Question

Integrais Triplas , me ajudem por gentileza. Obrigada!!

Answer 1

$I=\displaystyle\iiint\limits_{G}{\cos\left(\frac{z}{y} \right )\,d\mathbf{V}}$

Sendo $G$ a região definida como

$G=\left\{(x,\;y,\;z)\in\mathbb{R}^{3}\left|\,\dfrac{\pi}{6}\leq y \leq \dfrac{\pi}{2},\;y \leq x \leq\dfrac{\pi}{2},\;0\leq z \leq xy\right. \right \}$

$\bullet\;\;$ Os limites de integração já foram dados, então é só escrever a integral iterada:

$I=\displaystyle\iiint\limits_{G}{\cos\left(\frac{z}{y} \right )\,d\mathbf{V}}\\ \\ \\ =\int\limits_{\pi/6\,}^{\pi/2\,}\int\limits_{y}^{\,\pi/2\,}\int\limits_{0}^{xy}{\cos\left(\frac{z}{y} \right )\,dz\,dx\,dy}\\ \\ \\ =\int\limits_{\pi/6\,}^{\pi/2\,}\int\limits_{y}^{\,\pi/2\,}{y\left[\,\mathrm{sen}\left(\frac{z}{y} \right ) \right ]_{0}^{xy}\,dx\,dy}\\ \\ \\ =\int\limits_{\pi/6\,}^{\pi/2\,}\int\limits_{y}^{\,\pi/2\,}{y\left[\,\mathrm{sen}\left(\frac{x\diagup\!\!\!\! y}{\diagup\!\!\!\! y} \right )-\mathrm{sen}\left(\frac{0}{y} \right ) \right ]dx\,dy}\\ \\ \\ =\int\limits_{\pi/6\,}^{\pi/2\,}\int\limits_{y}^{\,\pi/2\,}{y\,\mathrm{sen\,}x\,dx\,dy}\\ \\ \\ =\int\limits_{\pi/6\,}^{\pi/2\,}{y\left.(-\cos x)\right|_{y}^{\pi/2}\,dy}\\ \\ \\ =\int\limits_{\pi/6\,}^{\pi/2\,}{y\cdot \left(-\cos \frac{\pi}{2}+\cos y \right )dy}\\ \\ \\ =\int\limits_{\pi/6\,}^{\pi/2\,}{y\cos y\,dy}$

A primitiva acima é encontrada pelo método de integração por partes:

$\begin{array}{ll} u=y\;\;&\;\;du=dy\\ \\ dv=\cos y\,dy\;\;&\;\;v=\mathrm{sen\,}y \end{array}\\ \\ \\ \\ \displaystyle\int{u\,dv}=uv-\int{v\,du}\\ \\ \\ \int\limits_{\pi/6}^{\pi/2}{y\cos y\,dy}=(y\,\mathrm{sen\,}y)|_{\pi/6}^{\pi/2}-\int\limits_{\pi/6}^{\pi/2}{\mathrm{sen\,}y\,dy}\\ \\ \\ =\left(\frac{\pi}{2}\,\mathrm{sen\,}\frac{\pi}{2}-\frac{\pi}{6}\,\mathrm{sen\,}\frac{\pi}{6} \right )+\cos y|_{\pi/6}^{\pi/2}\\ \\ \\ =\left(\frac{\pi}{2}\cdot 1-\frac{\pi}{6}\cdot \frac{1}{2} \right )+\left(\cos\frac{\pi}{2}-\cos \frac{\pi}{6} \right )\\ \\ \\ =\left(\frac{\pi}{2}-\frac{\pi}{12} \right )+\left(0-\frac{\sqrt{3}}{2} \right )\\ \\ \\ =\left(\frac{6\pi}{12}-\frac{\pi}{12} \right )-\frac{\sqrt{3}}{2}\\ \\ \\ =\frac{5\pi}{12}-\frac{\sqrt{3}}{2}.$