Question

Evaluate the following integrals. $$\iint_{R} \frac{x y}{1+x^{2}+y^{2}} d A ; R=\\{(x, y): 0 \leq y \leq x, 0 \leq x \leq 2\\}$$

Short answer

Question: Find the double integral of the function \(f(x, y) = \frac{xy}{1+x^2+y^2}\) over the region R, where \(0 \leq y \leq x\) and \(0 \leq x \leq 2\). Answer: The double integral of the function \(f(x, y)\) over the specified region R is given by: $$\frac{1}{2} \left[ 2^2\left(\ln(9) - \ln(5)\right) - \int_{0}^{2}\left(\frac{2x^3 dx}{1 + 2x^2} - \frac{2x^3 dx}{1 + x^2}\right)\right]$$

Step-by-step solution

Determine the limits of integration

Based on the given region R, we have the following inequalities for x and y: \(0 \leq y \leq x\) and \(0 \leq x \leq 2\). We will integrate first with respect to y and then with respect to x. So, the limits for y are from 0 to x, and the limits for x are from 0 to 2: $$\int_{0}^{2} \int_{0}^{x} \frac{x y}{1+x^{2}+y^{2}} dy dx$$

Compute the inner integral

Integrate the given function with respect to y: $$\int_{0}^{x} \frac{x y}{1+x^{2}+y^{2}} dy$$ We can use the substitution method: let \(u = 1 + x^2 + y^2\), then \(du = 2y dy\). So \(dy = \frac{du}{2y}\) Now, the integral becomes: $$\frac{x}{2} \int_{1+x^2}^{1+2x^2} \frac{1}{u} du$$ Solving this integral, we get: $$\frac{x}{2} \left[\ln{u}\right]_{1+x^2}^{1+2x^2}$$ $$= \frac{x}{2} (\ln{(1+2x^2)} - \ln{(1+x^2)})$$

Compute the outer integral

Next, we integrate the result with respect to x: $$\int_{0}^{2} \frac{x}{2} (\ln{(1+2x^2)} - \ln{(1+x^2)}) dx$$ We split the integral into two integrals: $$\frac{1}{2} \int_{0}^{2} x \ln{(1+2x^2)} dx - \frac{1}{2} \int_{0}^{2} x \ln{(1+x^2)} dx$$

Evaluate the integrals using integration by parts

Both integrals can be solved using integration by parts as follows: Integration by Parts Formula: \(\int udv = uv - \int v du\). For the first integral, let \(u = \ln(1 + 2x^2)\), \(dv = x dx\). Then, we have \(du = \frac{4x dx}{1 + 2x^2}\), \(v = \frac{x^2}{2}\). For the second integral, let \(u = \ln(1+x^2)\), \(dv = x dx\). Then, we have \(du = \frac{2x dx}{1 + x^2}\), \(v = \frac{x^2}{2}\). Evaluating the first integral using integration by parts: $$\frac{1}{2} \int_{0}^{2} x \ln{(1+2x^2)} dx =\frac{1}{4}(2^2 \ln(9)- \int_{0}^{2} \frac{2x^3 dx}{1 + 2x^2} )$$ Evaluating the second integral using integration by parts: $$\frac{1}{2}\int_{0}^{2} x \ln{(1+x^2)} dx = \frac{1}{4}(2^2 \ln(5) - \int_{0}^{2} \frac{2x^3 dx}{1 + x^2} )$$

Combine the results

Summing both integrals together: $$\frac{1}{2} \left[ 2^2\left(\ln(9) - \ln(5)\right) - \int_{0}^{2}\left(\frac{2x^3 dx}{1 + 2x^2} - \frac{2x^3 dx}{1 + x^2}\right)\right]$$ The result of this double integral is the solution to the exercise.

Key concepts

Limits of Integration

The limits of integration define the range over which you are integrating a function. For double integrals, these limits specify a region in the plane over which the integration occurs. In our exercise, the region \( R \) is defined by two conditions: \( 0 \leq y \leq x \) and \( 0 \leq x \leq 2 \). This means that for each slice of the function, defined by a constant \( x \), \( y \) will range from \( 0 \) to \( x \). After fully sweeping through all \( y \) given an \( x \), \( x \) itself will range from \( 0 \) to \( 2 \).
The limits allow you to set up the integral properly. Here, you first integrate with respect to \( y \), keeping \( x \) constant, then with respect to \( x \), analyzing each slice step by step, tracing the integral over the entire specified region.

Integration by Parts

Integration by parts is a technique derived from the product rule of differentiation. It's useful when integrating products of functions where direct integration isn't straightforward. The formula used is:\[\int u \ dv = uv - \int v \ du\]
In this exercise, the technique is applied twice, for two separate integrals, each involving a logarithmic and a polynomial function. When choosing \( u \) and \( dv \), it's often advantageous to let \( u \) be the function that becomes simpler when differentiated (like a logarithm), and \( dv \) be the function whose integral is elementary (like \( x \) dx).
For example:

• Let \( u = \ln(1 + 2x^2) \) and \( dv = x dx \)
• This means \( du = \frac{4x dx}{1 + 2x^2} \) and \( v = \frac{x^2}{2} \)

Using these, the integration by parts formula helps simplify the integral, reducing it to a solvable form.

Substitution Method

The substitution method is a key technique used to simplify integrals. It involves substituting a new variable to transform a complex expression into a simpler one. This is known as the "change of variables" technique.
In our problem, the substitution \( u = 1 + x^2 + y^2 \) is used during the integration with respect to \( y \). This substitution simplifies the function by turning it into a form where the variables separate, effectively reducing the interaction between \( x \) and \( y \).
Substitute and rearrange the derivative:

• Let \( u = 1 + x^2 + y^2 \)
• Then, \( du = 2y \, dy \), so \( dy = \frac{du}{2y} \)

This substitution converts the original function into a much easier log-integral, focusing more on integration limits, making computation straightforward.

Iterated Integrals

Iterated integrals allow you to compute double integrals by performing a sequence of single integrations. These are crucial for multi-variable calculus, helping tackle problems over two-dimensional domains by breaking them into manageable parts.
In our exercise, the iterated integral: \[\int_{0}^{2} \int_{0}^{x} \frac{x y}{1+x^{2}+y^{2}} \ dy \ dx\]first integrates with respect to \( y \), holding \( x \) constant. Once that's computed, it is then integrated with respect to \( x \).
The steps of an iterated integral can simplify complicated areas by:

• First addressing the inner integral over a smaller range
• Then expanding over a broader range for the outer integral

This ensures you precisely capture the total integration over the region \( R \), capturing all nuances of \( x \)-dependent changes.