Question

Necesito ayuda con estos ejercicios de integral

Answer 1

(b) D ={(x,y,z): 0 ≤ x ≤ 1 , 0 ≤ y ≤ 1 - x , 0 ≤ z ≤ 1 - x - y}

$\displaystyle I=\iiint\limits_{D}(1+x+y+z)^{-3}dV\\ \\ \\ I=\int_{0}^{1}\int_{0}^{1-x}\int_{0}^{1-x-y}(1+x+y+z)^{-3}~dz~dy~dx\\ \\ \\ I=\int_{0}^{1}\int_{0}^{1-x}-\dfrac{1}{4} \left.(1+x+y+z)^{-4}\right|_{z=0}^{z=1-x-y}~dy~dx\\ \\ \\ I=-\dfrac{1}{4} \int_{0}^{1}\int_{0}^{1-x}\dfrac{1}{16}-(1+x+y)^{-4}~dy~dx$

$\displaystyle I=-\dfrac{1}{4} \int_{0}^{1}\left.\left[\dfrac{y}{16}-\dfrac{(1+x+y)^{-5}}{-5}\right]\right|_{y=0}^{y=1-x}~dx\\ \\ \\ I=-\dfrac{1}{4} \int_{0}^{1}\dfrac{1-x}{16}+\dfrac{1}{160} -\dfrac{(1+x)^{-5}}{5}~dx\\ \\ \\ I=-\dfrac{1}{640} \int_{0}^{1}11-10x-32(1+x)^{-5}~dx\\ \\ \\ I=-\dfrac{1}{640} \left.\left[11x-5x^2+\dfrac{32}{6}(1+x)^{-6}\right]\right|_{x=0}^{x=1}\\ \\ \\ I=-\dfrac{1}{640}\left(11-5+\dfrac{16}{3}\times2^{-6}-\dfrac{16}{3}\right)\\ \\ \\ I=\dfrac{3}{128}$

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(c) D = {(x,y,z): 0 ≤ x ≤ 1 , 0 ≤ y ≤ 1- x , 0 ≤ z ≤ x² + y²}

$\displaystyle I=\iiint\limits_{D}dV\\ \\ \\ I=\int_{0}^{1}\int_{0}^{1-x}\int_{0}^{x^2+y^2}dz~dy~dx\\ \\ \\ I=\int_{0}^{1}\int_{0}^{1-x}x^2+y^2~dy~dx\\ \\ \\ I=\int_{0}^{1} \left.\left(x^2y+\dfrac{y^3}{3}\right)\right|_{y=0}^{y=1-x}dx\\ \\ \\ I=\int_{0}^{1} x^2(1-x)+\dfrac{(1-x)^3}{3}~dx\\ \\ \\ I=\int_{0}^{1} x^2-x^3+\dfrac{(1-x)^3}{3}~dx\\ \\ \\ I=\left.\left[\dfrac{x^3}{3}-\dfrac{x^4}{4}-\dfrac{(1-x)^4}{12}\right]\right|_{0}^{1}\\ \\ \\ I=\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{12}\\ \\ \\ I=\dfrac{1}{6}$