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Find the volume between the graph of the given function and the xy-plane over the specified rectangle in the xy-plane

f(x,y)=-2x2y3WhereR={(x,y)|-2≤x≤3,-1≤y≤5}

Short Answer

Expert verified

The value of the integral is1686.66cubicunits

Step by step solution

01

Given information

We are given a rectangular region

f(x,y)=-2x2y3WhereR={(x,y)|-2≤x≤3,-1≤y≤5}

02

Find the volume

We have,

f(x,y)>0whenR1={(x,y)|-2≤x≤0,-1≤y≤0}Andf(x,y)<0whenR2={(x,y)|0≤x≤3,0≤y≤5}

Also we have,

∫R-2x2y3dA=∫R1-2x2y3dA-∫R2-2x2y3dA=∫-10∫-20-2x2y3dxdy-∫03∫05-2x2y3dxdy=[-2x3y412]-[-2x3y412]=-1612+[2025012]=1686.66cubicunits

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