Question

The following integrals can be evaluated only by reversing the order of integration. Sketch the region of integration, reverse the order of integration, and evaluate the integral. $$\int_{0}^{1 / 2} \int_{y^{2}}^{1 / 4} y \cos \left(16 \pi x^{2}\right) d x d y$$

Short answer

Based on the given step by step solution, the value of the double integral: $$\int \int _R y \cos \left(16 \pi x^{2}\right) d y d x$$ where R represents the trapezoidal region with vertices at (1/4, 0), (1/4, 1/2), (1/2, 1/2), and (0, 0), is equal to 0.

Step-by-step solution

Sketch the Region of Integration

To sketch the region of integration, we need to understand the limits of integration. They are given by the following inequalities: $$0 \leq y \leq \frac{1}{2}$$ and $$y^2 \leq x \leq \frac{1}{4}$$ This region is a trapezoid in the xy-plane with vertices at (1/4, 0), (1/4, 1/2), (1/2, 1/2), and (0, 0).

Reverse the Order of Integration

Now we need to reverse the order of integration from dy dx to dx dy by changing the limits appropriately. For dx, the bounds can be determined from x = y^2 and x = 1/4, which yield the following expression after subtraction: $$0\leq x \leq \frac{1}{4}$$ For dy, the bounds can be determined from y = 0 and y = 1/2, which can be related to x as follows: $$0 \leq y \leq \sqrt{x}$$ The integral then becomes: $$\int_{0}^{1 / 4} \int_{0}^{\sqrt{x}} y \cos \left(16 \pi x^{2}\right) d y d x$$

Evaluate the Integral

Now, we evaluate the integral, first with respect to y: $$\int_{0}^{1 / 4} \left[\frac{1}{2} y^{2} \cos \left(16 \pi x^{2}\right)\right]_{0}^{\sqrt{x}} d x$$ $$\int_{0}^{1 / 4} \frac{1}{2} x \cos \left(16 \pi x^{2}\right) d x$$ Then we evaluate the integral with respect to x. To do this, let's define the following substitution: $$u = 16 \pi x^2$$ $$\frac{du}{dx} = 32\pi x$$ $$dx = \frac{du}{32\pi x}$$ Substituting these expressions into the integral, we obtain: $$\frac{1}{2} \int_{0}^{4 \pi } \frac{1}{32\pi} \cos(u) \ du$$ Now, we perform the integration and change the bounds to the new variable u: $$\frac{1}{64 \pi} \int_{0}^{4 \pi } \cos(u) du$$ $$\frac{1}{64 \pi} \left[ \sin(u) \right]_0^{4\pi}$$ $$\frac{1}{64 \pi} \left[ \sin(4\pi) - \sin(0) \right]$$ Since sin(4π) = 0 and sin(0) = 0, the final result is 0. So, the value of the given integral is 0.

Key concepts

Integration Techniques

In calculus, reversing the order of integration is an advanced technique that helps simplify the evaluation of double integrals, particularly when one order is complex to calculate directly. This method involves changing the sequence of integration variables and their limits, often making the calculation more manageable.

In the given problem, initially, the integral is set as \[ \int_{0}^{1 / 2} \int_{y^{2}}^{1 / 4} y \cos \left(16 \pi x^{2}\right) d x d y \]. By sketching the region, we can visualize the correct limits necessary for each variable. The original setup may not lead to an easy evaluation; hence, reversing the order of integration provides a more convenient route.

• Understanding the boundaries: Sketch the area defined by \(0 \leq y \leq \frac{1}{2}\) and \(y^2 \leq x \leq \frac{1}{4}\). This guides how to rearrange the limits for \(x\) and \(y\).
• Reversing the boundaries: Flip the limits from \(dx\,dy\) to \(dy\,dx\), resulting in \(0 \leq x \leq \frac{1}{4}\) and \(0 \leq y \leq \sqrt{x}\).

This technique highlights the versatility of multivariable integrals, making them accessible through strategic manipulation.

Definite Integrals

Definite integrals compute the accumulated area under the curve for a function within a given interval. In the context of double integrals, this extends to finding the volume under a surface defined by two variables.

In our problem, we take an integral over a defined range where\[ \int_{0}^{1/2} \int_{y^{2}}^{1 / 4} y \cos (16 \pi x^{2}) dx dy = 0. \]

Calculating definite integrals involves evaluating the upper and lower limits of the integration variables after setting up the integral correctly. Once the order is reversed, it becomes simpler:\[ \int_{0}^{1 / 4} \int_{0}^{\sqrt{x}} y \cos (16 \pi x^{2}) dy dx. \] The evaluation is done step-by-step:

• First, calculate the inner integral over \(y\).
• Substitute back to obtain a function of \(x\), and then integrate with respect to \(x\).

By systematically applying definite integral rules after rearranging, the complexity reduces, often resulting in simplified solutions.

Multivariable Calculus

Multivariable calculus extends the concepts of single-variable calculus to functions of several variables. It is essential for studying phenomena where multiple factors affect outcomes, such as in physics or engineering.

The integration of functions over a region defined by more than one variable involves calculating areas and volumes in multi-dimensional spaces. In this exercise, we handle a two-dimensional region defined by both \(x\) and \(y\). The process involves:

• Sketching the region: Identifying boundaries through inequalities helps visualize the space of integration, an essential step for successfully reversing integration order.
• Switching integration orders: Often necessary for dealing with complicated integrands or limits, aiding in simplifying the computations.
• Substituting variables: Introductory integration techniques in multivariable calculus include transforming the integrand by changing variables to facilitate easier computation.

Ultimately, multivariable calculus techniques provide a comprehensive toolkit for tackling complex problems by decomposing them into more manageable parts.