Problem 9 - University Calculus: Early Transcendentals

Question

Let \(R\) be the region in the first quadrant of the \(x y\) -plane bounded by the hyperbolas \(x y=1, x y=9\) and the lines \(y=x, y=4 x\) Use the transformation \(x=u / v, y=u v\) with \(u>0\) and \(v>0\) to rewrite $$ \iint_{R}(\sqrt{\frac{y}{x}}+\sqrt{x y}) d x d y $$ as an integral over an appropriate region \(G\) in the \(u v\) -plane. Then evaluate the \(u \boldsymbol{v}\) -integral over \(\boldsymbol{G}\)

Step-by-Step Solution

Answer

The integral evaluates to \(8 + \frac{52}{3} \ln(2)\).

Step 1: Identify the transformation

We are given the transformation \(x = \frac{u}{v}\) and \(y = uv\). This transformation will help us to re-express the region \(R\) in terms of \(u\) and \(v\).

Step 2: Determine the Jacobian

To transform the double integral, we need to find the Jacobian of the transformation. Calculate \(\frac{\partial(x,y)}{\partial(u,v)}\). Solve as follows:\[ x = \frac{u}{v}, \quad y = uv \]\[\frac{\partial x}{\partial u} = \frac{1}{v}, \quad \frac{\partial x}{\partial v} = -\frac{u}{v^2} \\frac{\partial y}{\partial u} = v, \quad \frac{\partial y}{\partial v} = u\]The Jacobian is:\[J = \begin{vmatrix} \frac{1}{v} & -\frac{u}{v^2} \ v & u \end{vmatrix} = \left(\frac{1}{v}\right)(u) - \left(-\frac{u}{v^2}\right)(v) = \frac{u}{v} + \frac{u}{v} = \frac{2u}{v}\]

Step 3: Transform the Region R to Region G

Using the transformation equations \(x = \frac{u}{v}\) and \(y = uv\), we translate the boundaries of region \(R\):1. \(xy = 1\) becomes \(u^2 = 1\) (so \(u = 1\)).2. \(xy = 9\) becomes \(u^2 = 9\) (so \(u = 3\)).3. \(y = x\) becomes \(v^2 = 1\) (so \(v = 1\)).4. \(y = 4x\) becomes \(v^2 = 4\) (so \(v = 2\)).Thus, region \(G\) in the \(uv\)-plane is bounded by \(1 \le u \le 3\) and \(1 \le v \le 2\).

Step 4: Rewrite the Integral

Substitute \(x = \frac{u}{v}\) and \(y = uv\) into the integral:\[\iint_{R}(\sqrt{\frac{y}{x}}+\sqrt{xy})\, dx \, dy = \iint_{G}\left(\sqrt{v^2} + \sqrt{u^2}\right) \left|\frac{2u}{v}\right| \, du \, dv\]Since \(u > 0\) and \(v > 0\), the absolute value is not needed:\[\int_{1}^{3}\int_{1}^{2} (v + u) \frac{2u}{v} \, dv \, du\]

Step 5: Separate and Simplify the Integral

Split the integral into two parts and simplify:\[\int_{1}^{3}\int_{1}^{2} 2u \, du \, dv + \int_{1}^{3}\int_{1}^{2} \frac{2u^2}{v} \, dv \, du\]Calculate each integral separately.

Step 6: Evaluate the Integral over v

Evaluate the inner integrals:1. \(\int_{1}^{2} 2u \, dv = 2u[v]_{1}^{2} = 2u(2 - 1) = 2u\)2. \(\int_{1}^{2} \frac{2u^2}{v} \, dv = 2u^2 \ln(v)\bigg|_{1}^{2} = 2u^2 \ln(2)\)

Step 7: Evaluate the Integral over u

Sum the evaluated integrals and compute the total:\[\int_{1}^{3} 2u \, du + \int_{1}^{3} 2u^2 \ln(2) \, du\]1. \(\int_{1}^{3} 2u \, du = [u^2]_{1}^{3} = 9 - 1 = 8\)2. \(2 \ln(2) \int_{1}^{3} u^2 \, du = 2 \ln(2)\left[\frac{u^3}{3}\right]_{1}^{3} = 2 \ln(2) \left(\frac{27}{3} - \frac{1}{3}\right) = 2 \ln(2) \times \frac{26}{3}\)Add the results to find:8 + \(\frac{52}{3} \ln(2)\).

Key Concepts

Change of Variables

In calculus, one powerful method to simplify the evaluation of a double integral is the 'change of variables' technique. This method involves transforming the variables of integration to make the region and the integrand more manageable. For instance, if an integral over variables \(x\) and \(y\) proves complex, changing these variables into new ones, such as \(u\) and \(v\), can simplify the problem. By doing so, we change the space in which we integrate.

• A common motivation for this method is to convert complex boundaries into more straightforward ones.
• Through substitutions, we often achieve integration over rectangular regions, which are easier to handle than curved regions.

In our exercise, we replaced variables \(x\) and \(y\) with \(u\) and \(v\), where \(x= \frac{u}{v}\) and \(y = uv\). This transformation shifted the problem from the \(xy\)-plane to the more tractable \(uv\)-plane.

Jacobian Determinant

The Jacobian determinant is a crucial component in the change of variables process. It serves as a scaling factor that adjusts the area element from the old coordinate system to the new one. Specifically, when moving from \((x, y)\) coordinates to \((u, v)\) coordinates, we need to determine how differentials such as \(dx\,dy\) change to \(du\,dv\).To compute the Jacobian of a transformation, first calculate the partial derivatives of the new variables with respect to the old ones. This involves the following steps:

• Organize these partial derivatives into a matrix.
• Take the determinant of this matrix.

In our example:\[J = \begin{vmatrix} \frac{\partial x}{\partial u} & \frac{\partial x}{\partial v} \ \frac{\partial y}{\partial u} & \frac{\partial y}{\partial v} \end{vmatrix} = \begin{vmatrix} \frac{1}{v} & -\frac{u}{v^2} \ v & u \end{vmatrix} = \frac{2u}{v}\]This determinant accounts for how areas transform between the regions \(R\) in the \(xy\)-plane and \(G\) in the \(uv\)-plane.

Region Transformation

Region transformation involves expressing the bounds of integration in the new coordinate system. This requires rewriting the constraints of the integral after transitioning to different variables.For the given problem, the integral was initially set over region \(R\) formed by hyperbolas and lines in the \(xy\)-plane. Once we made the change of variables, our task was to translate these bounds into terms of \(u\) and \(v\), thereby defining a new region \(G\).Key steps:

• Identify equations that define the region boundaries in old coordinates.
• Use the transformations relations (e.g., \(x=\frac{u}{v}\) and \(y=uv\)) to convert them into the new coordinates.
• Establish new limits of integration for \(u\) and \(v\).

In our transformation:

• The hyperbolas \(xy=1\) and \(xy=9\) translated into \(u^2=1\) and \(u^2=9\) respectively.
• The lines \(y=x\) and \(y=4x\) became \(v^2=1\) and \(v^2=4\) in the new coordinate system.

Thus, the region \(G\) is characterized by \(1 \le u \le 3\) and \(1 \le v \le 2\).

Hyperbolas

A hyperbola is a type of smooth curve lying in a plane, defined by its geometric and algebraic properties. In the problem, the hyperbolas \(xy=1\) and \(xy=9\) served as boundaries of the region \(R\) in the \(xy\)-plane. Upon transforming the problem, these hyperbolas translated into constant values for \(u\):

• \(xy = 1 \Rightarrow u^2 = 1\), meaning \(u = 1\).
• \(xy = 9 \Rightarrow u^2 = 9\), meaning \(u = 3\).

Understanding that hyperbolas are curves where the product of the coordinates is constant, helps to comprehend how they behave under transformation. Once these hyperbolas were expressed as lines in the \(uv\)-plane, it became simpler to integrate over the defined region \(G\). The transformation made it easier to work with these previously complex curves as they now became straightforward to integrate over.

Integral Evaluation

Evaluating the integral is the final step. After carefully defining the region and adjusting under the new variables, we rewrite the integrand using \(u\) and \(v\).In our transformed system, the original function within the integral \(\left(\sqrt{\frac{y}{x}}+\sqrt{xy}\right)\) simplifies and separates nicely after the change of variables:

• \(\sqrt{\frac{y}{x}} = \sqrt{v^2} = v\)
• \(\sqrt{xy} = \sqrt{u^2} = u\)

Thus, we arrive at: \[ \iint_{G}(v + u) \frac{2u}{v} \, du \, dv\]Breaking the integral into two separate, easier tasks and integrating over \(v\) and then \(u\) respectively, the result is found by summing up the evaluated individual parts. This allows us to determine the contribution of each section of region \(G\) to the total integral, culminating in the final solution of:8 + \(\frac{52}{3} \ln(2)\). This demonstrates the power of variable transformation in performing double integrations efficiently.