Question

Evaluate the double integral (9 - x^2 - y^2)^(1/2) where

R = {(x,y) : x^2 + y^2 < or = to 9 , x > or = to 0 }

Answer 1

ANSWER :9π

EXPLANATION

To calculate the integral , we transform the Cartesian coordinates into polar coordinates: $x=r cosθ, υ=r sinθ$ . The region of integration is a semicircle of radius $r=3, \theta \in \left[ -\frac { \pi }{ 2 } ,\frac { \pi }{ 2 } \right]$ . Since $x^2+y^2=9$ , then

$\iint _{ R }^{ \ }{ \sqrt { 9-{ x }^{ 2 }-{ y }^{ 2 } } dxdy\ =\int _{ -\frac { \pi }{ 2 } }^{ \ \frac { \pi }{ 2 } }{ \int _{ 0 }^{ 3 }{ r } \sqrt { 9-{ r }^{ 2 } } d\theta dr\ } }=$

$=\left( \int _{ -\frac { \pi }{ 2 } }^{ \ \frac { \pi }{ 2 } }{ d\theta } \right) \cdot \left( \int _{ 0 }^{ 3 }{ r } \sqrt { 9-{ r }^{ 2 } } \ dr \right) =$ $-\frac { \pi }{ 2 } \int _{ 0 }^{ 3 }{ \ } \sqrt { 9-{ r }^{ 2 } } \ d\left( 9-{ r }^{ 2 } \right) =-\frac { \pi }{ 2 } \cdot \frac { 2 }{ 3 } \ \left[ F(3)-F(0) \right]$ . Because$F(r)={ \left[ { \left( 9-{ r }^{ 2 } \right) }^{ \frac { 3 }{ 2 } } \right] }$, so $F(3)-F(0)=-27$ and $\iint _{ R }^{ \ }{ \sqrt { 9-{ x }^{ 2 }-{ y }^{ 2 } } dxdy\ =9\pi \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad }$