Volume with double integral












0















Find the volume of the region bounded by the planes $6x+5y+6z = 6$, $y=x$, $x=0$ and $z=0$.




From this, I got that the volume would simply consist of the region under $z = 1-x-(5/6)y$. And as $z=0$, the plane intersects the $x-y$ plane at $6x+5y=6$. Therefore, I thought the region was bounded by $y = x$, $6x+5y = 6$ and $x=0$. After rearranging the equations and drawing the diagrams, I got the following integral:



$$int _0^{ frac{6}{5}}int _x^{frac{6}{5}-frac{6}{5}x}1-x-frac{5}{6}y:dydx$$



This integral gave me a volume of $186/625$, but this was not correct.



Any help would be highly appreciated!










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  • What is the correct answer?
    – K Split X
    Nov 29 '18 at 2:29
















0















Find the volume of the region bounded by the planes $6x+5y+6z = 6$, $y=x$, $x=0$ and $z=0$.




From this, I got that the volume would simply consist of the region under $z = 1-x-(5/6)y$. And as $z=0$, the plane intersects the $x-y$ plane at $6x+5y=6$. Therefore, I thought the region was bounded by $y = x$, $6x+5y = 6$ and $x=0$. After rearranging the equations and drawing the diagrams, I got the following integral:



$$int _0^{ frac{6}{5}}int _x^{frac{6}{5}-frac{6}{5}x}1-x-frac{5}{6}y:dydx$$



This integral gave me a volume of $186/625$, but this was not correct.



Any help would be highly appreciated!










share|cite|improve this question
























  • What is the correct answer?
    – K Split X
    Nov 29 '18 at 2:29














0












0








0








Find the volume of the region bounded by the planes $6x+5y+6z = 6$, $y=x$, $x=0$ and $z=0$.




From this, I got that the volume would simply consist of the region under $z = 1-x-(5/6)y$. And as $z=0$, the plane intersects the $x-y$ plane at $6x+5y=6$. Therefore, I thought the region was bounded by $y = x$, $6x+5y = 6$ and $x=0$. After rearranging the equations and drawing the diagrams, I got the following integral:



$$int _0^{ frac{6}{5}}int _x^{frac{6}{5}-frac{6}{5}x}1-x-frac{5}{6}y:dydx$$



This integral gave me a volume of $186/625$, but this was not correct.



Any help would be highly appreciated!










share|cite|improve this question
















Find the volume of the region bounded by the planes $6x+5y+6z = 6$, $y=x$, $x=0$ and $z=0$.




From this, I got that the volume would simply consist of the region under $z = 1-x-(5/6)y$. And as $z=0$, the plane intersects the $x-y$ plane at $6x+5y=6$. Therefore, I thought the region was bounded by $y = x$, $6x+5y = 6$ and $x=0$. After rearranging the equations and drawing the diagrams, I got the following integral:



$$int _0^{ frac{6}{5}}int _x^{frac{6}{5}-frac{6}{5}x}1-x-frac{5}{6}y:dydx$$



This integral gave me a volume of $186/625$, but this was not correct.



Any help would be highly appreciated!







calculus integration multivariable-calculus volume






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edited Nov 29 '18 at 4:23









Key Flex

7,74741232




7,74741232










asked Nov 29 '18 at 2:24









sktsasussktsasus

1,010415




1,010415












  • What is the correct answer?
    – K Split X
    Nov 29 '18 at 2:29


















  • What is the correct answer?
    – K Split X
    Nov 29 '18 at 2:29
















What is the correct answer?
– K Split X
Nov 29 '18 at 2:29




What is the correct answer?
– K Split X
Nov 29 '18 at 2:29










1 Answer
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1














$6x+5y+6z = 6$, $y=x$, $x=0$ and $z=0$



The bounds for $z$ is from $0$ to $dfrac{6-6x-5y}{6}$



The solid region onto the $xy$ plane is



$y=x, x=0, 6x+5y=6$



The bounds for $y$ is from $x$ to $dfrac{6-6x}{5}$



The bounds for $x$ is from $0$ to $dfrac{6}{11}$



The Volume is $$int_{0}^{frac{6}{11}}int_{x}^{frac{6-6x}{5}}dfrac{6-6x-5y}{6} dy dx=dfrac{6}{55}$$






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    1














    $6x+5y+6z = 6$, $y=x$, $x=0$ and $z=0$



    The bounds for $z$ is from $0$ to $dfrac{6-6x-5y}{6}$



    The solid region onto the $xy$ plane is



    $y=x, x=0, 6x+5y=6$



    The bounds for $y$ is from $x$ to $dfrac{6-6x}{5}$



    The bounds for $x$ is from $0$ to $dfrac{6}{11}$



    The Volume is $$int_{0}^{frac{6}{11}}int_{x}^{frac{6-6x}{5}}dfrac{6-6x-5y}{6} dy dx=dfrac{6}{55}$$






    share|cite|improve this answer


























      1














      $6x+5y+6z = 6$, $y=x$, $x=0$ and $z=0$



      The bounds for $z$ is from $0$ to $dfrac{6-6x-5y}{6}$



      The solid region onto the $xy$ plane is



      $y=x, x=0, 6x+5y=6$



      The bounds for $y$ is from $x$ to $dfrac{6-6x}{5}$



      The bounds for $x$ is from $0$ to $dfrac{6}{11}$



      The Volume is $$int_{0}^{frac{6}{11}}int_{x}^{frac{6-6x}{5}}dfrac{6-6x-5y}{6} dy dx=dfrac{6}{55}$$






      share|cite|improve this answer
























        1












        1








        1






        $6x+5y+6z = 6$, $y=x$, $x=0$ and $z=0$



        The bounds for $z$ is from $0$ to $dfrac{6-6x-5y}{6}$



        The solid region onto the $xy$ plane is



        $y=x, x=0, 6x+5y=6$



        The bounds for $y$ is from $x$ to $dfrac{6-6x}{5}$



        The bounds for $x$ is from $0$ to $dfrac{6}{11}$



        The Volume is $$int_{0}^{frac{6}{11}}int_{x}^{frac{6-6x}{5}}dfrac{6-6x-5y}{6} dy dx=dfrac{6}{55}$$






        share|cite|improve this answer












        $6x+5y+6z = 6$, $y=x$, $x=0$ and $z=0$



        The bounds for $z$ is from $0$ to $dfrac{6-6x-5y}{6}$



        The solid region onto the $xy$ plane is



        $y=x, x=0, 6x+5y=6$



        The bounds for $y$ is from $x$ to $dfrac{6-6x}{5}$



        The bounds for $x$ is from $0$ to $dfrac{6}{11}$



        The Volume is $$int_{0}^{frac{6}{11}}int_{x}^{frac{6-6x}{5}}dfrac{6-6x-5y}{6} dy dx=dfrac{6}{55}$$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 29 '18 at 2:29









        Key FlexKey Flex

        7,74741232




        7,74741232






























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