The problem: Let $v,u\in \mathbb R^2$ be two vectors: $v=(v_1,v_2), u=(u_1,u_2)$, and consider the parallelogram P spanned by them (we also assume that all the points inside P $(x,y)$ are positive/equal to zero). Calculate the volume using Fubini's theorem.
My problem is defining the bounds of the integral: I know that the maximum y value of P is $v_2+u_2$, and the maximum x value is $u_1+v_1$. But the integral of that will be the volume of a rectangle and not a parallelogram.
I guess I need to subtract that area outside of P and the x and y axis but I am not sure how to do it.
Please help :)
Your $P$ has four vertices: $(0,0)$, $v$, $u$, and $u+v$. In order to apply Fubini's theorem you have to draw vertical lines through the vertices, i.e. at $\>x=0$, $\>x=v_1$, $\>x=u_1$, $\>x=v_1+u_1$. These lines partition $P$ (assumed in the first quadrant) into two triangles and a parallelogram in the middle. Now you have to apply Fubini to each part separately. Assuming $v_1<u_1$ you then get a sum $${\rm area}(P)=\int_0^{v_1}\int_\ldots^\ldots 1\>dy\>dx+\int_{v_1}^{u_1}\int_\ldots^\ldots 1\>dy\>dx+\int_{u_1}^{v_1+u_1}\int_\ldots^\ldots 1\>dy\>dx\ ,$$ and similarly if $u_1<v_1$. It's now up to you to fill in the $y$-boundaries $\ldots$ correctly. For this you need the $y=mx+c$ equations of the edges of $P$.