Change of Variables in Double Integrals Examples 2

# Change of Variables in Double Integrals Examples 2

Recall from the Change of Variables in Double Integrals page that if $z = f(x, y)$ is a continuous two variable real-valued function, $T : S \to R$ is a one-to-one transformation (except possibly on the boundary of $S$), the Jacobian Determinant $\frac{\partial (x, y)}{\partial (u, v)}$ is nonzero, and $x = x(u, v)$, $y = y(u, v)$, and their first partial derivatives with respect to $u$ and $v$ are continuous, then:

(1)
\begin{align} \quad \iint_R f(x, y) \: dA = \iint_S f(x(u, v), y(u, v)) \biggr \rvert \frac{\partial (x, y)}{\partial (u, v)} \biggr \rvert \: du \: dv \end{align}

We will now look at some more examples of evaluating double integrals using a change of variables.

## Example 1

Evaluate $\iint_D xy \: dA$ where $D$ is the region in the first quadrant bounded by the curves $y = x$, $y = 3x$, $y = \frac{1}{x}$, $y = \frac{3}{x}$ and using the transformation $x = \frac{u}{v}$ and $y = v$.

The region that we wish to integrate over is given below: We need to apply our transformation to each of the boundary curves. We will have that $y = x$ implies that $v^2 = u$, $y = 3x$ implies that $v = 3 \frac{u}{v}$ which implies that $v^2 = 3u$, $y = \frac{1}{x}$ implies that $v = \frac{v}{u}$ which implies that $u = 1$, and $y = \frac{3}{x}$ implies that $v = 3 \frac{v}{u}$ which implies that $u = 3$.

So our new boundary curves are $v^2 = u$, $v^2 = 3u$, $u = 1$ and $u = 3$: This region, call it $D_{uv}$ can be described as:

(2)
\begin{align} \quad D_{uv} = \{ (u, v) : 1 ≤ u ≤ 3, \sqrt{u} ≤ v ≤ \sqrt{3u} \} \end{align}

We now need to compute the Jacobian for this transformation:

(3)
\begin{align} \quad \frac{\partial (x, y)}{\partial (u, v)} = \begin{vmatrix} \frac{\partial x}{\partial u} & \frac{\partial x}{\partial v} \\ \frac{\partial y}{\partial u} & \frac{\partial y}{\partial v} \end{vmatrix} = \begin{vmatrix} \frac{1}{v} & - \frac{u}{v^2} \\ 0 & 1 \end{vmatrix} = \frac{1}{v} \end{align}

We are now ready to compute this integral.

(4)
\begin{align} \quad \iint_D xy \: dA = \iint_{D_{uv}} u \biggr \rvert \frac{\partial (x, y)}{\partial (u, v)} \biggr \rvert \: du \: dv \\ \quad \iint_D xy \: dA = \iint_{D_{uv}} \frac{u}{v} \: du \: dv \\ \quad \iint_D xy \: dA = \int_1^3 \int_{\sqrt{u}}^{\sqrt{3u}} \frac{u}{v} \: dv \: du \\ \quad \iint_D xy \: dA = \int_1^3 u \left [ \ln \mid v \mid \right ]_{v = \sqrt{u}}^{v = \sqrt{3u}} \: du \\ \quad \iint_D xy \: dA = \int_1^3 u [\ln \sqrt{3u} - \ln \sqrt{u} ] \: du \\ \quad \iint_D xy \: dA = \int_1^3 u \ln \frac{\sqrt{3u}}{\sqrt{u}} \: du \\ \quad \iint_D xy \: dA = \int_1^3 u \ln \sqrt{\frac{3u}{u}} \: du \\ \quad \iint_D xy \: dA = \int_1^3 u \ln \sqrt{3} \: du \\ \quad \iint_D xy \: dA = \ln \sqrt{3} \left [ \frac{u^2}{2} \right ]_{u=1}^{u=3} \\ \quad \iint_D xy \: dA = \ln \sqrt{3} \left [ \frac{9}{2} - \frac{1}{2} \right ] \quad \iint_D xy \: dA = 4 \ln \sqrt{3} \\ \quad \iint_D xy \: dA = \ln 9 \end{align}