Problem 21 - Advanced Engineering Mathematics

Question

Evaluate the given integral by means of the indicated change of variables. \(\iint_{R} y^{4} d A\), where \(R\) is the region in the first quadrant bounded by the graphs of \(x y=1, x y=4, y=x, y=4 x ; u=x y, v=y / x\)

Step-by-Step Solution

Answer

The integral evaluates to \( \frac{225}{4} \).

Step 1: Understand the Change of Variables

We are given the transformations \( u = xy \) and \( v = \frac{y}{x} \). This means we will convert the region and integrate in terms of \( u \) and \( v \). The region \( R \) is described by the boundaries \( xy = 1 \), \( xy = 4 \), \( y = x \), and \( y = 4x \) in terms of \( (x,y) \).

Step 2: Define the New Bounds

Using the transformations, the boundaries can be expressed as follows:- \( xy = 1 \) becomes \( u = 1 \)- \( xy = 4 \) becomes \( u = 4 \)- \( y = x \) becomes \( v = 1 \)- \( y = 4x \) becomes \( v = 4 \).Hence, the transformed region in \( uv \)-plane is defined by \( 1 \leq u \leq 4 \) and \( 1 \leq v \leq 4 \).

Step 3: Find the Jacobian of the Transformation

The Jacobian, \( J \), of the transformation is determined by:\[J = \begin{vmatrix} \frac{\partial x}{\partial u} & \frac{\partial x}{\partial v} \ \frac{\partial y}{\partial u} & \frac{\partial y}{\partial v} \end{vmatrix}.\]We solve for \( x \) and \( y \) in terms of \( u \) and \( v \):\[ x = \sqrt{\frac{u}{v}}, \quad y = \sqrt{uv}. \]

Step 4: Compute the Jacobian Determinant

Calculate the partial derivatives and form the Jacobian determinant:- \( \frac{\partial x}{\partial u} = \frac{1}{2} v^{-1/2} u^{-1/2} \)- \( \frac{\partial x}{\partial v} = -\frac{1}{2} u^{1/2} v^{-3/2} \)- \( \frac{\partial y}{\partial u} = \frac{1}{2} v^{1/2} u^{-1/2} \)- \( \frac{\partial y}{\partial v} = \frac{1}{2} u^{1/2} v^{-1/2} \)Hence,\[ J = \begin{vmatrix} \frac{1}{2} v^{-1/2} u^{-1/2} & -\frac{1}{2} u^{1/2} v^{-3/2} \ \frac{1}{2} v^{1/2} u^{-1/2} & \frac{1}{2} u^{1/2} v^{-1/2} \end{vmatrix}, \]\[ J = \frac{1}{2} \cdot \frac{1}{2} \cdot u^{-1} v^{-1} + \frac{1}{2} \cdot \frac{1}{2} \cdot u^{-1} v^{-1} = u^{-1} v^{-1}. \]

Step 5: Setup the Integral with the Jacobian

Substitute and incorporate the Jacobian in the integral:\[\iint_R y^4 \, dA = \int_{1}^{4} \int_{1}^{4} (uv)^2 \cdot u^{-1} v^{-1} \, dv \, du.\]

Step 6: Solve the Integral

After simplifying, the integral becomes:\[\int_{1}^{4} \int_{1}^{4} u v \, dv \, du = \int_{1}^{4} \left[ \frac{1}{2} u v^2 \right]_{1}^{4} \, du = \int_{1}^{4} \frac{15}{2} u \, du.\]Solving this gives:\[\left[ \frac{15}{4} u^2 \right]_{1}^{4} = 60 - \frac{15}{4} = \frac{225}{4}.\]

Step 7: Conclusion

The value of the integral \( \iint_R y^{4} \, dA \) over the specified region is \( \frac{225}{4} \).

Key Concepts

Jacobian Determinant

When we perform a change of variables in multiple integrals, it's crucial to account for how the transformation affects the area or volume elements. One essential tool for this is the Jacobian determinant. The Jacobian determinant helps us adjust the scale of integration from the original coordinate system to the new one.

For a transformation given by functions, like the ones in our problem, where \( (x, y) \) is transformed into \( (u, v) \), the Jacobian defines how these coordinates are interrelated. It's essentially determining the rate at which area (or possibly volume) changes under the transformation. Mathematically, the Jacobian is formed using the partial derivatives of the new coordinates with respect to the old ones.

The formula is given as:
\[J = \begin{vmatrix} \frac{\partial x}{\partial u} & \frac{\partial x}{\partial v} \ \frac{\partial y}{\partial u} & \frac{\partial y}{\partial v} \end{vmatrix}.\]

In our specific exercise, the calculations yield a Jacobian of \( u^{-1} v^{-1} \), which adjusts the integral to account for the distortion caused by the change of variables.

Integral Evaluation

Evaluating integrals often involves simplifying the domain and the function. By changing variables, things can become easier, but verifying the entire process involves recomputing the integral from the new perspective. For instance, in this exercise, once we apply the transformation, we integrate \( y^4 \) as \( (uv)^2 \) with respect to \( u \) and \( v \) over our specified new region.

This step involves substituting back into the integral formula, replacing \( y^4 \, dA \) with \( (uv)^2 \, u^{-1}v^{-1} \, dv \, du \). Simplifying this product and limits of integration is a key sub-task.

With the bounds of the transformation already defined, the next step in our integration process becomes straightforward, as we just follow the limits to calculate it step-by-step as:

• First integrate with respect to \( v \), holding \( u \) constant.
• Then, integrate the resulting expression with respect to \( u \).

The final result here confirms the value computed was correctly \( \frac{225}{4} \).

Transformations in Integral Calculus

Transformations are used in integral calculus to simplify complex regions in which we aim to evaluate an integral. This concept hinges on changing variables to new ones that make the bounds straightforward and easy to handle. The trick is to choose transformations that convert our original regions into standard shapes or simpler calculations.

The transformations in this exercise were carefully chosen:

• \( u = xy \) helped convert the hyperbolic boundaries \( xy = 1 \) and \( xy = 4 \) into simple vertical lines in the \( u \)-plane.
• \( v = \frac{y}{x} \) changed the bounding lines \( y = x \) and \( y = 4x \) into horizontal lines.

These changes make the region of integration a simple rectangular area in the new \( (u, v) \) plane, bounded by constant inequalities. This kind of simplification is why transformations are a powerful tool in integral calculus.

Boundaries in Coordinate Transformations

Setting boundaries in coordinate transformations can initially seem intricate but mastering it can dramatically simplify your integration tasks. The key lies in converting each of the initial region's descriptors into boundaries within the new system.

In the problem given, we shifted from \( (x, y) \) to \( (u, v) \):

• The line \( xy = 1 \) turns into \( u = 1 \).
• The line \( xy = 4 \) becomes \( u = 4 \).
• For \( y = x \), this simplifies to \( v = 1 \).
• Finally, \( y = 4x \) is transformed into \( v = 4 \).

The new bounds, beautifully, all align with a rectangle \( 1 \leq u \leq 4 \) and \( 1 \leq v \leq 4 \).

Understanding these transformations and the new boundaries helps streamline evaluations, as the region now lies within a simple shape that is much easier to work with than the original compound region.