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
Real Analysis
Integral Calculus
Triple Integral
But try the problem first...
Answer: $\textbf{(B)} \quad y$
IIT JAM 2016, Question No. 15
Differential and Integral Calculus: R Courant
First hint
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)
Second Hint
$2x+y+z=4$
$\Rightarrow z=4-2x-y$
$\Rightarrow 2x+y=4$ [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 ?
Final Step
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)
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
Real Analysis
Integral Calculus
Triple Integral
But try the problem first...
Answer: $\textbf{(B)} \quad y$
IIT JAM 2016, Question No. 15
Differential and Integral Calculus: R Courant
First hint
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)
Second Hint
$2x+y+z=4$
$\Rightarrow z=4-2x-y$
$\Rightarrow 2x+y=4$ [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 ?
Final Step
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)
