Let $\alpha=x\,dx+y\,dy+z\,dz, \gamma=xy\,dz$

Let $D$ be the square $0 \leq x \leq 1, 0 \leq y \leq 1, z=1$ oriented with the upward normal. Calculate $\iint_D \alpha \wedge \gamma$.

My professor have a totally different approach then using the formula that is commonly found over the internet, which seems to work on all cases.

My attempt:
parameterize the surface $\sigma(u,v)=(u,v,1)$ such that it is a transformation from $\mathbb{R}^2$ to $\mathbb{R}^3$
$\alpha \wedge \gamma=-x^2y\,dz\,dx+xy^2\,dy\,dz$

$$\begin{align} \iint_D \alpha \wedge \gamma & = \iint_{D} -x^2y\,dz\,dx+xy^2\,dy\,dz \\ & = \int_0^1\int_0^1 -u^2v\,dZ\,dX+uv^2\,dY\,dZ\\ &= \int_0^1\int_0^1-u^2v(0)(1du)+uv^2(1dv)(0)\\ &= \int_0^1\int_0^1 0 \\ &= 0 \end{align}$$

Is it correct?


1 Answer 1


Let's start at the beginning. We need to compute $\alpha\wedge\gamma,$ which you've almost done correctly: the answer should be $$\alpha\wedge\gamma = x^{2}y\,dx\,dz+xy^{2}\,dy\,dz,$$ using the fact that $dz\wedge dz=0.$ So we want to compute the surface integral $$\iint_{D}\alpha\wedge\gamma = \iint_{\{0\leq x\leq 1,\,0\leq y\leq 1, \,z=1\}}x^{2}y\,dx\,dz+xy^{2}\,dy\,dz.$$ The answer is plainly $0$, since $z$ is constant over the domain of integration. More explicitly, we can parameterise $D$ via $$\sigma(u,v)=(u,v,1),\qquad u,v\in[0,1].$$ Now we have $$\frac{\partial{\sigma}}{\partial{u}}=(1,0,0)\quad\text{ and }\quad\frac{\partial{\sigma}}{\partial{v}}=(0,1,0).$$ By definition, we have $$(\alpha\wedge\gamma)\left(\frac{\partial{\sigma}}{\partial{u}},\frac{\partial{\sigma}}{\partial{v}}\right)=u^{2}v\det{\begin{pmatrix}1 & 0 \\ 0 & 0\end{pmatrix}}+uv^{2}\det{\begin{pmatrix}0 & 0 \\ 1 & 0\end{pmatrix}}=0,$$ so the integral is $$\iint_{D}\alpha\wedge\gamma = \int_{0}^{1}\int_{0}^{1}(\alpha\wedge\gamma)\left(\frac{\partial{\sigma}}{\partial{u}},\frac{\partial{\sigma}}{\partial{v}}\right)\,du\,dv=0.$$

  • $\begingroup$ In answering this question, I found this pdf I found online quite helpful. It reminded me of what I had read in do Carmo's book (amazon link), which I also recommend. $\endgroup$
    – Will R
    Mar 18, 2018 at 21:21
  • $\begingroup$ Thank you for answering. Since you have answered, I have reversed my edit and made it clearer so that your answer would be relevant. Sorry for the inconvenience. $\endgroup$ Mar 18, 2018 at 21:35
  • $\begingroup$ @KatharineKim: I hope my answer and the links provided are helpful. $\endgroup$
    – Will R
    Mar 18, 2018 at 21:44
  • $\begingroup$ A very common approach to deal with surface integral indeed. Though, it was not what I was hoping, that it would explain all the exterior derivative and substitution method which I was taught. $\endgroup$ Mar 19, 2018 at 5:35

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