If the triple integral over the region bounded by the planes $2x+y+z=4$ $x=0$ $y=0$ $z=0$ is given by $\int\limits_0^2\int\limits_0^{\lambda(x)}\int\limits_0^{\mu(x,y)}\mathrm d z\mathrm d y\mathrm d x$ then the function $\lambda(x)-\mu(x,y)$ is

Here we are given with triple integral over the region bounded by the planes $2x+y+z=4, x=0, y=0$ and $z=0$

Now we our aim here is to find $\lambda (x) $ and $\mu(x,y)$. Now we will approach this problem by find the volume of $(x,y,z)$ based on $2x+y+z=4$ can you do this ??? (With the given information x=0, y=0, z=0)

$2x+y+z=4$

$\Rightarrow z=4-2x-y$

$\Rightarrow 2x+y=4$ [as $z=0$]

$\Rightarrow y=4-2x$

Again,

$2x+y+z=4$

Now as $y=z=0$ we have $2x=4$

Therefore $x=2$

Now can you use this to move forward with this problem ?

So our triple integral become,

$\int_0^2\int_0^{(4-2x)}\int_0^{(4-2x-y)}\mathrm d z\mathrm d y\mathrm dx$

On compairing $\lambda(x)=4-2x$ and $\mu(x,y)=4-2x-y$

Therefore $\lambda(x)-\mu(x,y)=4-2x-4+2x+y=y$ (ANS)