Do multivariate integrals take account of orientation?

Question

Suppose a well-constructed function: are those integrals equivalents? $$ \int_{-1}^{+1} \int_{-1}^{+1} f(x, y) \, dx \, dy \hspace{35pt} \int_{+1}^{-1} \int_{-1}^{+1} f(x, y) \, dx \, dy $$

If they are not, then why?

Answer 1

You have to consider here that $$g(y)=\int_{-1}^{+1}f(x,y)\ dx$$ thus $$\int_{-1}^{+1}\int_{-1}^{+1}f(x,y)\ dx\ dy=\int_{-1}^{+1}g(y)\ dy$$

And OTOH you should know that $$\int_{-1}^{+1}g(y)\ dy=-\int_{+1}^{-1}g(y)\ dy$$

Therefore $$\int_{-1}^{+1}\int_{-1}^{+1}f(x,y)\ dx\ dy=-\int_{-1}^{+1}\int_{+1}^{-1}f(x,y)\ dx\ dy=\\=-\int_{+1}^{-1}\int_{-1}^{+1}f(x,y)\ dx\ dy=\int_{+1}^{-1}\int_{+1}^{-1}f(x,y)\ dx\ dy$$

--- rk

Answer 2

Suppose, $$ \frac{d F(x,y)}{dy} = f(x,y). $$ Then, \begin{align*} \int_{-1}^{+1} \int_{-1}^{+1} f(x, y) \, dx \, dy &= \int_{-1}^{+1} \big[ F(x,1) - F(x,-1) \big] \, dx \\[10pt] &= -\int_{+1}^{-1} [F(x,1) - F(x,-1)] \, dx \\[10pt] &= -\int_{+1}^{-1} \int_{-1}^{+1} f(x, y) \, dx \, dy. \\[10pt] \end{align*}