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. $$\iint_{R} y^{3} \sin x y^{2} d A ; R=\\{(x, y): 0 \leq x \leq 2,0 \leq y \leq \sqrt{\pi / 2}\\}$$

Short answer

Question: Evaluate the double integral: $$\iint_R y^3 \sin(xy^2)dA,$$ where $$R = \{(x, y): 0 \le x \le 2, 0 \le y \le \sqrt{\frac{\pi}{2}} \}.$$ Answer: \(-\frac{\pi}{4}\).

Step-by-step solution

Write the integral in both orders

Consider the given integral in both orders: $$\int_0^2 \int_0^{\sqrt{\pi/2}} y^3\sin(xy^2) dy dx \quad \text{and} \quad \int_0^{\sqrt{\pi/2}}\int_0^2 y^3\sin(xy^2) dx dy.$$

Analyze the orders for ease of integration

If we consider integrating with respect to \(y\) first, we find that taking the integral of \(y^3 \sin(xy^2)\) with respect to \(y\) is difficult since the sine term involves a product of \(x\) and \(y^2\). However, if we integrate with respect to \(x\) first, we will isolate the sine function inside the integral making it easier to compute as the other terms involving \(y\) can be factored out. Therefore, we choose the second order: $$\int_0^{\sqrt{\pi/2}}\int_0^2 y^3\sin(xy^2) dx dy.$$

Integrate with respect to x

Now, integrate the inner integral with respect to \(x\): $$\int_0^{\sqrt{\pi/2}} \left[ \frac{y^3}{2y^2}\left(-\cos(xy^2)\right)\Biggl|_0^2 \right] dy.$$

Evaluate the limits of x

Evaluate the limits for \(x\): $$\int_0^{\sqrt{\pi/2}} \left[\frac{y^3}{2y^2}(-\cos(2y^2) - (-\cos(0)))\right] dy$$ Simplify the expression: $$\int_0^{\sqrt{\pi/2}} \left[\frac{1}{2} y (\cos(2y^2) - 1)\right] dy$$

Integrate with respect to y

Now, integrate the outer integral with respect to \(y\): $$\frac{1}{2} \left[\int_0^{\sqrt{\pi/2}} y \cos(2y^2) dy - \int_0^{\sqrt{\pi/2}} y dy \right]$$ For the first integral, let's find the antiderivative using substitution. Let \(u = 2y^2\), then \(\frac{du}{2} = y dy\). So the first integral becomes: $$\int_0^{\sqrt{\pi/2}}\frac{y}{2}\cos(2y^2) dy = \frac{1}{4}\int_0^{\pi} \cos(u) du $$ Now, let's determine the antiderivatives of both integrals: $$\frac{1}{2} \left[\frac{1}{4}\left(\sin(u)\Biggr|_0^{\pi}\right) - \frac{1}{2} \left(\frac{y^2}{2}\Biggr|_0^{\sqrt{\pi/2}}\right) \right].$$

Evaluate the limits of y

Finally, substitute the limits back into the integral and evaluate to find the final answer: $$\frac{1}{2} \left[\frac{1}{4} (\sin(\pi) - \sin(0)) - \frac{1}{2} \left(\frac{\pi}{2}-0\right)\right] = -\frac{1}{2}\cdot\frac{1}{2}\cdot\frac{\pi}{2} = -\frac{\pi}{4}$$ Therefore, the value of the given integral is \(-\frac{\pi}{4}\).

Key concepts

Iterated Integral

Understanding iterated integrals is key to evaluating double integrals more efficiently. When dealing with a double integral, you are essentially calculating the volume under a surface within a specified region of the xy-plane. An iterated integral allows you to break this calculation into manageable steps by integrating one variable at a time. Imagine peeling layers off an onion one by one rather than trying to slice through them all at once.

In the context of the exercise, we originally have a double integral over region \(R\). Transform this into two single integrals, performed in sequence, known as iterated integrals. Here, you would either integrate with respect to \(x\) first and then \(y\), or vice versa.

By evaluating one variable's integral at a time, iterated integrals simplify the process. This layered approach lets you break complex expressions into simpler parts, making them easier to solve step by step.

Integration Order

Choosing the right integration order simplifies the calculation of double integrals significantly. At a glance, each order may seem to lead to the same result, but their computational ease can vary greatly. For the given problem, deciding on whether to integrate \(x\) before \(y\) or vice versa affects the steps that follow.

In the provided example, starting with \(x\) was preferable. This choice was driven by the function inside the integral, \(y^3 \sin(xy^2)\). If \(x\) is integrated first, the sine term becomes easier to handle due to variable isolation, which does not occur if \(y\) is chosen first. This careful analysis helps avoid unnecessary complexity and simplifies the integration process.

Hence, observing the function and determining which variable can be integrated more readily is crucial. This strategic decision directly impacts the difficulty of solving the problem, leading to a more straightforward calculation.

Change of Variables

The substitution of variables can be a powerful technique when evaluating integrals, making seemingly intractable integrals more approachable. This change of variables shifts the problem into a new frame of reference or set of coordinates where the integration becomes simpler.

In our exercise, to integrate the term \(y \cos(2y^2)\), a substitution was used: let \(u = 2y^2\). Changing variables in this way transforms the integral into a more familiar and manageable form with respect to \(u\).

By doing this, you not only simplify the integration process but also ensure the limits of integration reflect the original variables' range appropriately. Thus, change of variables facilitates a smoother integration by transforming it into a form that offers an easier path to the solution.

Antiderivative

Finding the antiderivative is a core skill when dealing with integrals. It involves identifying a function whose derivative returns the original integrand. In layman's terms, it's like "reverse engineering" a function.

During the final steps of our solution, once the integration order is settled and substitutions applied, we need to calculate the antiderivative of each part of the integrand. For instance, in \(\int y \cos(2y^2) dy\), identifying the antiderivative helped us compute the integral to reach a solution.

Solving for antiderivatives allows you to evaluate the definite integral. This process often involves reversing differentiation rules and leveraging known antiderivative forms, offering a means to solve even the most complex integrals fully. Understanding antiderivatives is crucial to progress from a mere formulaic evaluation to a deep comprehension of the integral calculus.