Answer to Question #142084 in Calculus for Promise Omiponle

Question

Evaluate ∬∫D y dV, where the integration volume D is the region bounded by the 3 coordinate planes and 2x+ 2y+z= 4.

Answer 1

$\iiint_D y dV, D:2x + 2y +z = 4, x>0,y>0,z>0$

$x\in[0;2] ,\\y\in[0;2-x],\\ z\in[0;4-2x-2y]$

$\int_0^2dx\int_0^{2-x}dy\int_0^{4-2x-2y} ydz = \\= \int_0^2dx\int_0^{2-x}ydy z|_0^{4-2x-2y} =\\= \int_0^2dx \int_0^{2-x}y(4-2x-2y)dy =\\= \int_0^2dx\int_0^{2-x} (4y - 2xy -2y^2)dy =\\= \int_0^2dx (2y^2 - xy^2 - \frac{2y^3}{3})|_0^{2-x} =\\= \int_0^2(2(2-x)^2 - x (2-x)^2 - \frac{2(2-x)^3}{3})dx =\\= \int_0^2((2-x)^2(2-x-\frac{4-2x}{3}))dx =\\= \int_0^2((4-4x+x^2)\frac{2-x}{3})dx=\\= \int_0^2\frac{4(2-x)-4x(2-x) - x^2(2-x)}{3}dx =\\=\int_0^2\frac{8-4x-8x+4x^2-2x^2 + x^3}{3}dx = \\= \int_0^2(\frac{8}{3} - 4x + \frac{2}{3}x^2 + \frac{x^3}{3})dx = \\= \frac{8}{3}x - 2x^2 + \frac{2x^3}{6} + \frac{x^4}{12} |_0^2 = \\= \frac{16}{3} - 8 + \frac{8}{3} + \frac{16}{12} = \frac{4}{3}$