I'm going to rush through most of this stuff, as I feel I've got it correct. It's just the answer to the last question that's bugging me.

**Question**

Let *R *be the region in the *xy*-plane enclosed by the lines:

*x *+ *y *= 0, *x *+ *y *= 1, *x *– *y *= 1, *x *– *y *= 4.

(a) Draw a picture of the region *R*.

**answer:**

y= -x, y = 1-x, y=x-1 & y = x-4. It makes a rectangle with the corners (0.5,-0.5), (1,0), (2,-2), & (2.5,-1.5).

(b) Define a change-of-variables transformation by the equations

*u *= *x *+ *y* & *v *= *x *– *y*.

What is the image *S *in the *uv*-plane of *R *under this transformation?

**answer:**

solving for x and y:

x=1/2(u+v) & y=1/2(u-v)

Putting these into the above equations I get the lines u=0, u=1, v=1/2 and v=2. It makes a rectangle with the corners (0,1/2), (0,2), (1,1/2) & (1,2)

(c) Solve the transformation equations for *x *and *y *in terms of *u *and *v*, and hence find the determinant of the Jacobian matrix.

**answer:**

$\displaystyle \tfrac{\partial x}{\partial u} = \tfrac{1}{2}, \tfrac{\partial x}{\partial v} = \tfrac{1}{2}, \tfrac{\partial y}{\partial u} = \tfrac{1}{2}, \tfrac{\partial y}{\partial v} = \tfrac{\text{-}1}{2} $

$\displaystyle \frac{\partial(x,y)}{\partial(u,v)} =\begin{vmatrix}\frac{1}{2} & \frac{1}{2}\\ \frac{1}{2} & \frac{\text{-}1}{2}\end{vmatrix}=\frac{\text{-}1}{2}$

(d)Use this change of variables to evaluate

$\displaystyle \int \int_R (x - y)e^{x^2-y^2}dA$

**answer:**

simplifying

$\displaystyle x-y = \tfrac{1}{2}(u+v)-\tfrac{1}{2}(u-v) = v$

$\displaystyle x^2-y^2 = (\tfrac{1}{2}(u+v))^2 - \tfrac{1}{2}(u-v))^2 = \frac{uv}{2}$

makes the equation:

$\displaystyle \int_{\frac{1}{2}}^{2}\int_{0}^{1}v.e^{\frac{vu}{2 }}(\tfrac{\text{-}1}{2}). \partial u\partial v$

inner integral:

$\displaystyle \int_{\frac{1}{2}}^{2}\int_{0}^{1}v.e^{\frac{vu}{2 }}(\tfrac{\text{-}1}{2}). \partial u\partial v = -e^{\frac{uv}{2}}\bigg|_{0}^{1} = \int_{\frac{1}{2}}^{2}\text{-}e^{\frac{v}{2}}+1. \partial v$

outer integral:

$\displaystyle \int_{\frac{1}{2}}^{2}\text{-}e^{\frac{v}{2}}+1. \partial v = v-2e^{\frac{v}{2}}\bigg|_{\frac{1}{2}}^{2} = (2 - 2e^{1})-(\tfrac{1}{2} - 2e^{\frac{1}{4}})= 1\tfrac{1}{2} - 2(e^{1}-e^{\frac{1}{4}})$

So is my answer to (d) correct? It's seems so unwieldy and unpleasant an answer considering how nice and easy the rest of the Question was. If anyone could check my work and affirm/negate my answers, please do!

cheers