Question

For the below mentione figure ,conversion from cartesian coordinate  ∭R f(x,y,z)dx dy dz to spherical polar  with coordinates p(r,θ,∅) is given by ______

(a) ∭R^* f(r,θ,∅) sin⁡θ dr dθ d∅

(b) ∭R^* f(r,θ,∅) r^2 dr dθ d∅

(c) ∭R^* f(r,θ,∅) r^2  cos⁡θ dr dθ d∅

(d) ∭R^* f(r,θ,∅) r^2 sin⁡θ dr dθ d∅

This question was posed to me in an interview for job.

I'm obligated to ask this question of Change of Variables In a Triple Integral in portion Multiple Integrals of Engineering Mathematics

Answer 1

Right option is (d) ∭R^* f(r,θ,∅) r^2 sin⁡θ dr dθ d∅

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

now we know that during a 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 r,θ,∅ respectively

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

\frac{∂x}{∂r} &\frac{∂x}{∂θ} &\frac{∂x}{∂∅}\\

\frac{∂y}{∂r} &\frac{∂y}{∂θ}& \frac{∂y}{∂∅}\\

\frac{∂z}{∂r} &\frac{∂z}{∂θ} &\frac{∂z}{∂∅}\\

\end{vmatrix} = \begin{vmatrix}

sin θ cos ∅ &r cos θ cos ∅  &-r sin θ sin ∅\\

sin θ sin ∅ &r cos θ sin ∅ &r sin θ cos ∅\\

cos θ& -r sin θ &0\\

\end{vmatrix}\)

= sin θ cos ∅(r^2 sin^2 θ cos⁡∅) + r cos θ cos ∅(r sin θ cos ∅ cos θ) – r sin θ sin ∅

= (-r sin^2 θ sin⁡∅-r cos^2 θ sin⁡∅)……on solving we get r^2  sin⁡θ

thus ∭R f(x,y,z)dx dy dz = ∭R^* f(r,θ,∅)r^2  sin⁡θ  dr dθ d∅ where R^* is the new region.