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Evaluate the iterated integral :

โˆซ0ฯ€/2โ€Šโˆซ0cosโกyโ€Šโˆซ0sinโกyโ€Š(2x+y)dzdxdy

Short Answer

Expert verified

ฯ€8+13

Step by step solution

01

Step 1. Given information.

Given integral is :

โˆซ0ฯ€/2โ€Šโˆซ0cosโกyโ€Šโˆซ0sinโกyโ€Š(2x+y)dzdxdy

We have to evaluate it.

02

Step 2. Evaluate.

Given

โˆซ0ฯ€/2โ€Šโˆซ0cosโกyโ€Šโˆซ0sinโกyโ€Š(2x+y)dzdxdy

=โˆซ0ฯ€2โ€Šโˆซ0cosโกyโ€Šโˆซ0sinโกyโ€Š(2x+y)dzdxdy=โˆซ0ฯ€2โ€Šโˆซ0cosโกyโ€Š2xโˆซ0sinโกyโ€Šdz+yโˆซ0sinโกyโ€Šdzdxdy=โˆซ0ฯ€2โ€Šโˆซ0cosโกyโ€Š2x(z)0sinโกy+y(z)0sinโกydxdy=โˆซ0ฯ€2โ€Šโˆซ0cosโกyโ€Š[2xsinโกy+ysinโกy]dxdy

03

Step 3. Integrate with respect to x.

Integrate with respect to x

=โˆซ0ฯ€2โ€Šโˆซ0cosโกyโ€Š2xsinโกydx+โˆซ0cosโกyโ€Šysinโกydxdy=โˆซ0ฯ€2โ€Š2sinโกyโˆซ0cosโกyโ€Šxdx+ysinโกyโˆซ0cosโกyโ€Šdxdy=โˆซ0ฯ€2โ€Š2sinโกyx220cosโกy+ysinโกy(x)0cosโกydy=โˆซ0ฯ€2โ€Š2sinโกycos2โกy2+ysinโกy(cosโกy)dy

Integrate with respect to y

=โˆซ0ฯ€2โ€Šsinโกycos2โกy+ysinโกycosโกydy=โˆซ0ฯ€2โ€Šsinโกycos2โกy+12โˆซ0ฯ€2โ€Šysinโก2ydy=โˆ’cos3โกy30ฯ€2+12โˆ’ycosโก2y2+sinโก2y40ฯ€2=โˆ’13(0โˆ’1)+12โˆ’12ฯ€2cosโกฯ€โˆ’0+14(sinโกฯ€โˆ’cosโก0)=13+12ฯ€4=ฯ€8+13

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