Q. Test if the following equations are dimensionally correct (answer me in brief){I want detailed explanation}
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Answer:
fXY(x,y) =∫
∞
−∞
fXYZ(x,y,z)dz =∫
1
0
1
3
(x+2y+3z)dz =
1
3
[(x+2y)z+
3
2
z2]
1
0
=
1
3
(x+2y+
3
2
), for0≤x,y≤1.
Thus,
fXY(x,y)={
1
3
(x+2y+
3
2
) 0≤x≤1,0≤y≤1 0
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