Question

X’=x+y⁄√2 , Y’=x-y∕√2. The relation between the area element dx’dy’ and dxdy is given by dx’dy’=Jdxdy.the value of J is….(A)2, (B)1,(C)-1,(D)-2

Answer 1

What is the question?

You are using the Jacobian and change of variables to calculate some area or other ?

Answer 2

oh, i read this again on laptop, makes more sense now!

for the Jacobian here, you are looking for the determinant of the matrix \(\Large \left[\begin{matrix}\frac{\partial x'}{\partial x} & \frac{\partial x'}{\partial y} \\ \frac{\partial y'}{\partial x} & \frac{\partial y'}{\partial y}\end{matrix}\right]\)

if it's \(x' = x + \dfrac{y}{\sqrt{2}}\) and \(y' = x - \dfrac{y}{\sqrt{2}}\) -- your post is unclear -- then that is \[\Large \left|\begin{matrix}1 & \frac{1}{\sqrt{2}} \\ 1 & -\frac{1}{\sqrt{2}}\end{matrix}\right|\]

if not, you get the idea. :p

Answer 3

oh...sry its a question from questio paper.......so i didnt recogniz it....thnxxx..........

Answer 4

X'=\[\frac{ x+y }{ \sqrt{2} }\] & Y'=\[\frac{ x-y }{ \sqrt{2} }\]

Answer 5

in that case just recalculate the Jacobian \[\Large \left[\begin{matrix}\frac{\partial x'}{\partial x} & \frac{\partial x'}{\partial y} \\ \frac{\partial y'}{\partial x} & \frac{\partial y'}{\partial y}\end{matrix}\right]\]