Question

The volume of the region R defined by inequalities 0≤z≤1, 0≤y+z≤2,0≤x+y+z≤3 is given by ______

(a) 4

(b) 6

(c) 8

(d) 1

This question was addressed to me in an interview for internship.

The doubt is from Change of Variables In a Triple Integral topic in division Multiple Integrals of Engineering Mathematics

Answer 1

Right option is (b) 6

The explanation: It is observed from equations that the region is made of parallelepiped thus volume of parallelepiped is given by triple integral over the given region.

i.e by using substitutions as x+y+z=p, y+z=q, z=r the new region becomes R^* where p varies from 0 to 3, q varies from 0 to 2 & r varies from 0 to 1 jacobian of this transformation is given by

\(J\left(\frac{p,q,r}{x,y,z}\right) = \begin{vmatrix}

\frac{∂p}{∂x} & \frac{∂p}{∂y} &\frac{∂p}{∂z}\\

\frac{∂q}{∂x} &\frac{∂q}{∂y} &\frac{∂q}{∂z}\\

\frac{∂r}{∂x} &\frac{∂r}{∂y} &\frac{∂r}{∂z}\\

\end{vmatrix} = \begin{vmatrix}

1&1&1\\

0&1&1\\

0&0&1\\

\end{vmatrix} = 1(1) – 1(0) + 1(0) = 1\)

but we need \(J\left(\frac{x,y,z}{p,q,r}\right) \,w.k.t\, J\left(\frac{x,y,z}{p,q,r}\right) J\left(\frac{p,q,r}{x,y,z}\right) = 1 \,thus\, J\left(\frac{x,y,z}{p,q,r}\right)=1\)

now the volume is given by \(\int_ 0^1 \int_ 0^2\int_ 0^3 \,dp \,dq \,dr = \int_ 0^1 \int_ 0^2 3\, dq \,dr = \int_ 0^1 6dr = 6.\)