Question

2.   Evaluate ∭_T  x²  dV where T is a solid tetrahedron with vertices  (0,0,0),(1,0,0),(0,1,0) and

(0,0,1).

Answer 1

Solution:

The region can be written as

$E=\{(x, y, z) \mid 0 \leq x \leq 1,0 \leq y \leq 1-x, 0 \leq z \leq 1-x-y\} .$

Hence

$\begin{aligned} \iiint_{E} x^{2} d V &=\int_{0}^{1} \int_{0}^{1-x} \int_{0}^{1-x-y} x^{2} d z d y d x \\ &=\left.\int_{0}^{1} \int_{0}^{1-x} x^{2} z\right|_{z=0} ^{z=1-x-y} d y d x \\ &=\int_{0}^{1} \int_{0}^{1-x} x^{2}(1-x-y) d y d x \\ &=\int_{0}^{1} \int_{0}^{1-x}\left(x^{2}(1-x)-x^{2} y\right) d y d x \\ &=\left.\int_{0}^{1}\left(x^{2}(1-x) y-x^{2} \frac{y^{2}}{2}\right)\right|_{y=0} ^{y=1-x} d x \\ &=\int_{0}^{1}\left(x^{2}(1-x)^{2}-x^{2} \frac{(1-x)^{2}}{2}\right) d x \\ &=\int_{0}^{1} \frac{1}{2} x^{2}(1-x)^{2} d x \\ \end{aligned}$

$\begin{aligned} &=\int_{0}^{1} x^{2}\left(\frac{1+x^{2}-2 x}{2}\right) d x \\ &=\int_{0}^{1}\left(\frac{x^{2}}{2}+\frac{x^{4}}{2}-x^{3}\right) d x \\ &=\left[\frac{x^{3}}{6}+\frac{x^{5}}{10}-\frac{x^{4}}{4}\right]_{0}^{1} \\ &=\left[\left(\frac{1}{6}+\frac{1}{10}-\frac{1}{4}\right)-0\right] \\ &=\frac{10+6-15}{60} \\ &=\frac{1}{60} \end{aligned}$