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1/3 x + y 7. Consider dA where R is the region bounded by the triangle with vertices (0,0), (2,0), V= x+y X-y and (0,-2). The

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Answer #1

In this question we have to first find the set S it is nothing but we have to write in the stet notation then put the points one by one and hence again we have a triangle with different coordinates.

Now find all the partial derivatives as told and then find the determinant.

Then just take inverse and look at the region find the limits and integrate.

In he last result there is (-1)^(1/3) so it has he complex value if you want o see he real value of this then result is 0 but if complex then there is some value as (-1)^{1/3}=e^{i\pi/3} 0 10,-2) The charge of variabi Questo consider Ya Y3 x+y DA bounded the triangle SS (* where R x-Y 2 त्र R in the region withTo,o) (o tom, 0:0) = 10,0) 10,0) T10,0) = T12,0) (اوا) T12,0) = (,) T10, 3) = 100,-(-) 3 T10,-2) - (-1,!) All { 2,0), (2,05,て 3 (ninle Chele 아 아 2 NH 2 (ninle (hlade = ()(1)-(4-16) ) 2 s - 4 2 - 4 2 니 s) (^. nie (6“х). = --- 2 3 (c) Chale +/-)2 = =equation lineas и (0,0) to (1,1) slop » 1-0 So equation of the (V-1) =(1-0) VEU lles u varies from u=-v to UPV so -V to v and+3 11+1-193) !!:-) 9/3 () 813 - 3X (1+ (61)Y3) X3 8 q 16 (! + (143) Y3 Real Plane We can consider (1) -1 = of Y3 43 In That c

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