Question

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

${\int }_0^{\frac{\mathrm{\pi }}{2}}{\int }_0^1{\int }_0^{\sqrt{1-r^2}}f\left(r,\theta ,z\right)rdzdrd\theta$

Short answer

The region represents inside the circle with equation $r=1$bounded aabove by the sphere with radius 1 centered at the origin in the$r-\theta$plane

Step-by-step solution

Step 1:Given information 

The given expression is ${\int }_0^{\frac{\mathrm{\pi }}{2}}{\int }_0^1{\int }_0^{\sqrt{1-r^2}}f\left(r,\theta ,z\right)rdzdrd\theta$

Step 2:Simplification

Given

${\int }_0^{\frac{\mathrm{\pi }}{2}}{\int }_0^1{\int }_0^{\sqrt{1-r^2}}f\left(r,\theta ,z\right)rdzdrd\theta$

From the limits of z

$z=\sqrt{1-r^2}$

$z^2=1-r^2$

$z^2+r^2=1$

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

This equation represents the equation of sphere centered at origin

From the limits of r

$r=1$

$r^2=1$

$x^2+y^2=1$

This equation represents the equation of circle with radius 1

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

Hence, the region represents inside the circle with equation$r=1$bounded aabove by the sphere

with radius 1 centered at the origin in the $r-\theta$plane