Question

Estou precisando de ajuda:

Answer 1

Calcular a integral de superfície

$\displaystyle\iint_{S}\sqrt{x^2+y^2}\,dS$


onde $S$ é a porção do cone de equação

$\dfrac{x^2}{a^2}+\dfrac{y^2}{a^2}-\dfrac{z^2}{b^2}=0\,,~~~\text{com }0\le z\le b.$


Como $z\ge 0,$ podemos reescrever a equação do cone da seguinte forma: ($z$ em função de $x$ e $y$ )

$z=\dfrac{b}{a}\sqrt{x^2+y^2}\,,~~~\text{ com }0\le z\le b.$

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Parametrizando a superfície $S.$

Este é um cone circular, pois as seções paralelas ao plano $xy$ são circunferências. Além disso, a projeção do cone sobre o plano $xy$ é o disco com centro na origem e raio $a.$ Sendo assim, é conveniente usar coordenadas polares no plano $xy.$

O $z$ é obtido pela equação do cone.

$S:~~\begin{array}{cc} \left\{\begin{array}{l} x(r,\,\theta)=r\cos \theta\\\\ y(r,\,\theta)=r\,\mathrm{sen\,}\theta\\\\ z(r,\,\theta)=\dfrac{b}{a}\cdot r \end{array}\right.~~&~~\begin{array}{c} 0\le r\le a\\\\ 0\le \theta \le 2\pi \end{array} \end{array}$

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Encontrando os vetores $\mathbf{X}_r$ e $\mathbf{X}_\theta:$

$\bullet~~\mathbf{X}_r=\left(\dfrac{\partial x}{\partial r},\;\dfrac{\partial y}{\partial r},\;\dfrac{\partial z}{\partial r} \right )\\\\\\ \mathbf{X}_r=\left(\cos \theta,\;\mathrm{sen\,}\theta,\;\dfrac{b}{a} \right )\\\\\\\\ \bullet~~\mathbf{X}_\theta=\left(\dfrac{\partial x}{\partial \theta},\;\dfrac{\partial y}{\partial \theta},\;\dfrac{\partial z}{\partial \theta} \right )\\\\\\ \mathbf{X}_\theta=\left(-r\,\mathrm{sen\,} \theta,\;r\cos\theta,\;0 \right )$

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Achando um vetor normal à superfície (via produto vetorial):

$\overrightarrow{\mathbf{n}}=\mathbf{X}_r\times \mathbf{X}_\theta\\\\\\ =\det\!\left[ \begin{array}{ccc} \overrightarrow{\mathbf{i}}&\overrightarrow{\mathbf{j}}&\overrightarrow{\mathbf{k}}\\\\ \cos \theta&\mathrm{sen\,}\theta&\frac{b}{a}\\\\ -r\,\mathrm{sen\,}\theta&r\cos \theta&0 \end{array} \right]\\\\\\ =\left(\mathrm{sen\,}\theta\cdot 0-r\cos \theta\cdot \dfrac{b}{a}\right)\overrightarrow{\mathbf{i}}-\left(\cos \theta\cdot 0-(-r\,\mathrm{sen\,}\theta)\cdot \dfrac{b}{a} \right )\overrightarrow{\mathbf{j}}+(\cos \theta\cdot r\cos \theta-(-r\,\mathrm{sen\,\theta})\cdot \mathrm{sen\,}\theta )\overrightarrow{\mathbf{k}}\\\\\\ =-\dfrac{b}{a}\cdot r\cos \theta\overrightarrow{\mathbf{i}}-\dfrac{b}{a}\cdot r\,\mathrm{sen\,} \theta\overrightarrow{\mathbf{j}}+(r\cos^2\theta+r\,\mathrm{sen^2\,}\theta)\overrightarrow{\mathbf{k}}$

$=-\dfrac{b}{a}\cdot r\cos \theta\overrightarrow{\mathbf{i}}-\dfrac{b}{a}\cdot r\,\mathrm{sen\,} \theta\overrightarrow{\mathbf{j}}+r\overrightarrow{\mathbf{k}}\\\\\\ =(-r)\cdot \left(\dfrac{b}{a}\cos \theta,\;\dfrac{b}{a}\,\mathrm{sen\,}\theta,\;-1 \right )$


$\bullet~~$ Só nos interessa o módulo do vetor normal. Portanto,

$\|\overrightarrow{\mathbf{n}}\|=\|\mathbf{X}_r\times \mathbf{X}_\theta\|\\\\ =\left\|(-r)\cdot \left(\dfrac{b}{a}\cos \theta,\;\dfrac{b}{a}\,\mathrm{sen\,}\theta,\;-1 \right )\right\|\\\\\\ =r\cdot \sqrt{\left(\dfrac{b}{a}\cos \theta\right)^2+\left(\dfrac{b}{a}\,\mathrm{sen\,} \theta\right)^2+(-1)^2}\\\\\\ =r\cdot \sqrt{\dfrac{b^2}{a^2}\cos^2 \theta+\dfrac{b^2}{a^2}\,\mathrm{sen^2\,} \theta+1}\\\\\\ =r\cdot \sqrt{\dfrac{b^2}{a^2}\,(\cos^2 \theta+\mathrm{sen^2\,} \theta)+1}\\\\\\ =r\cdot \sqrt{\dfrac{b^2}{a^2}+1}\\\\\\ =r\cdot \sqrt{\dfrac{b^2+a^2}{a^2}}\\\\\\ \therefore~~\boxed{\begin{array}{c} \|\mathbf{X}_r\times \mathbf{X}_\theta\|=r\cdot \dfrac{\sqrt{b^2+a^2}}{a} \end{array}}$
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A função a ser integrada também sofrerá transformação:

$f(x,\;y)=\sqrt{x^2+y^2}\\\\\\f(x,\,y)=g(r,\;\theta)\\\\ \sqrt{x^2+y^2}=\sqrt{r^2\cos^2\theta+r^2\,\mathrm{sen^2\,}\theta}\\\\ \sqrt{x^2+y^2}=\sqrt{r^2}\\\\ \sqrt{x^2+y^2}=r~~\Rightarrow~~\boxed{g(r,\;\theta)=r}$

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Portanto a integral de superfície é

$\displaystyle\iint_{S}\sqrt{x^2+y^2}\,dS\\\\\\ =\iint_{S}f(x,\,y)\,dS\\\\\\=\int_0^{2\pi}\int_0^a g(r,\;\theta)\cdot\|\mathbf{X}_r\times\mathbf{X}_\theta \|\,dr\,d\theta\\\\\\ =\int_0^{2\pi}\int_0^a r\cdot\left(r\cdot \dfrac{\sqrt{b^2+a^2}}{a} \right)\,dr\,d\theta\\\\\\ =\dfrac{\sqrt{b^2+a^2}}{a}\cdot \int_0^{2\pi}\int_0^a r^2\,dr\,d\theta\\\\\\ =\dfrac{\sqrt{b^2+a^2}}{a}\cdot \int_0^{2\pi}\left.\left(\dfrac{r^3}{3} \right )\right|_0^a\,d\theta\\\\\\ =\dfrac{\sqrt{b^2+a^2}}{a}\cdot \int_0^{2\pi}\dfrac{a^3}{3}\,d\theta\\\\\\ =\dfrac{a^3}{3}\cdot \dfrac{\sqrt{b^2+a^2}}{a}\cdot \int_0^{2\pi}d\theta$

$=\displaystyle\dfrac{a^2}{3}\sqrt{b^2+a^2}\cdot \int_0^{2\pi}d\theta\\\\\\ =\dfrac{a^2}{3}\sqrt{b^2+a^2}\cdot (2\pi-0)\\\\\\ \therefore~~\boxed{\begin{array}{c} \displaystyle\iint_S\sqrt{x^2+y^2}\,dS=\dfrac{2\pi}{3}\,a^2\sqrt{b^2+a^2} \end{array}}$