Find the volume of the region above the triangle in the xy-plane with vertices (0,0),(1,0),(0,1) and below the surface z=6xy(1-x-y) This question is straightforward and my method is integrate $\int_0^1\int_0^{x-1} 6xy(1-x-y) \,\mathrm dy\,\mathrm dx$ and my answer is 1/4 but the answer is 1/20. Where I get wrong?
 A: Let us take a look at the region over which we will integrate:

We see that $x \in [0,1]$ and $y \in [0,1 - x]$. Indeed, $1-x$ is the line through the points $(1,0)$ and $(0,1)$. You can check this easily yourself ($\color{red}{\text{This is where you have made your mistake.}}$).
Moreover, the function $6xy(1-x-y)$ is positive on this region, since $x,y$ are positive and $x + y \leq 1$ (which follows form $y \leq 1-x$). Hence we compute the following double integral:
$$\int_{0}^{1}\int_{0}^{1-x}6xy(1-x-y)\text{d}y\text{d}x.$$
We make the following computations:

\begin{align} \int_{0}^{1}\int_{0}^{1-x}6xy(1-x-y)\text{d}y\text{d}x&= \int_{0}^{1}\int_{0}^{1-x}(6xy - 6x^2y - 6xy^2)\text{d}y\text{d}x\\ &= \int_{0}^{1}(3x(1-x)^2 - 3x^2(1-x)^2 - 2x(1-x)^3)\text{d}x\\&=\int_{0}^{1}(3x(1-x)^2 - 3x^2(1-x)^2 - 2x(1-x)^3)\text{d}x\\&= \int_{0}^{1}(x - 3x^2 + 3x^3 - x^4\text{d}x\\ &= \frac{1}{2} - 1 + \frac{3}{4} - \frac{1}{5}\\&= \frac{10}{20} - \frac{20}{20} + \frac{15}{20} - \frac{4}{20}\end{align}

which gives the desired answer $\frac{1}{20}$.
$\textbf{REMARK:}$ The part where I check that your function is positive on the considered region is to make sure that your volume is computed correctly: if your function would become negative on the considered region, you would have parts of the volume cancelling each other. Consider for example the one-dimensional case, where you compute the integral $\int_{-1}^{1}x\text{d}x$, which is equal to zero since $x$ is an odd function. 
