Question

Calcular ∫∫∫T((x^2)+y)dV, onde T é a região interior ao cilindro (x^2)+(y^2)-4 e z= 4-(x^2)-(y^2)

Answer 1

Calcular a integral tripla

$\displaystyle\iiint\limits_{T} {f(x,\,y,\,z)}\,dV$


onde $f(x,\,y,\,z)=x^{2}+y$

e $T$ é a região abaixo do paraboloide $z=4-x^{2}-y^{2}$
acima do plano $x0y,$ e interior ao cilindro $x^{2}+y^{2}=4$


$\bullet\;\;$ Vamos escrever a integral iterada em coordenadas cilindricas:

$\begin{array}{cc} \left\{ \begin{array}{l} x=r\cos\theta\\ y=r\,\mathrm{sen\,}\theta\\ z=z \end{array} \right.\;\;&\;\;\begin{array}{c} 0\leq \theta \leq 2\pi\\ 0\leq r \leq 2\\ 0\leq z\leq 4-r^{2} \end{array} \end{array}$


$\bullet\;\;$ O Jacobiano das coordenadas cilíndricas é $\text{Jac\,}\varphi=r.$


$\bullet\;\;$ A função integranda fica

$f(x,\,y,\,z)\\ \\ =f(r\cos \theta,\,r\,\mathrm{sen\,}\theta,\,z)\\ \\ =(r\cos \theta)^{2}+r\,\mathrm{sen\,}\theta\\ \\ =r^{2}\cos^{2}\theta+r\,\mathrm{sen\,}\theta$


$\bullet\;\;$ Escrevendo a integral iterada:

$\displaystyle\iiint\limits_{T} {f(x,\,y,\,z)}\,dV\\ \\ \\ =\displaystyle\int\limits_{0}^{2\pi}\int\limits_{0}^{2}\int\limits_{0}^{4-r^{2}}{f(r\cos \theta,\,r\,\mathrm{sen\,}\theta,\,z)\cdot |\mathrm{Jac\,}\varphi|\,dz\,dr\,d\theta}\\ \\ \\ =\displaystyle\int\limits_{0}^{2\pi}\int\limits_{0}^{2}\int\limits_{0}^{4-r^{2}}{(r^{2}\cos^{2}\theta+r\,\mathrm{sen\,}\theta)\,r\,dz\,dr\,d\theta}\\ \\ \\ =\displaystyle\int\limits_{0}^{2\pi}\int\limits_{0}^{2}\int\limits_{0}^{4-r^{2}}{(r^{3}\cos^{2}\theta+r^{2}\,\mathrm{sen\,}\theta)\,dz\,dr\,d\theta}\\ \\ \\ =\displaystyle\int\limits_{0}^{2\pi}\int\limits_{0}^{2}{(r^{3}\cos^{2}\theta+r^{2}\,\mathrm{sen\,}\theta)\,z\left|_{0}^{4-r^{2}}\right.\,dr\,d\theta}\\ \\ \\ =\displaystyle\int\limits_{0}^{2\pi}\int\limits_{0}^{2}{(r^{3}\cos^{2}\theta+r^{2}\,\mathrm{sen\,}\theta)\,(4-r^{2}) \,dr\,d\theta}$


$=\displaystyle\int\limits_{0}^{2\pi}\int\limits_{0}^{2}{(4r^{3}\cos^{2}\theta+4r^{2}\,\mathrm{sen\,}\theta-r^{5}\cos^{2}\theta-r^{4}\,\mathrm{sen\,}\theta)\,dr\,d\theta}\\ \\ \\ =\displaystyle\int\limits_{0}^{2\pi}\int\limits_{0}^{2}{[(4r^{3}-r^{5})\cos^{2}\theta+(4r^{2}-r^{4})\,\mathrm{sen\,}\theta\,]\,dr\,d\theta}\\ \\ \\ =\displaystyle\int\limits_{0}^{2\pi}{\left[\left(r^{4}-\dfrac{r^{6}}{6}\right)\cos^{2}\theta+\left(\dfrac{4r^{3}}{3}-\dfrac{r^{5}}{5}\right)\,\mathrm{sen\,}\theta\,\right]_{0}^{2}\,d\theta}\\ \\ \\ =\displaystyle\int\limits_{0}^{2\pi}{\left[\left(16-\dfrac{64}{6}\right)\cos^{2}\theta+\left(\dfrac{32}{3}-\dfrac{32}{5}\right)\,\mathrm{sen\,}\theta\,\right]\,d\theta}$


$=\displaystyle\int\limits_{0}^{2\pi}{\left[\left(\dfrac{16}{3}\right)\cos^{2}\theta+\left(\dfrac{64}{15}\right)\,\mathrm{sen\,}\theta\,\right]\,d\theta}\\ \\ \\ =\dfrac{16}{3}\displaystyle\int\limits_{0}^{2\pi}{\cos^{2}\theta\,d\theta}+\dfrac{64}{15}\displaystyle\int\limits_{0}^{2\pi}{\mathrm{sen\,}\theta\,d\theta}\\ \\ \\ =\dfrac{16}{3}\cdot \pi+\dfrac{64}{15}\cdot 0\\ \\ \\ =\dfrac{16\pi}{3}$