Q 44 - Calculus

Question

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

${\int }_0^{\frac{\mathrm{\pi }}{2}}{\int }_0^{\frac{\mathrm{\pi }}{2}}{\int }_0^{1}f\left(\rho ,\theta ,\varphi \right){\rho }^2\sin \varphi d\rho d\theta d\varphi$

Step-by-Step Solution

Answer

The region represents sphere with second quadrant with the hemisphere below the  $xy-$plane of the sphere with radius 1 centered at origin 

Step 1:Given information

The given expression is ${\int }_0^{\frac{\mathrm{\pi }}{2}}{\int }_0^{\frac{\mathrm{\pi }}{2}}{\int }_0^{1}f\left(\rho ,\theta ,\varphi \right){\rho }^2\sin \varphi d\rho d\theta d\varphi$

Step 2:Simplification

${\int }_0^{\frac{\mathrm{\pi }}{2}}{\int }_0^{\frac{\mathrm{\pi }}{2}}{\int }_0^{1}f\left(\rho ,\theta ,\varphi \right){\rho }^2\sin \varphi d\rho d\theta d\varphi$

According to the order of integration 

$\Rightarrow \rho =0 to \rho =1$

$\Rightarrow \sqrt{x^2+y^2+z^2}=1$

$\Rightarrow x^2+y^2+z^2=1^2$

This is a sphere equation with radius 1 centered at origin

Now,

$\Rightarrow \theta$ varies from $0 to \frac{\mathrm{\pi }}{2}$

It means it is in second quadrant

Now,

$\Rightarrow \vartheta$ varies from $0 to \frac{\mathrm{\pi }}{2}$

these the region represents the hemisphere below the xy -plane of the sphere with radius 1 centered at origin