Question

Use the given transformation to evaluate the integral. (x ? 10y) dA, R where R is the triangular region with vertices (0, 0), (9, 1), and (1, 9). x = 9u + v, y = u + 9v

Answer 1

The given change of coordinates has Jacobian

$\mathbf J=\begin{bmatrix}\dfrac{\partial x}{\partial u}&\dfrac{\partial x}{\partial v}\\\\\dfrac{\partial y}{\partial u}&\dfrac{\partial y}{\partial v}\end{bmatrix}=\begin{bmatrix}9&1\\1&9\end{bmatrix}$

so the area element is

$\mathrm dA=\mathrm dx\,\mathrm dy=|\det\mathbf J|\,\mathrm du\,\mathrm dv=80\,\mathrm du\,\mathrm dv$

The new region is a right triangle with vertices (0, 0), (1, 0), and (0, 1) in the $u,v$ plane.

Then the integral becomes

$\displaystyle\iint_R(x-10y)\,\mathrm dA=-80\int_{u=0}^{u=1}\int_{v=0}^{v=1-u}(u+89v)\,\mathrm dv\,\mathrm du=-1200$

Answer 2

Using the given transformation to evaluate the given integral ∫$_{R}$ (x - 10y) dA is; -1200

How to solve double integral?

We are given;

x = 9u + v

y = u + 9v

This is a matrix;

$\left[\begin{array}{ccc}9&1\\1&9\\\end{array}\right]$

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;

$J_{A} \int\limits^ {} \, \int\limitsx_{R} {(x - 10y)} \, dA$ = $-80\int\limitsx_{u = 0} ^{u = 1} \int\limitsx_{v = 0} ^{v = 1 - u} {u + 89v} \, dv.du$

Using online double integral calculator, we have;

Integral = -1200

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