Question

A integral $\int\int\int\ {x y^{2} } \, dx dy dz$, onde A é a região definida por $x^{2} + y^{2} = a^{2}, 0 \leq z \leq 5$, pode ser reescrita em coordenadas cilídricas como:

Answer 1

$x^2+y^2= a^2 \\\\ r^2=a^2\\r=a \to \boxed{ 0 \leq r \leq a}$


$\int \int \int xy^2\;dx\;dy\;dz\\\\ =\int \int \int r*cos(\theta)*(r*sen(\theta))^2rdrd\theta dz \\\\ = \int\limits^{2\pi}_0 \int\limits^{a}_0 \int\limits^{5}_0 {r^4cos(\theta)sen^2(\theta)dz\;dr\;d\theta}$

Answer 2

Lembre-se que...

$\displaystyle \iiint\limits_{R} f(x,y,z)\,dx\,dy\,dz = \iiint\limits_{R'} r\cdot g(r,\theta,z)\,dr\,d\theta\,dz \\ \\ \\ \texttt{Onde:}\\ g(r,\theta,z)=f(r\cos\theta,r\sin\theta,z)=f\circ T(r,\theta,z)\\ \\ T:R'\to R$
 
                             $r=\left| \begin{matrix} \dfrac{\partial}{\partial r}(r\cos\theta)&\dfrac{\partial}{\partial \theta}(r\cos\theta)&\dfrac{\partial}{\partial z}(r\cos\theta)\\ \\ \dfrac{\partial}{\partial r}(r\sin\theta)&\dfrac{\partial}{\partial \theta}(r\sin\theta)&\dfrac{\partial}{\partial z}(r\sin\theta)\\ \\ \dfrac{\partial}{\partial r}(z)&\dfrac{\partial}{\partial \theta}(z)&\dfrac{\partial}{\partial z}(z) \end{matrix}\right|$

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Para este caso

$\texttt{Seja}: f(x,y,z)=xy^2\\\\ g(r,\theta,z)=f(r\cos\theta,r\sin\theta,z)\\ \\ g(r,\theta,z)=r\cos\theta\cdot (r\sin\theta)^2\\ \\ g(r,\theta,z)=r^3\sin^2\theta\cos\theta\\ \\ \\ \texttt{Limites de integraci\'on: }\\ \\ x^2+y^2=a^2\to (r\cos\theta)^2+(r\sin\theta)^2=a^2\\ \\ x^2+y^2=a^2\to r^2=a^2\to r=a\\ \\ \texttt{ent\~ao: }R=\{(r,\theta,z):0\leq r\leq a\,;\, 0\leq \theta \ \textless \ 2\pi,0\leq z \leq 5\}$
 
          $\displaystyle \iiint\limits_{S}xy^2\,dx\,dy\,dz=\iiint\limits_{R}r^4\sin^2\theta\cos\theta\,dr\,d\theta\,dz\\ \\ \\ \boxed{\iiint\limits_{S}xy^2\,dx\,dy\,dz=\int\limits_{0}^{a}\int\limits_{0}^{2\pi}\int\limits_{0}^{5}r^4\sin^2\theta\cos\theta\,dz\,d\theta\,dr}$