Answer to Question #224465 in Calculus for hey

Question

Evaluate ∫ ∫ ∫ X 2YZ D X D Y D Z Over the Volume Bounded by Planes X=0,Y=0, Z=0 and X + Y + Z = 1

Answer 1

$$\displaystyle\int_{0}^{1}\displaystyle\int_{0}^{1-x}\displaystyle\int_{0}^{1-x-y}x^2yzdzdydx$$

$$=\displaystyle\int_{0}^{1}\displaystyle\int_{0}^{1-x}x^2y[\dfrac{z^2}{2}]\begin{matrix} 1-x-y\\ 0 \end{matrix}dydx$$

$$=\dfrac{1}{2}\displaystyle\int_{0}^{1}\displaystyle\int_{0}^{1-x}x^2y(1-x-y)^2dydx$$

$$=\displaystyle\int_{0}^{1}\displaystyle\int_{0}^{1-x}\dfrac{x^2y}{2}(1+x^2+y^2-2x-2y+2xy)dydx$$

$$=\displaystyle\int_{0}^{1}\dfrac{x^2}{2}[\dfrac{(1+x^2-2x)y^2}{2}+\dfrac{y^4}{4}+\dfrac{(-2+2x)y^3}{3}]\begin{matrix} 1-x \\ 0 \end{matrix}dx$$

$$=\displaystyle\int_{0}^{1}\dfrac{x^2}{2}(\dfrac{1}{2}+\dfrac{1}{4}-\dfrac{2}{3})(1-x)^4dx$$

$$=\dfrac{1}{24}\displaystyle\int_{0}^{1}x^2(1-4x+6x^2-4x^3+x^4)dx$$

$$=\dfrac{1}{24}[\dfrac{x^3}{3}-x^4+\dfrac{6x^5}{5}-\dfrac{2x^6}{3}+\dfrac{x^7}{7}]\begin{matrix} 1 \\ 0 \end{matrix}$$

$$=\dfrac{1}{2520}$$