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Let $S$ be that part of the surface of the paraboloid $z=x^2+y^2$ between the planes $z=1$ and $z=4$.

Now given $\vec{V}=x^3\hat j+z^3\hat k$ and evaluate the line integrals $\int_{C}{_1}\vec{V}.dr+\int_{C}{_2}\vec{V}.dr$ where $C_1$ and $C_2$ are the curves bounding $S$

I know the answer should be $\frac{45\pi}{4}$ since Let $\vec V=\hat y x^3+\hat z z^3$. Then, $\nabla \times \vec V=3\hat z x^2$.

A vector point on the surface can be written as $\vec r=\hat \rho\rho +\hat z\rho^2$, where $\hat \rho=\hat x\cos(\phi)+\hat y\sin(\phi)$.

So, the surface element is $\hat n\,dS=\left(\frac{\partial \vec r}{\partial \rho}\times\frac{\partial \vec r}{\partial \phi}\right)\,d\rho\,d\phi=\left(-2\hat\rho\rho^2+ \hat z \rho\right)\,d\rho\,d\phi$.

Therefore,

$$\int_S\nabla\times \vec V\cdot \hat n\,dS=\int_0^{2\pi}\int_1^2 3\rho^3\cos^2(\phi)\,d\rho\,d\phi=45\pi/4$$

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For this circles $dz=0$. The first integral $$\int_{C_{1}}{x^{3}}dy$$ Let $x=cos(t)$ and $y=sin(t)$ (which indeed parametrizes that circle), then $$\int_{C_{1}}{x^{3}}dy=\int_{2\pi}^{0}cos^{4}(t)dt=-\frac{3\pi}{4}$$ The second integral is exactly the same except the parametarization is $x=2cos(t)$ and $y=2\sin(t)$ $$\int_{C}{x^{3}}dy=\int_{C_{1}}{x^{3}}dy+\int_{C_{2}}{x^{3}}dy=\int_{2\pi}^{0}cos^{4}(t)dt+16\int_{0}^{2\pi}cos^{4}(t)dt$$

$$\int_{C}{x^{3}}dy=15\int_{0}^{2\pi}cos^{4}(t)dt=\frac{45\pi}{4}$$

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  • $\begingroup$ i first evaluated the surface integral $\iint_S \nabla \times V.ndS$ and calculated it as $\frac{45\pi}{4}$ and i'm assuming if the surface integral is equal to this then it is safe to say that the line integrals should also turn out to be $\frac{45\pi}{4}$ due to stokes' theorem. $\endgroup$ – user395952 Mar 8 '17 at 14:12
  • $\begingroup$ should it be $cos^3(t)$ since it is $x^3$? $\endgroup$ – user395952 Mar 8 '17 at 14:13
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    $\begingroup$ No, you have $dy=cos{(t)}dt$, so $x^{3}dy=cos^4(t)dt$ $\endgroup$ – Kiryl Pesotski Mar 8 '17 at 14:14
  • $\begingroup$ oh okay, could you maybe evaluate the surface integral, because if i've made a mistake somewhere i dont understand where. $\endgroup$ – user395952 Mar 8 '17 at 14:15
  • $\begingroup$ I still get 51. Maybe show your solution? $\endgroup$ – Kiryl Pesotski Mar 8 '17 at 14:27
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$$\int_C \textbf{V} \ d\textbf{r} = \int_{a}^b \left(\textbf{V} \circ \gamma\right)(t) \cdot \gamma'(t) \ dt$$

where $\gamma: I \to \mathbb{R}^n$ parametrizes $C$ on $[a,b]$. All you have to do is parametrize the curve given by $x^2+y^2 = 1$ and $x^2+y^2 = 4$ which are circles.

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  • $\begingroup$ could we let $x=rcos(\theta)$ and $y=rsin(\theta)$ $\endgroup$ – user395952 Mar 8 '17 at 14:01
  • $\begingroup$ Yes. You can use that parametrization, but you have to add a $z$-coordinate which gives the location of the circle. $\endgroup$ – Faraad Armwood Mar 8 '17 at 14:04
  • $\begingroup$ so could we set $z=r$ which would be the radius's of each circle? $\endgroup$ – user395952 Mar 8 '17 at 14:07
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    $\begingroup$ Yup! That is correct. I'm glad you figured it out by yourself. I think that way is more rewarding. $\endgroup$ – Faraad Armwood Mar 8 '17 at 14:15
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    $\begingroup$ ur Stokes have given ur the wrong result, (i guess) cause u have chosen the wrong orientation of the normal. $\endgroup$ – Kiryl Pesotski Mar 8 '17 at 23:01

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