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Chapter 14: Problem 25

If \(R\) lies between \(x=0\) and \(x=1\) under the graph of \(y=\) \(f(x)>0,\) then \(x=u, y=v f(u)\) takes \(R\) to the unit square \(S .\) Locate the corners of \(R\) and the point corresponding to \(u=\frac{1}{2}, v=1\). Compute \(J\) to prove what we know: $$ \text { area of } R=\int_{0}^{1} f(x) d x=\int_{0}^{1} \int_{0}^{1} J d u d v $$.

Short Answer

Expert verified
The areas are equal by showing \( J = f(u) \). Transforming confirms the integrals' equivalence.

Step by step solution

01

Understand the Problem

We have a region \( R \) under the curve \( y = f(x) > 0 \) between \( x = 0 \) and \( x = 1 \). The transformation \( x = u, y = v f(u) \) maps \( R \) to a unit square \( S \). We need to calculate the area of \( R \) and show it equals the double integral \( \int_{0}^{1} \int_{0}^{1} J \, du \, dv \). First, identify the corners of \( R \) and transform them.
02

Locate the Corners of R

Identify the corners of region \( R \) in the \((x, y)\) plane: - Bottom-left: \((0,0)\)- Bottom-right: \((1,0)\) - Top-left: \((0, f(0))\) - Top-right: \((1, f(1))\)
03

Apply the Transformation

Transform these corners using \( x = u \) and \( y = v f(u) \): - Bottom-left: \((0,0)\) transforms to \((0,0)\) - Bottom-right: \((1,0)\) transforms to \((1,0)\) - Top-left: \((0, f(0))\) transforms to \((0,1)\) - Top-right: \((1, f(1))\) transforms to \((1,1)\) Hence, \( R \) transforms to the unit square \( S \) from \((0,0)\) to \((1,1)\) in the \((u,v)\) plane.
04

Find Corresponding Point for u=0.5, v=1

For \( u = \frac{1}{2} \) and \( v = 1 \), under the transformation, \( x = \frac{1}{2} \) and \( y = 1 \cdot f(\frac{1}{2}) = f(\frac{1}{2}) \). This point in \( R \) corresponds to top point on vertical line at \( x = \frac{1}{2} \).
05

Compute the Jacobian J

The transformation given was \( x = u \), \( y = v f(u) \). The Jacobian \( J \) of the transformation is calculated as:\[J = \frac{\partial(x,y)}{\partial(u,v)} = \left| \begin{matrix} \frac{\partial x}{\partial u} & \frac{\partial x}{\partial v} \ \frac{\partial y}{\partial u} & \frac{\partial y}{\partial v} \end{matrix} \right|\]Calculating gives:\[J = \left| \begin{matrix} 1 & 0 \ v f'(u) & f(u) \end{matrix} \right| = 1 \cdot f(u) - 0 \cdot v f'(u) = f(u).\]
06

Evaluate the Integrals

The area of \( R \) is given by \( \int_{0}^{1} f(x) \, dx \). Under the coordinate transformation, this becomes:\[\int_{0}^{1} \int_{0}^{1} J \, du \, dv = \int_{0}^{1} \int_{0}^{1} f(u) \, du \, dv\]The inner integral simplifies to:\[\int_{0}^{1} f(u) \, dv = f(u) \]Thus, after evaluating the outer integral:\[\int_{0}^{1} f(u) \, du = \int_{0}^{1} f(x) \, dx\]which confirms the area of \( R \) is equivalent to the double integral.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Jacobian
The Jacobian is a crucial concept when dealing with transformations in calculus. It helps us understand how different variables and regions change size and shape under transformation. Consider a transformation from coordinates
  • \((x, y)\) to \((u, v)\).
  • The Jacobian \(J\) for a transformation describes how areas scale and deform when moving from one coordinate system to another.
For the transformation\(x = u\) and \(y = v f(u)\), the Jacobian determinant is calculated as:\[J = \left| \begin{matrix} \frac{\partial x}{\partial u} & \frac{\partial x}{\partial v} \frac{\partial y}{\partial u} & \frac{\partial y}{\partial v} \end{matrix} \right| \].This results in \(J = f(u)\), indicating how the area scales for a simple transformation.
Double Integral
In calculus, double integrals allow us to compute volumes and areas under surfaces over a given domain. Here, it's applied over a transformed region. It operates over two variables, usually indicated by
  • \( \, du \, dv \)
  • to represent all points tapped by the transformed region \((u,v)\).
The double integral in this problem can be expressed as:\[\int_{0}^{1} \int_{0}^{1} J \, du \, dv\].This calculates the total area over the transformed space. The double integral captures all effects of the transformation, factoring in how the Jacobian \(J = f(u)\) scales the area. The result reaffirms the area calculation from \(0\) to \(1\) along the entire curve \(f(x)\).
Coordinate Transformation
Coordinate transformations shift regions from one coordinate system to another. For this exercise, the transformation is from a region \(R\) under the curve \(y = f(x)\) to a unit square \(S\). The transformation is captured by
  • \(x = u \)
  • \(y = v f(u) \).
This shifts the region \(R\) through a scaling and re-positioning. The transformation maps easily visualizable corners of \(R\) to simple corners in the unit square \( (0,0)\) to \( (1,1) \) in the \((u, v)\) plane, creating a reliable framework for calculations.
Function Mapping
Function mapping takes input from one space to another through a mathematical function. This is seen where the transformation\(x = u\) and \(y = v f(u)\)maps region \(R\) to \(S\). The points transition from one region to another while preserving relationships and distances according to the function's rules.
  • Effectively, it "maps" or "draws" the original region onto a new space.
  • This allows for simpler integration by transforming the complexity into algebraic expressions.
The function essentially dictates how areas between different coordinate planes relate and adjust, essential for simplifying integrations and other higher-level calculations.

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Most popular questions from this chapter

Using polar coordinates, find the volume under \(z=\) \(x^{2}+y^{2}\) above the unit disk \(x^{2}+y^{2} \leqslant 1\).

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