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Evaluate the iterated integral :

0π/20cosy0siny(2x+y)dzdxdy

Short Answer

Expert verified

π8+13

Step by step solution

01

Step 1. Given information.

Given integral is :

0π/20cosy0siny(2x+y)dzdxdy

We have to evaluate it.

02

Step 2. Evaluate.

Given

0π/20cosy0siny(2x+y)dzdxdy

=0π20cosy0siny(2x+y)dzdxdy=0π20cosy2x0sinydz+y0sinydzdxdy=0π20cosy2x(z)0siny+y(z)0sinydxdy=0π20cosy[2xsiny+ysiny]dxdy

03

Step 3. Integrate with respect to x.

Integrate with respect to x

=0π20cosy2xsinydx+0cosyysinydxdy=0π22siny0cosyxdx+ysiny0cosydxdy=0π22sinyx220cosy+ysiny(x)0cosydy=0π22sinycos2y2+ysiny(cosy)dy

Integrate with respect to y

=0π2sinycos2y+ysinycosydy=0π2sinycos2y+120π2ysin2ydy=cos3y30π2+12ycos2y2+sin2y40π2=13(01)+1212π2cosπ0+14(sinπcos0)=13+12π4=π8+13

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