# Line and surface integrals $R^{3}$

So i actually missed the class where this material was covered so plaese bear with me if my understanding is not so good. one of the problems in my textbook is as follow's.

Prove the following integration by parts formula for triple integrals.

$\int \int \int_{R} f (\partial g/ \partial x) dV = - \int \int \int_{R} g (\partial f/ \partial x) dV + \int \int_{\partial R} fgn_{x} dA$

where $n_{x}$ is the x-component of the uit outward normal to $\partial R$

Other then the notation and the fact that i have never seen an integration by parts on more then one step this seems very intuitive, though i don't like how its written.

Firstly id liek to rewrite it as $\int \int \int_{R} f (\partial g/ \partial x) dzdydx =\int \int_{\partial R} fgn_{x} dA - \int \int \int_{R} g (\partial f/ \partial x) dV$

then pick f= u and $(\partial g/ \partial x)$ =dv then we have $uv-\int v du$

where $du = (\partial f/ \partial x)$, $v=g$ which leads us to

$\int \int_{R2} f g dzdy - \int \int \int_{R} g (\partial f/ \partial x) dV = \int \int \int_{R} f (\partial g/ \partial x) dV$

this doesnt make any sence but im gunna tyr and fudge this to

$\int \int_{R2} f g dzdy = \int \int \int_{R} f (\partial g/ \partial x) dV + \int \int \int_{R} g (\partial f/ \partial x) dV$

$\int \int_{R2} f g dzdy = \int \int \int_{R} [f (\partial g/ \partial x) + g (\partial f/ \partial x)] dV$
i was hoping that green's theorem would lend itself to me at this point but i have done somthing incorrectly i believe on both sides firstly im missing a dot product of $n_{x}$ on the left hand side secondly i don't have any gradients there's something im missing crucially on the right hand side as well perhaps i missed something when derivating f with respect to x? as well integrating $\partial g/\partial x$ perhaps i missed a term form the Chain rule here that will fix this. that or my intuition has led me completely astray with this problem!