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If $F = 6z\mathbf i + (2x+y)\mathbf j -x\mathbf k$, evaluate $\int_S \mathbf F \cdot \mathbf n ds$ over the surface bounded by the cylinder $x^2 + z^2 = 9. x=0, y=0, z=0 $ and $ y= 6$

Okay, so I know this is a quarter cylinder in the first octant. Now assuming the surface they're talking about is $z = \sqrt{ 9 - x^2 } $, here's what I've got so far.

$ n = \nabla f $ where $f(x,y,z)= x^2 + z^2 - 9 = 0$ which yields $<2x, 0, 2z>$

now for $\hat n$ $$ \hat n = \frac {<2x, 0, 2z>} {\sqrt {4x^2 + 4z^2 }}$$

and do get $ds$, I used $g(x,y) = \sqrt{ (9 - x^2) } $ where $g_x = \frac {-x}{\sqrt{9-x^2}}$ and $g_y= 0 $

therefore $$ds = \sqrt{ \frac {9}{9-x^2} } = \frac {3}{\sqrt{ {9-x^2} }} dA$$

Finally $$\int_S\mathbf F\cdot \hat n ds = <6z, 2x+4, -x> \cdot \frac {<2x, 0, 2z>} {\sqrt {4x^2 + 4z^2 }}\frac {3}{\sqrt{ {9-x^2} } } $$

$$ = \int\int_R \frac {10x \sqrt{ {9-x^2} } } {\sqrt {4x^2 + 4z^2 }}\frac {3}{\sqrt{ {9-x^2} } }dA $$ OR

$$ = \int\int_R \frac {15x } {\sqrt {x^2 + z^2 } }dxdy = \int_0^6\int_0^3 5x dx dy$$

My question is, is this setup correct or is any of this wrong?

Dimitri
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1 Answers1

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For a direct computation, note that your surface is the union of the five surfaces parameterized by \begin{align*} \mathbf r(x,y)&=\left\langle x,y,\sqrt{9-x^2}\right\rangle & (x,y)&\in[0,3]\times[0,6] \\ \mathbf s(x,y) &= \langle x,y,0\rangle & (x,y)&\in[0,3]\times[0,6] \\ \mathbf t(y,z) &=\langle0,y,z\rangle & (y,z)&\in[0,6]\times[0,3] \\ \mathbf u(r,\theta) &= \langle r\cos\theta,0,r\sin\theta\rangle & (r,\theta)&\in[0,3]\times[0,\pi/2] \\ \mathbf v(r,\theta) &= \langle r\cos\theta,6,r\sin\theta\rangle & (r,\theta)&\in[0,3]\times[0,\pi/2] \end{align*} Our computation is then \begin{align*} \iint_S(\mathbf F\cdot\mathbf n)\,dS &= \int_0^6\int_0^3\mathbf F\bigl(\mathbf r(x,y)\bigr)\cdot(\mathbf r_x\times\mathbf r_y)\,dx\,dy \\ &\qquad+\int_0^6\int_0^3\mathbf F\bigl(\mathbf s(x,y)\bigr)\cdot(\mathbf s_y\times\mathbf s_x)\,dx\,dy \\ &\qquad\qquad+\int_0^3\int_0^6\mathbf F\bigl(\mathbf t(y,z)\bigr)\cdot(\mathbf t_z\times\mathbf t_y)\,dy\,dz \\ &\qquad\qquad\qquad+\int_0^{\pi/2}\int_0^3\mathbf F\bigl(\mathbf u(r,\theta)\bigr)\cdot(\mathbf u_r\times\mathbf u_\theta)\,dr\,d\theta \\ &\qquad\qquad\qquad\qquad+\int_0^{\pi/2}\int_0^3\mathbf F\bigl(\mathbf v(r,\theta)\bigr)\cdot(\mathbf v_\theta\times\mathbf u_r)\,dr\,d\theta \\ &= \int_0^6\int_0^3\left\langle6\sqrt{9-x^2},2\,x+y,-x\right\rangle\cdot\left\langle \frac{x}{\sqrt{9-x^2}},0,1\right\rangle\,dx\,dy \\ &\qquad+\int_0^6\int_0^3\langle0,2\,x+y,-x\rangle\cdot\langle0,0,-1\rangle\,dx\,dy \\ &\qquad\qquad+\int_0^3\int_0^6\langle6\,z,y,0\rangle\cdot\langle-1,0,0\rangle\,dy\,dz \\ &\qquad\qquad\qquad+\int_0^{\pi/2}\int_0^{3}\langle6r\sin\theta,2r\cos\theta,-r\cos\theta\rangle\cdot\langle0,-r,0\rangle\,dr\,d\theta \\ &\qquad\qquad\qquad\qquad+\int_0^{\pi/2}\int_0^{3}\langle6r\sin\theta,2r\cos\theta+6,-r\cos\theta\rangle\cdot\langle0,r,0\rangle\,dr\,d\theta \\ &= \underbrace{5\int_0^6\int_0^3x\,dx\,dy+\int_0^6\int_0^3x\,dx\,dy-6\int_0^3\int_0^6z\,dy\,dz}_{=0} \\ &\qquad+\int_0^{\pi/2}\int_0^3-2r^2\cos\theta\,dr\,d\theta+\int_0^{\pi/2}\int_0^3(2 r^2\cos\theta+6r)\,dr\,d\theta \\ &= 6\int_0^{\pi/2}\int_0^3r\,dr\,d\theta \\ &= 6\left(\frac{9}{2}\right)\int_0^{\pi/2}d\theta \\ &= 6\left(\frac{9}{2}\right)\left(\frac{\pi}{2}\right) \\ &=\frac{27}{2}\pi \end{align*}

The problem is considerably easier if we use the divergence theorem. We are asked to compute $$ \iint_{S}(\mathbf F\cdot\mathbf n)\,dS $$ where $\mathbf F(x,y,z)=\langle 6\,z,2\,x+y,-x\rangle$ and $S$ is the quarter cylinder you describe. The divergence theorem states that $$ \iint_{S}(\mathbf F\cdot\mathbf n)\,dS=\iiint_T(\nabla\cdot\mathbf F)\,dV $$ where $T$ is the solid enclosed by $S$ and $\displaystyle\nabla=\left\langle\frac{\partial}{\partial x},\frac{\partial}{\partial y},\frac{\partial}{\partial z}\right\rangle$. Our computation is then \begin{align*} \iint_{S}(\mathbf F\cdot\mathbf n)\,dS &=\iiint_T(\nabla\cdot\mathbf F)\,dV \\ &=\iiint_T\left\langle\frac{\partial}{\partial x},\frac{\partial}{\partial y},\frac{\partial}{\partial z}\right\rangle\cdot\langle 6\,z,2\,x+y,-x\rangle\,dV \\ &= \iiint_T(0+1+0)\,dV \\ &= \iiint_TdV \\ &= \text{volume of }T \\ &= \frac{1}{4}\pi\cdot 3^2\cdot 6 \\ &= \frac{27}{2}\pi \end{align*} This verifies that our direct computation was, indeed, correct.

Brian Fitzpatrick
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    I have studied divergence theorem, but the question asked me not to use divergence.theorem, though it would be a lot easier. – Dimitri Jun 21 '15 at 20:02
  • In fact, my setup gives 405 as an answer, which seems to be wrong in my case... – Dimitri Jun 21 '15 at 20:05
  • @Dimitri I updated my answer to better address your question. – Brian Fitzpatrick Jun 21 '15 at 21:45
  • I was expecting it to be all five surfaces but I was not expecting it to be a question to be asked on a final exam though. I thought I misinterpreted the question entirely when it asked "over the surface bounded by..." which I assumed was the surface pointing outwards in the z-axis (what I wrote above). This is very tedious, especially for a final exam. – Dimitri Jun 21 '15 at 21:52
  • @Dimitri Perhaps I mis-interpreted the surface we are working with. If the surface is just the part described by $\mathbf r(x,y)$, then the answer would be $5\int_0^6\int_0^3x,dx,dy=135$. – Brian Fitzpatrick Jun 21 '15 at 21:59
  • Yes, which is the setup I had (I had forgotten to divide by 3) if that is really the surface, which answers my question. However, this question itself I find ambiguous as to what surface they're referring to... – Dimitri Jun 21 '15 at 22:03
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    @Dimitri In other words, your computation is correct assuming that we only care about the "top part" of the region I described in the answer. – Brian Fitzpatrick Jun 21 '15 at 22:03