Problem 1
1. Let \(R\) be the triangular region with vertices \((0,0)\), \((1,1)\), \((1,2)\) in the \(xy\)-plane. Find the linear transformation that takes the triangle \(R_0\) with vertices \((0,0)\), \((0,1)\), \((1,0)\) in the \(uv\)-plane to \(R\). Then use the change of variables formula to find the area of \(R\). (5 points)
Solution. Recall that a linear transform from the \(uv\)-plane to the \(xy\)-plane is of the form \((x(u,v),y(u,v)) = T(u,v) = (au+bv, cu+dv)\). Using this as a base, we plug in the requested points to come up with a transformation that works. Note that \(T(0,0) = (0,0)\) guarantees the transformation is linear and we have no constant term. Then, \(T(0,1) = (b,d)\), and we want this vertex to go to \((1,1)\), so \(b=1\) and \(d=1\). Similarly, \(T(1,0) = (a,c) = (1,2)\), so our final linear transformation is \(T(u,v) = (u+v, 2u+v)\). Taking the Jacobian gives
\[ J = \begin{vmatrix} 1 & 1 \\ 2 & 1 \end{vmatrix} = |-1| = 1. \]
Lastly, we calculate the area with the change of variables formula. \[ \int\!\!\int_R dA = \int\!\!\int_{R_0} 1\, dA = \text{Area}(R_0) = 1/2. \]
Problem 2
2. Let \(R\) denote the parallelogram in the \(xy\)-plane with vertices \((0,0)\), \((1,0)\), \((2,1)\), \((1,1)\). Set up but do not calculate the double integral \(\displaystyle\int\!\!\int_R (y-x)\, dA\) using the change of variables theorem. Hint: Using the inverse transform \(u=y-x\), \(v=y\) will help find the corresponding region in the \(uv\)-plane. (5 points)
Solution. As before, we already know the form of a linear transformation. In addition, we have the inverse already, and rearranging tells us that the forward transform must be \(T(u,v) = (v-u, v)\). Next, to find the region to integrate over, use the inverse transform on each of the points. Let \(S(x,y) = (y-x, y)\). Then \(S(0,0) = (0,0)\), \(S(1,0) = (-1,0)\), \(S(2,1) = (-1,1)\), and \(S(1,1) = (0,1)\). That is, the shape in the \(uv\)-plane is a square with vertices \((0,0)\), \((0,1)\), \((-1,0)\), \((-1,1)\). Lastly, we take the Jacobian of the transform, which is
\[ J = \begin{vmatrix} -1 & 1 \\ 0 & 1 \end{vmatrix} = |-1| = 1. \]
Now we set up the integral. \[ \int\!\!\int_R y-x\, dA = \int_{-1}^0\!\int_0^1 u\, dv\, du. \]