Calculate the iterated integral integra20 integrapi0 5r sin2

Calculate the iterated integral integra^2_0 integra^pi_0 5r sin^2 theta dtheta dr

Solution

[ 0 to 2   (5 r) dr] [ 0 to ( sin²) d]

= [ 0 to 2   (5 r) dr] [ 0 to ( 1-cos2 /2) d]

= 5 [r²/2]₀² . (1/2) [ () - (sin2/2) ]₀

= 5 [2²/2 - 0] . (1/2) [ (-0) - ( sin2/2 - 0 )]

= 5.[2].(1/2) [ -0]                                   [ sin = 0 ]

= 5