Compute the Jacobian of the transformation T given by $x = \sqrt { 2 } u - \frac { 1 } { \sqrt { 2 } } v$ , $y = \frac { 1 } { \sqrt { 2 } } u + \sqrt { 2 } v$ . Compute the area of the image of $S = \{ ( u , v ) \mid 0 \leq u \leq 1,0 \leq v \leq 1 \}$ and compare it to the area of S.

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The Answer of Compute the Jacobian of the transformation T given by $x = \sqrt { 2 } u - \frac { 1 } { \sqrt { 2 } } v$ , $y = \frac { 1 } { \sqrt { 2 } } u + \sqrt { 2 } v$ . Compute the area of the image of $S = \{ ( u , v ) \mid 0 \leq u \leq 1,0 \leq v \leq 1 \}$ and compare it to the area of S.

Compute the Jacobian of the Transformation T Given By x=2u−12vx = \sqrt { 2 } u - \frac { 1 } { \sqrt { 2 } } vx=2​u−2​1​v

Compute the Jacobian of the Transformation T Given By $x = \sqrt { 2 } u - \frac { 1 } { \sqrt { 2 } } v$