# Volume with double integral

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!

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

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