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

Marks : 6

Year : MAY 2015

Tetrahedron $x+y+z=a \\ \therefore Here \space A(9,0,0) , B(0,a,0) , C(0,0,a)\\ Volume =\int\limits^a_{z=0}\int\limits_{y=0}^{a-z}\int\limits_{x=0}^{a-y-z}dx\space dy\space dz$

$=\int\limits_{z=0}^a\int\limits^{a-z}_{y=0}\int\limits_{x=z}^{a-y-z}dx\space dy\space dz\\ =\int\limits_0^a\int\limits_0^{a-z}[x]_0^{a-y-z}dy\space dz\\ =\int\limits_{z=0}^{a}\int\limits_{y=0}^{a-z}(-y-z)dy\space dz\\ =\int\limits_0^a\Bigg[ay-\dfrac {y^2}2-zy\Bigg]_0^{a-z}dz\\ Volume= \int\limits_0^a\Bigg[a(a-z)-\dfrac {(a-z)^2}2-z(a-z)\Bigg] dz\\ =\int\limits_0^a\Bigg(a^2-za-\dfrac {a^2}2+\dfrac {2az}z-\dfrac {z^2}2-az+z^2\Bigg)dz \\ =\int\limits_0^a\Bigg[\dfrac {a^2}2-az+\dfrac {z^2}2\Bigg]dz\\ =\Bigg[\dfrac {a^2z}2-\dfrac {az^2}z+\dfrac {z^3}{2\times 3}\Bigg]_0^a\\ =\Bigg[\dfrac{a^3}2-\dfrac {a^3}2+\dfrac {a^3}{2\times 3}\Bigg]-[0]\\ Volume =\dfrac {a^3}6$