Chapter 13 - Multiple Integration - 13.2 Double Integrals over General Regions - 13.2 Exercises - Page 982: 65

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\[\begin{align} & \int_{0}^{1/2}{\int_{{{y}^{2}}}^{1/4}{y\cos \left( 16\pi {{x}^{2}} \right)}dxdy} \\ & x={{y}^{2}}\to y=\sqrt{x} \\ & \text{Using the graph to switch the order of integration} \\ & R=\left\{ \left( x,y \right):0\le y\le \sqrt{x},\text{ 0}\le x\le \frac{1}{4}\text{ } \right\} \\ & \text{Then}, \\ & \int_{0}^{1/2}{\int_{{{y}^{2}}}^{1/4}{y\cos \left( 16\pi {{x}^{2}} \right)}dxdy}=\int_{0}^{1/4}{\int_{0}^{\sqrt{x}}{y\cos \left( 16\pi {{x}^{2}} \right)}dydx} \\ & \text{Integrating} \\ & =\int_{0}^{1/4}{\left[ \frac{1}{2}{{y}^{2}}\cos \left( 16\pi {{x}^{2}} \right) \right]_{0}^{\sqrt{x}}dx} \\ & =\int_{0}^{1/4}{\left[ \frac{1}{2}{{\left( \sqrt{x} \right)}^{2}}\cos \left( 16\pi {{x}^{2}} \right) \right]dx} \\ & =\frac{1}{2}\int_{0}^{1/4}{x\cos \left( 16\pi {{x}^{2}} \right)dx} \\ & =\frac{1}{64\pi }\int_{0}^{1/4}{\left( 32\pi x \right)\cos \left( 16\pi {{x}^{2}} \right)dx} \\ & =\frac{1}{64\pi }\left[ \sin \left( 16\pi {{x}^{2}} \right) \right]_{0}^{1/4} \\ & =\frac{1}{64\pi }\left[ \sin \left( 16\pi {{\left( \frac{1}{4} \right)}^{2}} \right)-\sin \left( 16\pi {{\left( 0 \right)}^{2}} \right) \right] \\ & =\frac{1}{64\pi }\left[ \sin \left( \pi \right)-\sin \left( 0 \right) \right] \\ & =0 \\ \end{align}\]