Question

Evaluate the triple integral ∭E xy dV where E is the solid tetrahedon with vertices (0,0,0),(2,0,0),(0,10,0),(0,0,1).

Answer 1

$\iiint_E xy dV$ , where $E = \{x > 0,y>0,z>0, 5x+y+10z = 10\}$

So we have:

$x\in[0;2]\\y\in[0;10-5x]\\z\in[0;1-\frac{x}{2}-\frac{y}{10}]$

$\iiint_E xy dV =\int_0^2dx\int_0^{10-5x}dy\int_0^{1-\frac{x}{2}-\frac{y}{10}}xydz= \\= \int_0^2dx\int_0^{10-5x}(xy - \frac{x^2y}{2} - \frac{xy^2}{10})dy=\\= \int_0^2dx(\frac{xy^2}{2} - \frac{x^2y^2}{4} - \frac{xy^3}{30}|_0^{10-5x}) =\\= \int_0^2(10-5x)^2(\frac{x}{2} - \frac{x^2}{4} - \frac{10x-5x^2}{30})dx =\\= \int_0^2(100-100x+25x^2)(\cfrac{-5x^2 + 10x}{60})dx =\\= \int_0^2(\cfrac{-125x^2+750x^3-1500x^2 + 1000x}{60})dx=\\= (-\cfrac{25x^5}{60} + \cfrac{750x^4}{240} -\cfrac{500x^3}{60} + \cfrac{500x^2}{60})|_0^2=\\= \cfrac{-800 -4000+2000}{60} +\cfrac{12000}{240} = \cfrac{3000+2000-4000-800}{60}= \cfrac{10}{3}$