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Currently I am studying vector calculus at my university, and I came across a question that I was having problem in solving. The question is this

Question

Evaluate $$\iint\limits_S \vec{A} \cdot \hat{n}\,dS$$ over the entire surface $S$ of the region bounded by the cylinder $x^2 +z^2 =9$, $x=0$, $y=0$, $y=8$, $z=0$, if $\vec{A}= 6z\,\hat{i}+(2x+y)\,\hat{j}-x\,\hat{k} $

The given answer is 18$\pi$.

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Solution

Approach 1: Surface Integrals


Sketch of the problem

For $S_1$, the curved surface projected over $xy$ plane

$$dS=\frac{dx\,dy}{\hat{n}\cdot\,\hat{k}}$$

and

$$\hat{n}=\frac{\nabla\cdot S}{|\nabla\cdot S|}=\frac{2x\,\hat{i}+2z\,\hat{k}}{\sqrt{4x^2+4z^2}}=\frac{2x\,\hat{i}+2z\,\hat{k}}{2\cdot3}=\frac{2x\,\hat{i}+2z\,\hat{k}}{6}=\frac{1}{3}x\,\hat{i}+\frac{1}{3}z\,\hat{k}$$

$$\hat{n}\cdot\hat{k}=\frac{1}{3}z$$

So,

$$dS=\frac{dx\,dy}{\frac{1}{3}z}=\frac{3\,dx\,dy}{z}$$

Now evaluating $\vec{A}\cdot \hat{n}$

$$\vec{A}\cdot \hat{n}=\left(6z\,\hat{i}+(2x+y)\,\hat{j}-x\,\hat{k}\right)\cdot\left(\frac{1}{3}x\,\hat{i}+\frac{1}{3}z\,\hat{k}\right) $$

$$\vec{A}\cdot \hat{n}=2zx-\frac{xz}{3}=\frac{5zx}{3}$$

$$S_1=\iint\limits_S \vec{A} \cdot \hat{n}\,dS=\frac{5xz}{3}\times \frac{3\,dy\,dx}{z}=\int\limits_0^8\int\limits_0^3 5x\, dx\, dy=180$$

$$S_1=180$$

For $S_2$ horizontal surface on $xy$ plane

$$\hat{n}=\hat{-k}$$ $$dS=dx\,dy$$ $$\vec{A}\cdot\hat{n}=x$$ $$S_2=\int\limits_0^8\int\limits_0^3 x\, dx\, dy=36$$ $$S_2=36$$

For $S_3$ flat surface on $yz$ plane

$$\hat{n}=\hat{-i}$$ $$dS=dy\,dz$$ $$\vec{A}\cdot\hat{n}=-6z$$ $$S_3=\int\limits_0^3\int\limits_0^8 6z\, dy\, dz=-216$$ $$S_3=-216$$

For $S_4$ flat 'quarter circle' surface on $xz$ plane

$$\hat{n}=\hat{-j}$$ $$dS=dx\,dz$$ $$\vec{A}\cdot\hat{n}=-(2x+y)$$ $$S_4=\int\limits_0^{\sqrt{9-x^2}}\int\limits_0^3 -(2x+y)\, dx\, dz$$

Here $y=0$

$$S_4=\int\limits_0^{\sqrt{9-x^2}}\int\limits_0^3 -2x\, dx\, dz$$ $$S_4=\int\limits_0^{\pi/2}\int\limits_0^3 -2r^2\cos{\theta} \, dr\, d\theta$$

Conversion to Polar coordinates

$dx\,dz=r\,dr\,d\theta$, $x=r\cos{\theta}$, $z=r\sin{\theta}$

$$S_4=\int\limits_0^{\pi/2}\int\limits_0^3 -2r^2\cos{\theta} \, dr\, d\theta$$

$$S_4=-18$$

For $S_5$ flat 'quarter circle' surface on $y=8$ plane

$$\hat{n}=\hat{j}$$ $$dS=dx\,dz$$ $$\vec{A}\cdot\hat{n}=(2x+y)$$ $$S_5=\int\limits_0^{\sqrt{9-x^2}}\int\limits_0^3 (2x+y)\, dx\, dz$$

Here $y=8$

$$S_5=\int\limits_0^{\sqrt{9-x^2}}\int\limits_0^3 (2x+8)\, dx\, dz$$

Conversion to Polar coordinates

$dx\,dz=r\,dr\,d\theta$, $x=r\cos{\theta}$, $z=r\sin{\theta}$

$$S_5=\int\limits_0^{\pi/2}\int\limits_0^3 (2r\cos{\theta}+8)\, r \, dr\, d\theta$$

$$S_5=\int\limits_0^{\pi/2}\int\limits_0^3 (2r^2\cos{\theta}+8r) \, dr\, d\theta$$

$$S_5=18(1+\pi)$$

Total

$$S=S_1+S_2+S_3+S_5=18\pi$$

Approach 2: Divergence Theorem


The Divergence Theorem

$$\iint\limits_S \vec{A}\cdot n\, dS=\iiint\limits_V \nabla\cdot\vec{A}\,\, dV$$

$$\nabla\cdot\vec{A}=1$$ $$dV=dx\,dy\,dz$$

$$S=\iiint_0^8\, dy\,dx\, dz=\iint 8\, dx\,dz$$

Conversion to Polar Coordinates

$$S=\int_0^{\pi/2}\int_0^3 8r\,dr\,d\theta=\int_0^{\pi/2}36\,d\theta=\left.36\,\theta\right|_0^{\pi/2}$$

$$S=18\pi$$

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  • $\begingroup$ Can u tell me a book which describes method 1 of yours $\endgroup$ – Aladdin May 6 '20 at 11:27

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