Question

When converted to an iterated integral, the following double integrals are easier to evaluate in one order than the other. Find the best order and evaluate the integral. $$\iint_{R} x \sec ^{2} x y d A ; R=\\{(x, y): 0 \leq x \leq \pi / 3,0 \leq y \leq 1\\}$$

Short answer

Question: Find the result of the double integral $$\iint_R x\sec^2(xy) dA$$ where \(R\) is defined by \((x,y) : 0 \leq x \leq \pi/3, 0 \leq y \leq 1\). Answer: The result of the double integral is \(\tan(\pi^2) \ln(\pi)\).

Step-by-step solution

Analyze the given integral

The given integral is $$ \iint_R x \sec ^{2} x y d A = \int_{0}^{\pi/3} \int_{0}^1 x\sec^2 (xy) \, dy \, dx $$ or $$ \iint_R x \sec ^{2} x y d A= \int_{0}^1 \int_{0}^{\pi/3} x\sec^2 (xy) \, dx \, dy. $$ We need to evaluate which order simplifies the problem the most.

Consider integrating with respect to y first

If we integrate with respect to y first, we have $$ \int_{0}^1 x\sec^2 (xy)\, dy. $$ Here, the integrand does not simplify directly and requires integration by substitution.

Consider integrating with respect to x first

If we integrate with respect to x first, we have $$ \int_{0}^{\pi/3} x\sec^2 (xy) \, dx. $$ In this case, we can use the substitution \(u = xy\) to simplify the integrand.

Evaluate the integral with respect to x first

Let \(u = xy\). Then, \(du = y\hspace{0.5em}dx\). So, the integral becomes $$ \int_{0}^1 \int_{0}^{\pi/3} x\sec^2 (xy) \, dx \, dy = \int_{0}^1\left(\int_0^{u(y)} \sec^2(u) \frac{du}{y} \right) dy. $$ Now, integrate with respect to \(u\): $$ \int_{0}^1\left[\frac{\tan(u)}{y}\Big|_0^{u(y)}\right] dy, $$ where \(u(0) = 0\) and \(u(1) = x\).

Substitute u = xy back

Replace \(u = xy\) back into the integral: $$ \int_{0}^1\left[\frac{\tan(xy)}{y}\Big|_0^\pi\right] dy = \int_{0}^1 \frac{\tan(x\pi)}{y} dy. $$

Evaluate the final integral

Notice that since \(0\leq x\leq \pi/3\), we have \(0\leq x\pi\leq \pi^2/3\). Now, integrate with respect to y: $$ \int_{0}^1 \frac{\tan(x\pi)}{y} dy = \tan(x\pi) \ln(x) |_{0}^{\pi} = \tan(\pi^2) \ln(\pi) - \lim_{x\to 0} \tan(x\pi) \ln(x). $$ The limit \(\lim_{x\to 0} \tan(x\pi) \ln(x) = 0\) since it is a product of an indeterminate form \(\lim_{x\to 0}\tan(x\pi)=0\) and \(\lim_{x\to 0}\ln(x)=-\infty\). Apply L'Hopital's rule to prove this. Thus, the final result of the given integral is: $$ \iint_R x \sec ^{2} x y d A = \tan(\pi^2) \ln(\pi). $$

Key concepts

Integration by Substitution

Integration by Substitution is an essential technique used to simplify the process of integrating complex functions. The main idea is to transform a difficult integral into a simpler one by making a substitution that simplifies the integrand.\\For instance, in the problem provided, integrating with respect to \(x\) first offered a convenient substitution: \(u = xy\). This effectively converted the integrand \(x \sec^2(xy)\) into \(\sec^2(u)\ \frac{du}{y}\), which is much simpler to handle.\\**Key Steps in Integration by Substitution**:\- **Identify a part of the integrand** that, when substituted, simplifies the integral.\- **Substitute**: Let \(u = g(x)\), and determine \(du\) by differentiating \(g(x)\).\- **Change the limits**: If you are changing the variable of integration, also change the limits accordingly.\- **Integrate** the simpler function and finally\- **Substitute back** the original variables to obtain the solution in terms of the original notation.\\Using substitution effectively can make a complex double integral problem much more approachable.

Double Integrals

Double integrals allow us to extend the concept of integration to functions of two variables, essentially summing these functions over a region in the plane. They're useful for finding volumes under surfaces or evaluating areas in the plane. In practice, selecting the best order of integration — either \(dx\) first or \(dy\) first — can greatly simplify calculations.\\**Understanding Double Integrals**:\- Consider the region \(R\) over which you're integrating, defined by its limits, such as \(0 \leq x \leq \pi/3\) and \(0 \leq y \leq 1\).\- Choose the integration order that makes the computation easier. Converting a double integral into an iterated integral in the more favorable order can lead to a simpler form of the integral.\\In the presented problem, evaluating \(x\) first proved simpler because it allowed for an efficient substitution. Depending on the problem, switching the order of integration can turn a virtually impossible integral into a straightforward task. Experimenting with both orders when possible is generally a good strategy to find the less cumbersome path.

L'Hôpital's Rule

L'Hôpital's Rule comes in handy when you're dealing with limits that result in indeterminate forms like \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\). It provides a method for evaluating these by differentiating the numerator and the denominator separately, and then taking the limit again.\\In the final step of the solved exercise, L'Hôpital's Rule is applied to evaluate the limit \(\lim_{x \to 0} \tan(x\pi) \ln(x)\), which initially appears as \(0 \cdot (-\infty)\). By re-expressing it as \(\frac{\tan(x\pi)}{1/\ln(x)}\), we reduce it to the form \(\frac{0}{0}\), perfect for L'Hôpital's technique.\\**Steps of Applying L'Hôpital's Rule**:\- **Rewrite** the function to a \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\) form.\- **Differentiate** both the numerator and the denominator separately.\- **Re-evaluate the limit** on the new fraction. If another indeterminate form emerges, L'Hôpital's Rule can be applied again.\\This approach effectively handles tricky limits that arise in the evaluation of integrals and ensures calculations are precise and correct.