The given change of coordinates has Jacobian
[tex]\mathbf J=\begin{bmatrix}\dfrac{\partial x}{\partial u}&\dfrac{\partial x}{\partial v}\\\\\dfrac{\partial y}{\partial u}&\dfrac{\partial y}{\partial v}\end{bmatrix}=\begin{bmatrix}9&1\\1&9\end{bmatrix}[/tex]
so the area element is
[tex]\mathrm dA=\mathrm dx\,\mathrm dy=|\det\mathbf J|\,\mathrm du\,\mathrm dv=80\,\mathrm du\,\mathrm dv[/tex]
The new region is a right triangle with vertices (0, 0), (1, 0), and (0, 1) in the [tex]u,v[/tex] plane.
Then the integral becomes
[tex]\displaystyle\iint_R(x-10y)\,\mathrm dA=-80\int_{u=0}^{u=1}\int_{v=0}^{v=1-u}(u+89v)\,\mathrm dv\,\mathrm du=-1200[/tex]
Using the given transformation to evaluate the given integral ∫[tex]_{R}[/tex] (x - 10y) dA is; -1200
We are given;
x = 9u + v
y = u + 9v
This is a matrix;
[tex]\left[\begin{array}{ccc}9&1\\1&9\\\end{array}\right][/tex]
The determinant of this matrix is;
J_a = (9 * 9) - (1 * 1)
J_a = 80
We are given that we are to evaluate the integral, (x − 10y)dA, over R.
Let us put 9u + v for x and u + 9v for y to get;
x - 10y = 9u + v - 10(u + 9v)
⇒ -u - 89v = -(u + 89v)
Thus, by the jacobian method, the integral is;
[tex]J_{A} \int\limits^ {} \, \int\limitsx_{R} {(x - 10y)} \, dA[/tex] = [tex]-80\int\limitsx_{u = 0} ^{u = 1} \int\limitsx_{v = 0} ^{v = 1 - u} {u + 89v} \, dv.du[/tex]
Using online double integral calculator, we have;
Integral = -1200
Read more about double integral at; https://brainly.com/question/19818792
