Question

Math

Posted 4 months ago

```
Suppose we wanted to evaluate the double integral
$S=\iint_{D} \sqrt{x^{2}+y^{2}} d x d y$
by first applying a change of variables from $D$ to $R$ :
$\begin{array}{l}
x=X_{1}(u, v)=-u^{2}+v \\
y=X_{2}(u, v)=u^{2}+2 u-v
\end{array}$
What is $S$ under the change of variables?
If you know an expression within absolute value is non-negative, do not use absolute value at all.
$S=\iint_{R} \square \quad \overline{=x} d u d v$
```

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Answer from Sia

Posted 4 months ago

Solution by Steps

step 1

To evaluate the double integral with a change of variables, we first need to find the Jacobian determinant of the transformation

step 2

The Jacobian matrix is given by $\begin{bmatrix} \frac{\partial x}{\partial u} & \frac{\partial x}{\partial v} \\ \frac{\partial y}{\partial u} & \frac{\partial y}{\partial v} \end{bmatrix}$, which for our transformation is $\begin{bmatrix} -2u & 1 \\ 2u+2 & -1 \end{bmatrix}$

step 3

The Jacobian determinant, $J(u,v)$, is the determinant of the Jacobian matrix: $J(u,v) = (-2u)(-1) - (1)(2u+2)$

step 4

Simplifying the Jacobian determinant gives us $J(u,v) = 2u - 2u - 2 = -2$

step 5

Since we know that $J(u,v)$ is non-negative, we can omit the absolute value and use $J(u,v) = 2$ for the change of variables

step 6

The double integral $S$ under the change of variables is then $S = \iint_{R} \sqrt{(-u^2+v)^2+(u^2+2u-v)^2} \cdot |J(u,v)| \, du \, dv$

step 7

Substituting the Jacobian determinant, we get $S = \iint_{R} \sqrt{(-u^2+v)^2+(u^2+2u-v)^2} \cdot 2 \, du \, dv$

step 8

Simplify the integrand to find the expression in terms of $u$ and $v$

Answer

The expression for $S$ under the change of variables is $S = \iint_{R} \sqrt{(-u^2+v)^2+(u^2+2u-v)^2} \cdot 2 \, du \, dv$.

Key Concept

Change of Variables in Double Integrals

Explanation

When changing variables in a double integral, the integrand and the differential area element $dx \, dy$ are transformed using the new variables and the absolute value of the Jacobian determinant of the transformation.

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