∬ysin(xy)dA=∫0π∫12ysin(xy)dxdy∫12ysin(xy)dx=−yycos(xy)∣12=−cos(xy)∣12=−(cos2y−cosy)=−cos2y+cosy∫0π−cos2y+cosydy=−12sin2y+siny∣0π=(−12sin2π+sinπ)−(−12sin0+sin0)=0The answer is a\begin{aligned} &\iint y \sin (xy)dA\textcolor{blue}{=\int_{0}^{\pi}\textcolor{red}{ \int_{1}^{2} y \sin (xy) dx}dy }\\[1 em] \textcolor{red}{ \int_{1}^{2} y \sin (xy)dx}& =\frac{-y}{y}\cos (xy)\mid _{1}^{2} =-\cos (xy)\mid _{1}^{2}\\[1 em] &=-(\cos 2y -\cos y) =-\cos 2y +\cos y\\[1 em] \textcolor{blue}{ \int_{0}^{\pi} -\cos 2y +\cos y dy} &=\frac{-1}{2}\sin 2y+\sin y\mid _{0}^{\pi}\\[1 em] &=(\frac{-1}{2}\sin 2\pi+\sin \pi)-(\frac{-1}{2}\sin 0+\sin 0)=\fcolorbox{red}{aqua}{0}\\[1 em] \text{The answer is }\fcolorbox{red}{aqua}{a} \end{aligned}∫12ysin(xy)dx∫0π−cos2y+cosydyThe answer is a∬ysin(xy)dA=∫0π∫12ysin(xy)dxdy=y−ycos(xy)∣12=−cos(xy)∣12=−(cos2y−cosy)=−cos2y+cosy=2−1sin2y+siny∣0π=(2−1sin2π+sinπ)−(2−1sin0+sin0)=0
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