# Relations between curl and grad

I want to prove the following two equalities: $$\oint_{\sigma}(f\nabla g)\cdot T\ dS=\iint_{\Sigma}(\nabla f\times \nabla g)\cdot N\ dA \\ \oint_{\sigma}(f\nabla g+g\nabla f)\cdot T\ dS=0$$



For the first one:

From the Stokes theorem we have the following: $$\oint_{\sigma}(f\nabla g)\cdot T\ dS=\iint_{\Sigma}(\nabla \times (f\nabla g))\cdot N\ dA$$ So, we have to show that $$\nabla \times (f\nabla g)=\nabla f\times \nabla g$$

We have that $$\nabla \times (f\nabla g)=\text{curl} (f\nabla g)=\text{curl} \left (f\left (g_x, g_y, g_z\right )\right )$$ How can we continue?

• hint: product rule for the second equality – Secret Dec 6 '17 at 11:16
• You mean $\nabla (fg)=f(\nabla g)+g(\nabla f)$, right? So we get $\oint_{\sigma}(f\nabla g+g\nabla f)\cdot T\ dS=\oint_{\sigma}(\nabla (fg))\cdot T\ dS=\iint_{\Sigma}(\nabla \times (\nabla (fg))\cdot N\ dA$, right? @Secret – Mary Star Dec 6 '17 at 11:35
• right, now what is curl of grad (for anything that is continuous in first derivatives) ? – Secret Dec 6 '17 at 12:37
• It is equal to 0, right? @Secret – Mary Star Dec 6 '17 at 13:37
• you answered your own question – Secret Dec 6 '17 at 15:05