$\int^\frac{1}{2}_0 \int^{y^2}_\frac{1}{4} y \cos (16 \pi x^2) \,dxdy$

Question

Evaluate by first changing the order of integration:

$$\int^\frac{1}{2}_0 \int^{y^2}_\frac{1}{4} y \cos (16 \pi x^2) \,dxdy$$

How to solve this problem?

Answer 1

$\def\d{\mathrm{d}}$\begin{align*} &\mathrel{\phantom{=}} \int_0^{\frac{1}{2}} \int_{\frac{1}{4}}^{y^2} y \cos(16 πx^2) \,\d x \d y = -\int_0^{\frac{1}{2}} \int_{y^2}^{\frac{1}{4}} y \cos(16 πx^2) \,\d x \d y\\ &= -\iint\limits_{\substack{0 \leqslant y \leqslant \frac{1}{2} \\ y^2 \leqslant x \leqslant \frac{1}{4}}} y \cos(16 πx^2) \,\d x \d y = -\iint\limits_{\substack{0 \leqslant x \leqslant \frac{1}{4} \\ 0 \leqslant y \leqslant \sqrt{x}}} y \cos(16 πx^2) \,\d x \d y\\ &= -\int_0^{\frac{1}{4}} \int_0^{\sqrt{x}} y \cos(16 πx^2) \,\d y \d x = -\int_0^{\frac{1}{4}} \cos(16 πx^2) \,\d x \int_0^{\sqrt{x}} y \,\d y\\ &= -\frac{1}{2} \int_0^{\frac{1}{4}} x\cos(16 πx^2) \,\d x = -\frac{1}{2} \cdot \frac{1}{32π} \sin(16 πx^2) \left.\vphantom{\intop}\right|_0^{\frac{1}{4}} = 0. \end{align*}

Answer 2

Note that when $0 \le y \le \frac{1}{2}$ we have $y^2 \le \frac{1}{4}$, so your integral is in fact equal to $$-\int_0^{\frac{1}{2}} \int_{y^2}^{\frac{1}{4}} y\cos(16\pi x^2)\,dx\,dy$$ and, as such, the region of integration consists of those points $(x,y)$ for which $$0 \le y \le \frac{1}{2} \quad \text{and} \quad y^2 \le x \le \frac{1}{4}$$ When $y=0$ we have $y^2=0$ and when $y=\frac{1}{2}$ we have $y^2=\frac{1}{4}$, so you can check that this consists exactly of the points $(x,y)$ for which $$0 \le x \le \frac{1}{4} \quad \text{and} \quad 0 \le y \le \sqrt{x}$$ So your integral is equal to $$-\int_0^{\frac{1}{4}} \int_0^{\sqrt{x}} y\cos(16\pi x^2)\,dy\,dx$$ Evaluating this integral is straightforward.