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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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  • $\begingroup$ What is the correct answer? $\endgroup$ – K Split X Nov 29 '18 at 2:29
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$6x+5y+6z = 6$, $y=x$, $x=0$ 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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