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Triple Integral | IIT JAM 2016 | Question 15

Question 15 - Triple Integral (IIT JAM 2016)


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

  • x+y
  • x-y
  • x
  • y

Key Concepts


Real Analysis

Integral Calculus

Triple Integral

Check the Answer


Answer: \textbf{(B)} \quad y

IIT JAM 2016, Question No. 15

Differential and Integral Calculus: R Courant

Try with Hints


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)

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Question 15 - Triple Integral (IIT JAM 2016)


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

  • x+y
  • x-y
  • x
  • y

Key Concepts


Real Analysis

Integral Calculus

Triple Integral

Check the Answer


Answer: \textbf{(B)} \quad y

IIT JAM 2016, Question No. 15

Differential and Integral Calculus: R Courant

Try with Hints


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)

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