Q 45 - Calculus

Question

The iterated integrals use spherical coordinates. Describe the solids determined by the limits of integration. 

${\int }_0^{\frac{\mathrm{\pi }}{4}}{\int }_0^{2\mathrm{\pi }}{\int }_0^{3sec\theta }f\left(\rho ,\theta ,\varphi \right){\rho }^2\sin \varphi d\rho d\theta d\varphi$

Step-by-Step Solution

Answer

The region represents inside the upward opening right circular cone that has z-axis as its axis of symetry height 3 and radius 3

Step 1:Given information

The given expression is ${\int }_0^{\frac{\mathrm{\pi }}{4}}{\int }_0^{2\mathrm{\pi }}{\int }_0^{3sec\theta }f\left(\rho ,\theta ,\varphi \right){\rho }^2\sin \varphi d\rho d\theta d\varphi$

Step 2:Simplification

Given 

${\int }_0^{\frac{\mathrm{\pi }}{4}}{\int }_0^{2\mathrm{\pi }}{\int }_0^{3sec\theta }f\left(\rho ,\theta ,\varphi \right){\rho }^2\sin \varphi d\rho d\theta d\varphi$

According to order of integration 

$\rho =0 to \rho =3sec\theta$

$\Rightarrow \sqrt{x^2+y^2+z^2}=3sec\theta$

$\Rightarrow x^2+y^2+z^2=3^{2}se{c}^2\theta$

This represents equation of right circular cone

And $\theta$ varies from $\theta =0 to \theta =2\mathrm{\pi }$

Also $\varphi$varies from $\varphi =0 to \varphi =\frac{\mathrm{\pi }}{4}$

From these the region represents inside the upward opening right circular cone that has z-axis as its axis of symetry height 3 and radius 3