr 18 '19 at 00:45

1 Answers1

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The surface is parameterized as $(x, y, z) = (u+v, uv, u^2-v^2)$ where $u,v \in [0,1]$ giving $$ dx = du + dv, \quad dy = u \, dv + v \, du, \quad dz = 2u \, du - 2v \, dv $$

Let us take one term separately.

For the first term the differential form becomes $$ x \, dx \wedge dy = (u+v) \, (du+dv) \wedge (u \, dv + v \, du) = (u+v)(u-v) \, du \wedge dv = (u^2-v^2) \, du \wedge dv $$ resulting in the integral $$ \iint_\phi x \, dx \wedge dy = \iint_{[0,1]^2} (u^2-v^2) \, du \wedge dv = \int_0^1 \left( \int_0^1 (u^2-v^2) \, du \right) dv \\ = \int_0^1 \left[ \frac13 u^3-uv^2 \right]_{u=0}^1 dv = \int_0^1 \left( \frac13 - v^2 \right) dv = \left[ \frac13 v - \frac13 v^3 \right]_0^1 = 0. $$

For the second term the differential form becomes $$ y \, dy \wedge dz = uv \, (u \, dv + v \, du) \wedge (2u \, du - 2v \, dv) = uv \, (2u^2 \, dv \wedge du - 2v^2 \, du \wedge dv) \\ = -2uv (u^2+v^2) \, du \wedge dv = -2(u^3v + uv^3) \, du \wedge dv $$ resulting in the integral $$ \iint_\phi y \, dy \wedge dz = \iint_{[0,1]^2} -2(u^3v + uv^3) \, du \wedge dv = -2 \int_0^1 \left( \int_0^1 (u^3v + uv^3) \, du \right) \, dv \\ = -2 \int_0^1 \left[ \frac14 u^4v + \frac12 u^2v^3 \right]_0^1 \, dv = -2 \int_0^1 \left( \frac14 v + \frac12 v^3 \right) \, dv = -2 \cdot 0 = 0. $$

For the third term the differential form becomes $$ z \, dz \wedge dx = (u^2-v^2) (2u \, du - 2v \, dv) \wedge (du+dv) = (u^2-v^2) (2u \, du \wedge dv - 2v \, dv \wedge du) = (u^2-v^2) 2(u+v) \, du \wedge dv = 2(u^3+u^2v-uv^2-v^3) \, du \wedge dv $$ resulting in the integral $$ \iint_{\phi} z \, dz \wedge dx = \iint_{[0,1]^2} 2(u^3+u^2v-uv^2-v^3) \, du \wedge dv = \int_0^1 \left( \int_0^1 2(u^3+u^2v-uv^2-v^3) \, du \right) \, dv = \int_0^1 \left[ 2(\frac14 u^4 + \frac13 u^3v - \frac12 u^2v^2 - uv^3) \right]_0^1 \, dv = \int_0^1 2(\frac14 + \frac13 v - \frac12 v^2 - v^3) \, dv = \left[ \frac14 v + \frac16 v^2 - \frac16 v^3 - \frac14 v^4 \right]_0^1 \\ = \frac14 + \frac16 - \frac16 - \frac14 = 0. $$

Thus, the value of the integral is $0$.

md2perpe
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