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!
$6x+5y+6z = 6$, $y=x$, $x=0$ 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 $$\int_{0}^{\frac{6}{11}}\int_{x}^{\frac{6-6x}{5}}\dfrac{6-6x-5y}{6}\ dy\ dx=\dfrac{6}{55}$$