Problem 16

Question

\begin{equation} \begin{array}{c}{\text { Use the transformation } x=u^{2}-v^{2}, y=2 u v \text { to evaluate the }} \\ {\int_{0}^{1} \int_{0}^{2 \sqrt{1-x}} \sqrt{x^{2}+y^{2}} d y d x} \\ {\text { (Hint: Show that the image of the triangular region } G \text { with vertices }} \\ {(0,0),(1,0),(1,1) \text { in the } u v-\text { plane is the region of integration } R \text { in }} \\\ {\text { the } x y \text { -plane defined by the limits of integration.) }}\end{array} \end{equation}

Step-by-Step Solution

Verified

Answer

The integral evaluates to $ \frac{8}{5} $.

1Step 1: Understand the Problem

We need to evaluate the integral $ \int_{0}^{1} \int_{0}^{2 \sqrt{1-x}} \sqrt{x^{2}+y^{2}} \ dy \ dx $ by transforming coordinates using the given transformations: - $ x = u^2 - v^2 $,- $ y = 2uv $.

2Step 2: Describe the Region in the UV-plane

Given points in the $uv$-plane are triangle vertices $(0,0),(1,0),(1,1)$. Sketch this triangle and understand it represents the integration area in the $uv$-plane. The region is bounded by curves $v = u$, $u = 0$ and $v = 0$.

3Step 3: Transformation to XY-plane

Apply transformations to convert this triangular region to the $xy$-plane. 1. The point $(0,0)$ in the $uv$-plane maps to $(0,0)$ in the $xy$-plane.2. The point $(1,0)$ in the $uv$-plane maps to $(1,0)$ in the $xy$-plane.3. The point $(1,1)$ in the $uv$-plane maps to $(0,2)$ in the $xy$-plane.

4Step 4: Show Image is R from Limits of Integration

Verify by image points that the transformations map the triangular region in the $uv$-plane to the region in the $xy$-plane defined by: - $x $ from 0 to 1 - $y $ from 0 to $2\sqrt{1-x}$. This confirms the limits describe the mapped region.

5Step 5: Change of Variables Formula

Use the change of variables formula:\[ dx \ dy = |J| du \ dv \].Find the Jacobian determinant $J$ of the transformation:\[ J = \begin{vmatrix} 2u & -2v \ 2v & 2u \end{vmatrix} = 4u^2 + 4v^2 = 4(u^2 + v^2) \].Thus, $dx \ dy = 4(u^2 + v^2) du \ dv $.

6Step 6: Express $\sqrt{x^2 + y^2}$ in Terms of $u$ and $v$

Using $ x = u^2 - v^2 $ and $ y = 2uv $:\[ x^2 + y^2 = (u^2 - v^2)^2 + (2uv)^2 = u^4 - 2u^2v^2 + v^4 + 4u^2v^2 = (u^2 + v^2)^2 \].Therefore, $ \sqrt{x^2 + y^2} = u^2 + v^2 $.

7Step 7: Evaluate the Integral in UV-plane

Substitute back into the integral:\[ \int_{0}^{1} \int_{0}^{u} (u^2 + v^2)^2 \times 4(u^2 + v^2) dv \ du \].Simplify and compute:\[ 4 \int_{0}^{1} \int_{0}^{u} (u^2 + v^2)^3 dv \ du \].Evaluate the inner integral, then the outer integral to find the final result.

Key Concepts

Jacobian DeterminantCoordinate TransformationDouble IntegralIntegration Techniques

Jacobian Determinant

The Jacobian determinant is a crucial concept when performing a change of variables in multiple integrals. It allows us to transform an integral from one coordinate system to another. In this problem, we are given the transformations $ x = u^2 - v^2 $ and $ y = 2uv $. These transformations change our coordinate system from $xy$ to $uv$. To properly convert the area of integration, we need to use the Jacobian determinant.

The Jacobian matrix is constructed using partial derivatives of the transformations with respect to each variable.
For this transformation, the matrix is \[ J = \begin{pmatrix} \frac{\partial x}{\partial u} & \frac{\partial x}{\partial v} \\frac{\partial y}{\partial u} & \frac{\partial y}{\partial v} \end{pmatrix} = \begin{pmatrix} 2u & -2v \2v & 2u \end{pmatrix} \].
The Jacobian determinant is then calculated as: \[ |J| = \begin{vmatrix} 2u & -2v \2v & 2u \end{vmatrix} = 4(u^2 + v^2). \]

This determinant not only scales the function being integrated but also represents the factor needed to adjust the scale of integration from one coordinate system to another, ensuring consistent area or volume measurements.

Coordinate Transformation

Coordinate transformation is the method used to convert a given problem from one set of coordinates to another which can simplify the problem or fit it to the region of interest. In this exercise, the transformation $ x = u^2 - v^2 $ and $ y = 2uv $ re-describes the area from the $xy$-plane into the $uv$-plane.By choosing this transformation, specific points or regions in the $uv$-space are mapped onto points or similar regions in the $xy$-space. Knowing the points in the transformation:

Point $(0,0)$ in $uv$-plane maps to $(0,0)$ in $xy$-plane.
Point $(1,0)$ in $uv$-plane maps to $(1,0)$ in $xy$-plane.
Point $(1,1)$ in $uv$-plane maps to $(0,2)$ in $xy$-plane.

This process helps to illustrate how integration limits in one system describe specific boundaries in the other system. Confirming this mapping is crucial to ensuring that transformations align correctly, allowing integration to be performed accurately.

Double Integral

Double integrals extend the concept of a single integral into two dimensions, allowing you to compute the volume under a surface z = f(x, y) over a given region in the $xy$-plane. In this exercise, the integral $ \int_{0}^{1} \int_{0}^{2 \sqrt{1-x}} \sqrt{x^2 + y^2} \, dy \, dx $ needs evaluating.

The region of integration is typically described by the limits provided for $x$ and $y$. Here, $x$ ranges from 0 to 1 and, for each $x$, $y$ ranges from 0 to $2\sqrt{1-x}$.
These limits describe a particular region in the domain that determines where the function will be summed to find the total integral value.

By transforming the coordinates, the double integral's complexity reduces, often resulting in easier calculations. A correct application of the transformation allows the new region to be more aligned geometrically and simplifies both the region and function defined over it, easing the computation process.

Integration Techniques

Integration techniques are varied approaches used to evaluate integrals. They can vary from substitution to more advanced methods like integration by parts. In this problem, using coordinate transformation, specifically with the substitution method, is essential.Through the substitution via $ x = u^2 - v^2 $ and $ y = 2uv $,

We express the magnitude function $ \sqrt{x^2 + y^2} $ in terms of $u$ and $v$: $(u^2 + v^2)^2$.
The double integral is then rephrased with the Jacobian factor into the $uv$-plane: \[ 4 \int_{0}^{1} \int_{0}^{u} (u^2 + v^2)^3 \, dv \, du. \]

These substitutions simplify the original integral into a form more straightforward to compute. This transformation ensures the range aligns with the simpler integration limits of the new coordinate system, and computations can then proceed smoothly, evaluating one integral at a time to reach the final solution.

Problem 15

Problem 16

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