# Let R be the region in the first quadrant of the xy-plane bounded by the graphs of y = fraction...

## Question:

Let {eq}R {/eq} be the region in the first quadrant of the {eq}xy-plane {/eq} bounded by the graphs of {eq}y = \frac {1}{x}, \ \ y = \frac {2}{x}. \ \ y = 2x. \ \ and \ \ y = \frac {x}{2} {/eq}. Use the transformation {eq}u = xy {/eq} and {eq}v = \frac {y}{x} {/eq} to evaluate {eq}\displaystyle \int \int_R xy dA. {/eq}

## Applying the Jacobian:

The Jacobian is a useful tool when we wish to change variables in a double integral. To apply the Jacobian, follow these steps:

Step 1. Express the region of integration in terms of {eq}u {/eq} and {eq}v. {/eq}

Step 2. Solve the coordinate transformation for {eq}x {/eq} and {eq}y. {/eq}

Step 3. Calculate the Jacobian using the formula

{eq}\displaystyle\frac{\partial (x, y)}{\partial (u, v)} = \left| \begin{array}{cc} \displaystyle\frac{\partial x}{\partial u} & \displaystyle \frac{\partial x}{\partial v} \\ \displaystyle \frac{\partial y}{\partial u} & \displaystyle \frac{\partial y}{\partial v} \end{array} \right| {/eq}

Step 4. Use the formula

{eq}\displaystyle\iint_R f(x, y) \: dA= \displaystyle\int_a^b \int_c^d f(x(u, v), y(u, v)) \left| \displaystyle\frac{\partial(x, y)}{\partial(u, v)} \right| \: dv \: du {/eq}

to rewrite and evaluate the new integral.

## Answer and Explanation:

We are given the region in the first quadrant of the {eq}xy-plane {/eq} bounded by the graphs of {eq}y = \displaystyle\frac {1}{x}, \: y = \frac {2}{x}, \: y = 2x, {/eq} and {eq}y = \displaystyle\frac {x}{2} {/eq}. We can rewrite these equations as {eq}xy = 1, \: xy = 2, \: \displaystyle\frac{y}{x} = \frac12, \: \frac{y}{x} = 2, {/eq} so {eq}1 \leq u \leq 2, \: \displaystyle\frac12 \leq v \leq 2. {/eq}

In order to calculate the Jacobian, we need to solve these equations for {eq}x {/eq} and {eq}y. {/eq} Since {eq}uv = y^2, {/eq} we have {eq}y = u^{1/2} v^{1/2}. {/eq} Since {eq}\displaystyle\frac{u}{v} = x^2, {/eq} we have {eq}x = u^{1/2} v^{-1/2}. {/eq} Therefore the Jacobian of the coordinate transformation is

{eq}\displaystyle\frac{\partial (x, y)}{\partial (u, v)} = \left| \begin{array}{cc} \displaystyle\frac12 u^{-1/2} v^{1/2} & \displaystyle \frac12 u^{1/2} v^{-1/2} \\ \displaystyle \frac12 u^{-1/2} v^{-1/2} & -\displaystyle \frac12 u^{1/2} v^{-3/2} \end{array} \right| = -\frac12 v^{-1}. {/eq}

The double integral becomes

{eq}\begin{eqnarray*}\iint_R xy dA & = & \int_1^2 \int_{1/2}^2 u \cdot \displaystyle\frac12 v^{-1} \: dv \: du \\ \\ & = & \displaystyle\frac12 \int_1^2 u \ln v \biggr|_{1/2}^2 \: du \\ \\ & = & \displaystyle\frac12 \int_1^2 u \left( \ln 2 - \ln \displaystyle\frac12 \right) \: du \\ \\ & = & \displaystyle\frac12 \int_1^2 2u \ln 2 \: du \\ \\ & = & \ln 2 \int_1^2 u \: du \\ \\ & = & \displaystyle\frac{u^2 \ln 2}2 \biggr|_1^2 \\ \\ & = & 2 \ln 2 \end{eqnarray*} {/eq}

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from AP Calculus AB & BC: Help and Review

Chapter 12 / Lesson 14