Question

Transform the integrals given, using the substitutions indicated: a) \(\int_{0}^{1} \int_{0}^{x} \log \left(1+x^{2}+y^{2}\right) d y d x, x=u+v, \quad y=u-v\) b) \(\int_{0}^{1} \int_{1-x}^{1+x} \sqrt{1+x^{2} y^{2}} d y d x, x=u, y=u+v\)

Short answer

Question: Using the given substitutions, transform the following integrals: a) $$\int_{0}^{1} \int_{0}^{x} \log \left(1+x^{2}+y^{2}\right) dydx$$ Substitutions: \(x = u+v\), \(y = u-v\) b) $$\int_{0}^{1} \int_{1-x}^{1+x} \sqrt{1+x^{2}y^{2}} dydx$$ Substitutions: \(x = u\), \(y = u+v\) Answer: a) $$2 \int_{0}^{1} \int_{-u}^{u} \log \left(1+(u+v)^{2}+(u-v)^{2}\right) dvdu$$ b) $$\int_{0}^{1} \int_{0}^{2u} \sqrt{1+u^{2}(u+v)^{2}} dvdu$$

Step-by-step solution

Identify the given substitutions and differential

For the first integral, we have the substitutions: $$x = u+v$$ $$y = u-v$$ We also need to find the differential for this transformation: $$dxdy = \left|\frac{\partial(x, y)}{\partial(u, v)}\right| dudv$$

Compute the Jacobian

To compute the Jacobian, find the partial derivatives of \(x\) and \(y\) with respect to \(u\) and \(v\): $$\frac{\partial x}{\partial u} = 1 \qquad \frac{\partial x}{\partial v} = 1$$ $$\frac{\partial y}{\partial u} = 1 \qquad \frac{\partial y}{\partial v} = -1$$ Now, find the determinant of this matrix: $$\left|\frac{\partial(x, y)}{\partial(u, v)}\right| = \left|\begin{matrix} 1 & 1 \\ 1 & -1 \end{matrix}\right| = (-1) - (1) = -2$$ We should take the absolute value of the Jacobian when transforming the differential, so \(dxdy = 2 dudv\).

Find new limits of integration

Now, we will find the new limits of integration for u and v using our substitutions. For \(0 \leq x \leq 1\): $$0 \leq u+v \leq 1$$ For \(0 \leq y \leq x\): $$0 \leq u-v \leq u+v$$ From these inequalities, we can get the limits for \(u\) and \(v\): $$0 \leq u \leq 1$$ $$-u \leq v \leq u$$

Transform the integral

Now, we can transform the original integral using our substitutions, new limits, and differential: $$\int_{0}^{1} \int_{0}^{x} \log \left(1+x^{2}+y^{2}\right) dydx = 2 \int_{0}^{1} \int_{-u}^{u} \log \left(1+(u+v)^{2}+(u-v)^{2}\right) dvdu$$ #b) Integral transformation#

Identify the given substitutions and differential

For the second integral, we have the substitutions: $$x = u$$ $$y = u+v$$ We also need to find the differential for this transformation: $$dxdy = \left|\frac{\partial(x, y)}{\partial(u, v)}\right| dudv$$

Compute the Jacobian

To compute the Jacobian, find the partial derivatives of \(x\) and $$ with respect to \(u\) and \(v\): $$\frac{\partial x}{\partial u} = 1 \qquad \frac{\partial x}{\partial v} = 0$$ $$\frac{\partial y}{\partial u} = 1 \qquad \frac{\partial y}{\partial v} = 1$$ Now, find the determinant of this matrix: $$\left|\frac{\partial(x, y)}{\partial(u, v)}\right| = \left|\begin{matrix} 1 & 0 \\ 1 & 1 \end{matrix}\right| = 1 - 0 = 1$$ So, \(dxdy = dudv\).

Find new limits of integration

Now, we will find the new limits of integration for u and v using our substitutions. For \(0 \leq x \leq 1\): $$0 \leq u \leq 1$$ For \(1-x \leq y \leq 1+x\): $$1-u \leq u+v \leq 1+u$$ From this inequality, the limits for v are: $$0 \leq v \leq 2u$$

Transform the integral

Now, we can transform the original integral using our substitutions, new limits, and differential: $$\int_{0}^{1} \int_{1-x}^{1+x} \sqrt{1+x^{2}y^{2}} dydx = \int_{0}^{1} \int_{0}^{2u} \sqrt{1+u^{2}(u+v)^{2}} dvdu$$

Key concepts

Jacobian Determinant

The Jacobian determinant is a crucial concept when changing variables during the process of integration by substitution. It helps in transforming the differential elements from the original variables to new ones. When you are given a transformation such as

• \( x = u + v \)
• \( y = u - v \)

, you will need to compute the Jacobian determinant to find how the differential areas change under this transformation.
To compute the Jacobian determinant, create a matrix of the partial derivatives for the transformations. For example:

• Find \( \frac{\partial x}{\partial u} \), \( \frac{\partial x}{\partial v} \), \( \frac{\partial y}{\partial u} \), and \( \frac{\partial y}{\partial v} \).
• Arrange these derivatives into a matrix.
• Calculate the determinant.

In our first example, the determinant is calculated as follows:\[\left| \begin{matrix} 1 & 1 \ 1 & -1 \end{matrix} \right| = (1 \cdot (-1)) - (1 \cdot 1) = -2\,\]which reflects the change in area. Remember, we often take the absolute value when considering integration: \( | -2 | = 2 \).
Jacobian determinant plays a key role in correctly computing the new differential \(dxdy\) as \(2 dudv\).

Transformation of Integrals

When integrating using substitution, transforming integrals is a systematic process that allows us to solve more complex integrals efficiently. The essential steps involve:

• Identifying the substitution provided or choosing one that simplifies the integral.
• Replacing variables in the integrand and limits of integration using the new substitution.

In the exercise provided, integrations were transformed based on substitutions like \( x = u + v \) and \( y = u - v \) for the first problem. This substitution simplifies the form of the expression within the integral, possibly into a form that is easier to integrate.
This process is pivotal in converting a difficult integral involving possibly complex limits and expressions into a simpler form that leverages symmetry or product-rule-friendly structures, thereby making integrations more straightforward. In practice, transforming integrals with correct replacements in the limits and differential terms is what allows us to evaluate them analytically, sometimes directly leading to elementary applications of calculus rules.

Limits of Integration

Determining the new limits of integration is crucial after transforming variables. When you substitute variables, the limits themselves change and need to be recalculated for the new variables.For the two variable problems:

• For example, when using substitutions like \( x = u+v \) and \( y = u-v \), originally \( 0 \leq x \leq 1 \) and \( 0 \leq y \leq x \) change to:
  • \( 0 \leq u+v \leq 1 \)
  • \( 0 \leq u-v \leq u+v \)

This requires solving inequalities to express the bounds for the new variables \( u \) and \( v \).
The new region of integration is determined by solving these inequalities carefully.In the exercise example, these constraints led us to limits:

• \( 0 \leq u \leq 1 \)
• \( -u \leq v \leq u \)

It's necessary to carefully evaluate these new limits, ensuring they fully encapsulate the original geometric region being integrated over. Thus, when transforming integral limits, always focus on the relationship dictated by your substitution to correctly represent the problem domain in new variables.