Question: Evaluate $\iiint (x+y+z) \,dx\,dy\,dz$ over the tetrahedron bounded by the planes $x=0,y=0,z=0$ and $x+y+z=1$
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Subject : Applied Mathematics 2

Topic : Triple integration and Applications of Multiple integrals

Difficulty : Medium

am2(82) • 358 views
$I= \int^1_{x=0}\int^{1-x}_{y=0} \int^{1-x-y}_{z=0}(x+y+z) dx \hspace{0.1cm}dy \hspace{0.1cm}dz\\ \hspace{0.2cm}= \int^{1}_{x=0}\int_{y=0}^{1-x}\Big[\frac{(x+y+z)^2}{2}\Big]^{1-x-y}_0 dx\hspace{0.1cm}dy\\ \hspace{0.2cm}= \frac{1}{2}\int^1_{x=0}\int^{1-x}_{y=0}[1- (x+y)^2]dx \hspace{0.1cm}dy\\ \hspace{0.2cm}= \frac{1}{2}\int_0^1\big[y-\frac{(x+y)^3}{3}\big]^{1-x}_0 dx\\ \hspace{0.2cm} = \frac{1}{2}\int_0^1\Big((1-x)-\frac{1}{3} + \frac{x^3}{3}\Big)dx\\ \hspace{0.2cm} = \frac{1}{2}\Big[\frac{2}{3}x-\frac{x^2}{2}+\frac{x^4}{12}\Big]_0^1\\ \hspace{0.2cm} = \frac{1}{2}\Big[\frac{2}{3}-\frac{1}{2}+\frac{1}{12}\Big]\\ \hspace{0.2cm} = \frac{1}{2} . \frac{3}{12} = \frac{1}{8}$