# Stokes theorem for Cuboid

I need to proof stokes theorem $\int_Qd\omega=\int_{\partial Q}\omega\;$ for a 2-form and $Q\subset \mathbb R^3 \;$a cuboid.

Since $\omega \;$ is a two form it can be written as $$\omega =f_1(x,y,z)dx\wedge dy+f_2(x,y,z)dy\wedge dz+f_3(x,y,z)dz\wedge dx$$

Hence: $$d\omega =(\frac{\partial f_1}{\partial z}+\frac{\partial f_2}{\partial x}+\frac{\partial f_3}{\partial y})dx\wedge dy \wedge dz$$

Let $Q =[x_1,x_2]\times [y_1,y_2] \times [z_1,z_2]$

Then I get the following: $$\int_Q d\omega =\int_{z_1}^{z_2}\int_{y_1}^{y_2}\int_{x_1}^{x_2}\frac{\partial f_1}{\partial z}+\frac{\partial f_2}{\partial x}+\frac{\partial f_3}{\partial y}dxdydz =\int_{z_1}^{z_2}\int_{y_1}^{y_2}f_2(x_2,y,z)-f_2(x_1,y,z)dydz-\int_{z_1}^{z_2}\int_{x_1}^{x_2}f_3(x,y_2,z)-f_3(x,y_1,z)dxdz+\int_{z_1}^{z_2}\int_{y_1}^{y_2}f_1(x,y,z_2)-f_1(x,y,z_1)dxdy$$

However, integration over $\partial Q\;$ gets me.

$$\int_{\partial Q} \omega =\int_{z_1}^{z_2}\int_{y_1}^{y_2}f_2(x_2,y,z)-f_2(x_1,y,z)dydz+\int_{z_1}^{z_2}\int_{x_1}^{x_2}f_3(x,y_2,z)-f_3(x,y_1,z)dxdz+\int_{z_1}^{z_2}\int_{y_1}^{y_2}f_1(x,y,z_2)-f_1(x,y,z_1)dxdy$$ given the orientation of $\partial Q$

I think I am wrong with the second part (Integration over $\partial Q$)

Can someone explain to me where I am wrong and how would one fix that. Thanks

## 1 Answer

First, you'll make your life a minus sign simpler if you use the order $dz\wedge dx$ rather than $dx\wedge dz$ :) Second, you also have various typos in your limits of integration, as well as a sign error in the integrand in your second term (integrating over $\partial Q$). Get the typos fixed and I think your problems will disappear.