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

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

Step-by-step solution

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.

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: $$\int_{0}^{1/4} \int_{\sqrt{x}}^{1/2} y \cos(16\pi x^2) dy dx$$

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: $$0$$