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Let $S$ be the surface of the cone $z=2-\sqrt{x^2 +y^2}$ above $xy$-plane. Evaluate $\iint_S \mathrm{curl}\space\mathbf F \cdot d\mathbf S$ where the vector field $\mathbf F$ is defined by:

$$\mathbf F=(x-z)\space\mathbf i + (x^3+yz)\space\mathbf j - (3xy^2)\space \mathbf k$$

I tried to let $x=r \cos\theta$, $y =r \sin\theta$, but I am confused on integration.

If I let $x=r \cos\theta$, $y =r \sin\theta$, my parametric equation will be $\vec{r}(r,\theta)=<r\cos\theta,r\sin\theta,2-r>$, but now the problem is about $\text{curl }\mathbf F$.

I change the $x, y$ in $\text{curl }\mathbf F$ also into $(r,\theta)$ form, but when I multiply $\text{curl }\mathbf F$ with $\vec{r}(r)\times \vec{r}(\theta)$, there is where I confused. It is hard to integral. Is that I make some mistakes?

I need help, thanks.

coreyman317
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murtabak
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  • Welcome to MSE. Your question is phrased as an isolated problem, without any further information or context. This does not match many users' quality standards, so it may attract downvotes, or be closed. To prevent that, please [edit] the question. This will help you recognize and resolve the issues. Concretely: please provide context, and include your work and thoughts on the problem. These changes can help in formulating more appropriate answers. – José Carlos Santos Dec 12 '22 at 07:01

2 Answers2

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but when I multiply $\text{curl }\mathbf F$ with $\vec{r}(r)\times\vec r(\theta)$, there is where I confused. It is hard to integral. Is that I make some mistakes?

No, assuming you did your algebra correctly there, this is a correct way to find the surface integral $\iint_S \mathrm{curl}\space\mathbf F \cdot d\mathbf S$. However, it may be a difficult integral to solve in that form. Because of the presence of the $\text{curl }\mathbf F$, you should recall Stokes' Theorem:

If $S$ is an oriented smooth surface that is bounded by a closed boundary curve $C$ with positive orientation, then $$\iint_S\text{curl }\mathbf F \cdot d\mathbf S=\int_C\mathbf F\cdot d\mathbf r$$

Thus to evaluate your surface integral $\iint_S\text{curl }\mathbf F \cdot d\mathbf S$, we may instead evaluate the line integral $\int_C\mathbf F\cdot d\mathbf r$ where $C$ is the circle $x^2+y^2=2^2$ in the $xy$-plane centered at the origin.

Anne Bauval
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coreyman317
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I would understand this to be a surface integral of the flux of the curl through the cone you described (not including the flat bottom of the cone in the x, y plane). This is an integral subject to Stoke's theorem. You can replace it with the work of the original field around the boundary curve, which is a circle in the x, y plane. When z = 0, the radius of the circle is 2.

$$\iint_{S} \mathrm{curl} \ \vec{F} \cdot \mathrm dS = \int_{C} \vec{F} \cdot\ \mathrm d\vec{r}$$.

Our curve $\vec{r}$ is parameterized as $\vec{r} = 2\left<\cos\theta, \sin\theta, 0\right>$ and these spatial coordinates are put into the field as $\vec{F} = \left<2\cos\theta, 8\cos^{3}\theta,24\cos\theta\sin\theta\right>$. The differential of $\vec{r}$ is $d\vec{r}=2\left<-\sin\theta,\cos\theta,0\right>d\theta$. If I did all my math right, you get

$$\int_{0}^{2\pi} \left( -4\sin\theta\cos\theta+16\cos^{4}\theta\right)\mathrm d\theta$$

The first term integrates to $0$. The second can be integrated as

$$16\int_{0}^{2\pi} \left( \cos^{2}\theta-(\cos\theta\sin\theta)^{2}\right) \mathrm d \theta$$

The first term should integrate to $\pi$ and the second to $\frac{\pi}{4}$, for a total of $16\pi - 4\pi = 12\pi$

Poisson Aerohead
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