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Volume of a tetrahedron using integrals

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Find the volume of the tetrahedron bounded by the coordinate planes $x=0$, $y=0$ and $z=0$ and the plane $\\frac{x}{\\var{a}} + \\frac{y}{\\var{b}} + \\frac{z}{\\var{c}} =1$.

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The coordinate surfaces $x=0$, $y=0$ and $z=0$ are given. The plane $\\frac{x}{\\var{a}} + \\frac{y}{\\var{b}} + \\frac{z}{\\var{c}} = 1$ can be rewritten as a functions as follows 

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\\[z = f(x,y) = \\var{c}\\left( 1 - \\frac{x}{\\var{a}} - \\frac{y}{\\var{b}}\\right).\\]

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The surface of $f(x,y)$ intersects the axes at $x=\\var{a}, y=\\var{b}$ and $z=\\var{c}$.

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Let us integrate $y$-firts. Then the triangular region the tetrahedron sits is bounded by the lines $x=0, y=0$ and $\\frac{x}{\\var{a}} + \\frac{y}{\\var{b}} = 1$. Then we have the following bounds for integration 

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\\[0<x<\\var{a} \\quad \\mbox{and} \\quad 0 < y<  \\frac{\\var{b}}{\\var{a}}(\\var{a} -x).\\]

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Hence, we can compute the volume of the tetrahedron as 

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\\[V = \\int_0^\\var{a}\\int_0^{\\frac{\\var{b}}{\\var{a}}(\\var{a} -x)}f(x,y) \\,dy\\, dx = \\int_0^\\var{a}\\int_0^{\\frac{\\var{b}}{\\var{a}}(\\var{a} -x)} \\var{c}\\left( 1 - \\frac{x}{\\var{a}} - \\frac{y}{\\var{b}}\\right) \\,dy\\, dx =\\\\[3mm]
= \\int_0^\\var{a} \\var{c}\\left[ y - \\frac{xy}{\\var{a}} - \\frac{y^2}{\\simplify{2*{b}}}\\right]_0^{\\frac{\\var{b}}{\\var{a}}(\\var{a} -x)}\\, dx = \\int_0^\\var{a} \\frac{\\simplify{{b*c}}}{\\simplify{{2*a^2}}}(\\var{a}-x)^2 \\, dx = \\frac{\\simplify{{a*b*c}}}{6}\\]

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