Question

For the below-mentioned figure, conversion from cartesian coordinate ∭R f(x,y,z)dx dy dz to cylindrical polar with coordinates p(ρ,∅,z) is given by ______

(a) ∭R^* f(ρ,∅,z) ρ dρ d∅ dz

(b) ∭R f(ρ,∅,z)  dρ d∅ dz

(c) ∭R^*f(ρ,∅,z) ρ∅ dρ d∅ dz

(d) ∭R f(ρ,∅,z) ρ^2 dρ d∅ dz

I had been asked this question during an interview.

My enquiry is from Change of Variables In a Triple Integral in section Multiple Integrals of Engineering Mathematics

Answer 1

Right answer is (a) ∭R^* f(ρ,∅,z) ρ dρ d∅ dz

For explanation I would say: From the figure we can write x=ρ cos ∅, y=ρ sin ∅, z=z

now we know that during change of variables f(x,y,z) is replaced by

\(f(ρ,∅,z)*J\left(\frac{x,y,z}{ρ,∅,z}\right)\) with limits in functions of  x,y,z to functions of ρ,∅,z respectively

\(J\left(\frac{x,y,z}{ρ,∅,z}\right)= \begin{vmatrix}

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

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

\frac{∂z}{∂p} &\frac{∂z}{∂∅} &\frac{∂z}{∂z}\\

\end{vmatrix}

= \begin{vmatrix}

cos⁡∅ &-p sin⁡∅ &0\\

sin⁡∅ &p cos⁡∅ &0\\

0 &0 &1\\

\end{vmatrix} = cos ∅(ρ cos ∅) + ρ sin ∅ (sin ∅)\)

= ρ, thus ∭R f(x,y,z)dx dy dz = ∭R^* f(ρ,∅,z) ρ dρ d∅ dz where R^* is the new region.