Find the volume of the tetrahedron bounded by the planes $x=0, y=0, z=0, \; z = 2-2x-y$.
Solution
We easily check that the volume is given by:
$$V=\iint \limits_{D} f(x, y)\, {\rm d}(x, y)= \int_0^2 \int_0^{2-2x} (2- 2x-y)\, {\rm d}y \, {\rm d}x$$.
So, we have to evaluate the last integral which is trivial. It evaluates to $\displaystyle \frac{4}{3}$.
Hence the volume of the tetrahedron is $V=\frac{4}{3}$.
Solution
We easily check that the volume is given by:
$$V=\iint \limits_{D} f(x, y)\, {\rm d}(x, y)= \int_0^2 \int_0^{2-2x} (2- 2x-y)\, {\rm d}y \, {\rm d}x$$.
So, we have to evaluate the last integral which is trivial. It evaluates to $\displaystyle \frac{4}{3}$.
Hence the volume of the tetrahedron is $V=\frac{4}{3}$.
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