We first calculate the integral
0∫2exyzdz=={xyexyz∣02,forxy=0z∣02,otherwise={xye2xy−xy1,forxy=02,otherwise.
Then find
0∫2{xye2xy−xy1,forxy=02,otherwisedy==⎩⎨⎧0∫2xye2xy−1dy,forxy=00∫22dy,otherwise==⎩⎨⎧0∫2xye2xy−1dy,forxy=02y∣02,otherwise==⎩⎨⎧0∫2xye2xy−1dy,forxy=04,otherwise.
Then find
0∫2⎩⎨⎧0∫2xye2xy−1dy,forxy=04,otherwisedx==⎩⎨⎧0∫20∫2xye2xy−1dxdy,forxy=00∫24dx,otherwise==⎩⎨⎧0∫20∫2xye2xy−1dxdy,forxy=04x∣02,otherwise==⎩⎨⎧0∫20∫2xye2xy−1dxdy,forxy=08,otherwise.
Consider the function xye2xy−1 and find limit
xy↦0limxye2xy−1=2
because
t↦0limtet−1=1 .
This means that the function is integrable and the integral is convergent.
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