Question #224465

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


1
Expert's answer
2021-08-09T14:49:34-0400
0101x01xyx2yzdzdydx\displaystyle\int_{0}^{1}\displaystyle\int_{0}^{1-x}\displaystyle\int_{0}^{1-x-y}x^2yzdzdydx

=0101xx2y[z22]1xy0dydx=\displaystyle\int_{0}^{1}\displaystyle\int_{0}^{1-x}x^2y[\dfrac{z^2}{2}]\begin{matrix} 1-x-y\\ 0 \end{matrix}dydx

=120101xx2y(1xy)2dydx=\dfrac{1}{2}\displaystyle\int_{0}^{1}\displaystyle\int_{0}^{1-x}x^2y(1-x-y)^2dydx

=0101xx2y2(1+x2+y22x2y+2xy)dydx=\displaystyle\int_{0}^{1}\displaystyle\int_{0}^{1-x}\dfrac{x^2y}{2}(1+x^2+y^2-2x-2y+2xy)dydx

=01x22[(1+x22x)y22+y44+(2+2x)y33]1x0dx=\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

=01x22(12+1423)(1x)4dx=\displaystyle\int_{0}^{1}\dfrac{x^2}{2}(\dfrac{1}{2}+\dfrac{1}{4}-\dfrac{2}{3})(1-x)^4dx

=12401x2(14x+6x24x3+x4)dx=\dfrac{1}{24}\displaystyle\int_{0}^{1}x^2(1-4x+6x^2-4x^3+x^4)dx

=124[x33x4+6x552x63+x77]10=\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}

=12520=\dfrac{1}{2520}


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