Change of Variables in Double Integrals Examples 1

# Change of Variables in Double Integrals Examples 1

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 examples of evaluating double integrals using a change of variables.

## Example 1

Let $z = f(x, y)$ be a a continuous two variable real-valued function and consider the transformation $x = r \cos \theta$ and $y = r \sin \theta$. Use a change of variables to find a formula for $\iint_D f(x, y) \: dA$.

We must first compute the Jacobian $\frac{\partial (x, y)}{\partial (r, \theta)}$ as follows:

(2)
\begin{align} \quad \frac{\partial (x, y)}{\partial (r, \theta)} = \begin{vmatrix} \frac{\partial x}{\partial r} & \frac{\partial x}{\partial \theta} \\ \frac{\partial y}{\partial r} & \frac{\partial y}{\partial \theta} \end{vmatrix} = \begin{vmatrix} \cos \theta & -r \sin \theta\\ \sin \theta & r \cos \theta \end{vmatrix} = r\cos^2 \theta + r \sin^2 \theta = r(\cos^2 \theta + \sin^2 \theta) = r \end{align}

Therefore $dA = r \: dr \: d \theta$ and so:

(3)
\begin{align} \quad \iint_R f(x, y) \: dA = \iint_S f(r\cos \theta, r \sin \theta) r \: dr \: d \theta \end{align}

## Example 2

Evaluate $\iint_D x^2 \: dA$ where $D$ is ellipse $9x^2 + 4y^2 = 36$ and using the transformation $x = 2u$ and $y = 3v$.

We must first find our new boundary under the transformation given. If we take $9x^2 + 4y^2 = 36$ and plug in $x = 2u$ and $y = 3v$ then:

(4)
\begin{align} \quad 9(2u)^2 + 4(3v)^2 = 36 \\ \quad 9(4u^2) + 4(9v^2) = 36 \\ \quad 36u^2 + 36v^2 = 36 \\ \quad u^2 + v^2 = 1 \end{align}

Thus out region of integration becomes the unit circle in the $uv$-plane - call this region $D_{uv}$. We can more nicely represent this region with polar coordinates. Let $u = r \cos \theta$ and $v = r \sin \theta$ and then $D_{uv} = \{ (r, \theta) : 0 ≤ r ≤ 1, 0 ≤ \theta ≤ 2\pi \}$.

We now compute the Jacobian $\frac{\partial (x, y)}{\partial (u, v)}$:

(5)
\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} 2 & 0 \\ 0 & 3 \end{vmatrix} = 6 \end{align}
(6)
\begin{align} \quad \iint_D x^2 \: dA = \iint_{D_{uv}} 4u^2 \biggr \rvert \frac{\partial (x, y)}{\partial (u, v)} \biggr \rvert \: du \: dv \\ \quad \iint_D x^2 \: dA = \iint_{D_{uv}} 24u^2 \: du \: dv \\ \quad \iint_D x^2 \: dA = \int_0^{2 \pi} \int_0^1 24 r^3 \cos^2 \theta \: dr \: d \theta \\ \quad \iint_D x^2 \: dA = 24 \int_0^{2\pi} \cos^2 \theta \left [ \frac{r^4}{4} \right ]_{r=0}^{r=1} \: d \theta \\ \quad \iint_D x^2 \: dA = 6 \int_0^{2\pi} \cos^2 \theta \: d \theta \\ \quad \iint_D x^2 \: dA = 3 \int_0^{2\pi} (1 + \cos 2\theta) \: d \theta \\ \quad \iint_D x^2 \: dA = 3 \left [ \theta + \frac{1}{2} \sin 2 \theta \right ]_{\theta = 0}^{\theta = 2\pi} \\ \quad \iint_D x^2 \: dA = 6 \pi \end{align}