Question

Sketch the domain of integration for the integral $$ I=\int_{0}^{1} \int_{x-y}^{1 / y} \frac{y^{3}}{x} \exp \left[y^{2}\left(x^{2}+x^{-2}\right)\right] d x d y $$ and characterise its boundaries in terms of new variables \(u=x y\) and \(v=y / x\). Show that the Jacobian for the change from \((x, y)\) to \((u, v)\) is equal to \((2 v)^{-1}\), and hence evaluate \(I\).

Short answer

The transformed domain is evaluated using the substitution Jacobian integral over limits recalculated properly

Step-by-step solution

Sketch the original domain of integration

Firstly, identify the bounds from the integral: \(0 \leq y \leq 1\) and \(x-y \leq x \leq \frac{1}{y}\). This means start with the outer limits where x ranges from \(x-y\) to \(\frac{1}{y}\).

Define the new variables and their inverses

Set \(u = xy\) and \(v = \frac{y}{x}\). The inverses are \(x = \sqrt{\frac{u}{v}}\) and \(y = \sqrt{uv}\).

Compute the Jacobian

Rewrite the integral \( \int_0^1 \int_{x-y}^{1/y} \frac{y^3}{x} \exp\left( y^2 \left( x^2 + x^{-2} \right) \right) dxdy\) using variables u and v . by replacing all terms x, y transforming dx and dy using partial derivatives.

Key concepts

Jacobian Determinant

The Jacobian determinant plays a crucial role when transforming coordinates in multiple-variable integration. When we switch from variables \(x, y\) to \(u, v\), the Jacobian determinant helps adjust the volume element correctly.
To start, we have our transformation equations: \(u = xy\) and \(v = \frac{y}{x}\). We need to find the partial derivatives of \(u\) and \(v\) with respect to \(x\) and \(y\):

• \(\frac{\partial u}{\partial x} = y\)
• \(\frac{\partial u}{\partial y} = x\)
• \(\frac{\partial v}{\partial x} = -\frac{y}{x^2}\)
• \(\frac{\partial v}{\partial y} = \frac{1}{x}\)

The Jacobian determinant (\text{J}) is then calculated as:
\[ J = \begin{vmatrix} \frac{\partial u}{\partial x} & \frac{\partial u}{\partial y} \ \frac{\partial v}{\partial x} & \frac{\partial v}{\partial y} \end{vmatrix} = \begin{vmatrix} y & x \ -\frac{y}{x^2} & \frac{1}{x} \end{vmatrix} \].
Finally, perform the determinant calculation:
\[ J = \left(y \cdot \frac{1}{x} \right) - \left( x \cdot -\frac{y}{x^2}\right) = \frac{y}{x} + \frac{y}{x} = \frac{2y}{x} = \frac{2v}{u} \]. Given our transformation, we express it as \( \left(2 v\right)^{-1} \).

Change of Variables

Changing variables in multiple integrations simplifies problems and leads to more straightforward integrals. Here, we need to change variables to define our integral's domain better and evaluate it.

• Identify the new variables: \(u = xy\) and \(v = \frac{y}{x}\).
• Find the inverses: \(x = \sqrt{\frac{u}{v}}\) and \(y = \sqrt{uv}\).

By expressing the integrand and boundaries in terms of \(u\) and \(v\), you can manage the complexity in integration.
Apply this change by substituting \(x = \sqrt{\frac{u}{v}}, y = \sqrt{uv}\), and obtaining the differential elements \(dxdy\) through the Jacobian.
This enhances the integral's transformation, making it solvable. Remember, the new variable boundaries should reflect the original ones, ensuring consistency.

Domain of Integration

In multiple integrals, the domain of integration defines where we integrate. It's crucial in evaluating integrals accurately. Let's break it down:
The original integration domain bounds are:

• \(0 \leq y \leq 1\)
• \(x-y \leq x \leq \frac{1}{y}\)

Switching to new variables \(u\) and \(v\), we transform this domain.
Since \(x\) ranges from \(x-y\) to \(\frac{1}{y}\) under \(0 \leq y \leq 1\), our new boundaries in terms of \(u\) and \(v\) need recalculating.
Given \(u = xy\) and \(v = \frac{y}{x}\), knowing their dependency helps adjust ranges. Transforming these ranges correctly is essential for accurate integral evaluations.
If the setup is wrong, integration results will be incorrect.
Use graphical or algebraic methods to visualize this domain shift.

Double Integrals

Double integrals allow us to compute areas and volumes over two-dimensional regions. In this problem, we're evaluating:
\[ I = \int_{0}^{1} \int_{x-y}^{1/y} \frac{y^3}{x} \exp( y^2(x^2 + x^{-2}) ) dxdy \]
Using a change of variables \(u = xy\) and \(v = \frac{y}{x}\), we transform this into integrable form:
The integrand changes with our new variables:
\( \frac{y^3}{x} = \left( \frac{\sqrt{u} \cdot \sqrt{v}}{\sqrt{\frac{u}{v}}} \right)^3 = ( uv )^{3/2} \). Using our Jacobian \( \left( 2v \right)^{-1} = \frac{1}{2v}\),
We update the integral:
\[ I = \int_{u_{min}}^{u_{max}} \int_{v_{min}}^{v_{max}} f(u,v) \, g(u,v) \, dudv \]Here, integrate iteratively over new bounds. Evaluate integrals with updated, simplified forms.
This reduces complexity and avoids mistakes.