Mumbai University > First Year Engineering > sem 2 > Applied Maths 2

Marks : 8

Year : DEC 2014

Tetrahedron $\dfrac x2+\dfrac y3+\dfrac z4=1$

Here A$(2, 0, 0)$; B $(0, 3, 0)$; C$(0, 0, 4)$

Equation of co-ordinate plane are $x = 0, y = 0, z = 0\\ \therefore volume=\int\int\int dx\space dy\space dz$

Put $x = 2u ,y = 3v ,z = 4w \\ dx = 2du, dy = 3dv ,dz = 4dw$

Equation of tetrahedron becomes $u + v + w = 1$ and Equation of co-ordinate plane are $u = 0, v = 0, w = 0$

$\therefore v=\int\limits_0^1\int\limits_0^{1-w}\int\limits_0^{1-v-w}2du\times 3dv\times 4dw \\ \therefore v=24 \int\limits^1_{w=0}\int\limits^{1-w}_{v=0}\int\limits_{u=0}^{1-v-w} du\space dv\space dw$

Integrating w.r.t u

$\therefore v=24 \int\limits_0^1\int\limits_0^{1-w}[u]_0^{1-v-w} dv\space dw\\ =24\int\limits_0^1\int\limits_0^{1-w}(1-v-w)dv\space dw$

Integrating w.r.t v we get

$v=24\int\limits_0^1\Bigg[v-\dfrac {v^2}2-wv\Bigg]_0^{1-w}dw\\ =24\int\limits_0^1\Bigg[(1-w)-\dfrac {(1-w)^2}2-w(1-w)\Bigg] dw\\ =24\int\limits_0^1\Bigg(1-w-\dfrac 12+w-\dfrac {w^2}2-w+w^2\Bigg) dw\\ I=24\int\limits\_0^1\Bigg(\dfrac 12-w+\dfrac {1w^2}2\Bigg)dw\\ I= 24\Bigg[\dfrac 12w-\dfrac {w^2}2+\dfrac {w^3}{2\times 3}\Bigg]_0^1\\ I=\Bigg[\dfrac 12-\dfrac 12+\dfrac 16\Bigg]24 \\ I=\dfrac {24}6=4$