Problem 65 - Calculus Early Transcendentals

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. $$\int_{0}^{1 / 2} \int_{y^{2}}^{1 / 4} y \cos \left(16 \pi x^{2}\right) d x d y$$

Step-by-Step Solution

Answer

After reversing the order of integration and evaluating the given double integral, we found that its value is 0.

Step 1: Sketch the Region of Integration

The region of integration is defined by the given limits: $$0 \le y \le \frac{1}{2}$$ $$y^{2} \le x \le \frac{1}{4}$$ Plotting these inequalities on a coordinate system (xy-plane), we get a region bounded by the x-axis (y=0), the line y=1/2, curve x=y^2 and a vertical line x=1/4.

Step 2: Reverse the Order of Integration

To reverse the order of integration, we need to change the limits of integration. Currently, the outside integral is with respect to y and the inside integral is with respect to x. We will switch the order so that the outside integral is with respect to x and the inside integral with respect to y. For x: From the sketch, we can see that the lower limit is x=0 and the upper limit is x=1/4. For y: The lower limit is given by the curve x=y^2, which means y=sqrt(x). Since the upper limit is a horizontal line y=1/2, the limits for y are sqrt(x) to 1/2. The reversed integral becomes: $$\int_{0}^{1/4} \int_{\sqrt{x}}^{1/2} y \cos(16\pi x^2) dy dx$$

Step 3: Evaluate the Resulting Integral

Now, we will evaluate the integral: $$\int_{0}^{1/4} \int_{\sqrt{x}}^{1/2} y \cos(16\pi x^2) dy dx$$ First, integrate with respect to y: $$\int_{0}^{1/4} \left[\frac{y^2}{2} \cos(16\pi x^2)\right]_{\sqrt{x}}^{1/2} dx$$ Now, we will plug in the limits for y: $$\int_{0}^{1/4} \left[\frac{1}{4}\cos(16\pi x^2) - \frac{x}{2}\cos(16\pi x^2)\right] dx$$ Now, integrate with respect to x: $$\left[\frac{1}{32\pi}\sin(16\pi x^2) - \frac{x^2}{4}\frac{1}{32\pi}\sin(16\pi x^2)\right]_0^{1/4}$$ Finally, plug in the limits for x and evaluate: $$\left[\frac{1}{32\pi}\sin(4\pi) - \frac{1}{16}\frac{1}{32\pi}\sin(4\pi)\right] - \left[\frac{1}{32\pi}\sin(0) - \frac{0}{4}\frac{1}{32\pi}\sin(0)\right]$$ Since both sin(4π) and sin(0) are equal to 0, the value of the integral is: $$0$$