0
$\begingroup$

Let D be bounded by $x+y=0, x+y=2, x-y=0, x-y=0$ Compute $\iint ((x+y)e^{(x^2+y^2)} \ dx \ dy$ by using change of variables $u=x+y, v=x-y$

I did $-1< x <1 $and $-x < y < -x+2$ dydx and got a Jacobian value of $2$ with the 2x2 matrix $$\begin{bmatrix}1 && 1 \\ 1 && -1\end{bmatrix}$$ is that right? and where do i go from there?

  • 0
    You wrote $x-y=0$ twice, is that a typo? – 2017-02-04

2 Answers 2

2

We have an inverse transformation,

$$T^{-1} : (x,y) \to (u,v)$$

So the determinant of the inverse jacobian matrix is,

$$J_{T^{-1}}=\det \begin{bmatrix}1 && 1 \\ 1 && -1\end{bmatrix}$$

Using,

$$1=\det{A^{-1} A}=\det{A^{-1}}\det{A}$$

We have,

$$|J_{T}|=\frac{1}{|J_{T^{-1}}|}$$

Now note,

$$(x+y)^2+(x-y)^2=2x^2+2y^2$$

$$\frac{u^2+v^2}{2}=x^2+y^2$$

Then proceed, with a substitution.

  • 0
    why 1/Jac? dont we just multply by Jac? – 2017-02-04
  • 0
    Because when we talk about Jacobian we usually have $x=f(u,v)$ and $y=g(u,v)$. Here we have the opposite, inverse of that. $u=..$ and $v=...$. @user3427042 Alternatively you can solve for $x$ and $y$ in terms of $u$ and $v$ the construct the Jacobian but you will see it is the same as taking $1/J_{T^{-1}}$ – 2017-02-04
  • 0
    gotcha thanks!! – 2017-02-04
1

Hint

We know that $u=x+y,v=x-y.$ Since our original region was $0

Now our integrand becomes $$u e^{\frac{u^2+v^2}{2}}$$ because $u^2+v^2=(x+y)^2+(x-y)^2=2x^2+2y^2.$ We multiply the integrand by $$\left |\frac{\partial(x,y)}{\partial(u,v)} \right |= \frac{1}{2}$$ to get the double integral:

$$\int_{0}^{2} \int_{0}^{2} \frac{u e^{\frac{u^2+v^2}{2}}}{2} \ du \ dv,$$ which is our new integral.