How can solve this surface integral without stokes theorem?

Question

A 3D field vector field $F$ for a dynamic system is given as $$\vec F(x,y,z)=x\hat i+y\hat j+z^4\hat k$$ Find $$\iint \vec F\cdot d\vec S$$ where $S$ is the part of the cone $z=\sqrt{x^2+y^2}$ below the plane $z=1$ and is oriented downward.

My attempt:

Equation of the curve of intersection of surface and plane $z=1$ is $$1=x^2+y^2$$ whose parametric equation is $x=\cos t$ and $y=\sin t$ where, $0\le t\le 2\pi$. Now, using Stokes theorem $$\begin{align} \iint \vec F\cdot d\vec S&=\int_C \vec F\cdot d\vec r\\ &=\int_C (x, y, z^4)\cdot (dx, dy, dz)\\ &=\int_C (xdx+ ydy+z^4dz) \end{align}$$ plugging in, $x=\cos t, \ dx=-\sin t\ dt$, $y=\sin t, \ dy=\cos t\ dt$, $z=1, \ dz=0$, one should get $$\begin{align} &=\int_0^{2\pi} (\cos t (-\sin t)dt+\sin t(\cos t)dt+1^4 (o))\\ &=\int_0^{2\pi} (\cos t (-\sin t)dt+\sin t(\cos t)dt)\\ &=\int_0^{2\pi} (\sin t \cos t-\sin t\cos t )dt\\ &=0 \end{align}$$ I am not sure if I am correct. How, can I solve this using surface integral? Please help me solve this. Thanks

Answer 1

You cannot apply Stokes' theorem as you are not asked to find surface integral of the curl of a vector field. Stokes' theorem says -

$ \displaystyle \iint_S (\nabla \times \vec F) \cdot dS = \int_C \vec F \cdot dr $

Where $S$ is a surface with boundary $C$.

As you want to directly compute surface integral of vector field $\vec F$ over surface of the cone $z = \sqrt{x^2+y^2}, 0 \leq z \leq 1$,

The normal vector to the surface is,

$ \displaystyle \vec n = (f_x, f_y, -1) = \left(\frac{x}{\sqrt{x^2+y^2}}, \frac{y}{\sqrt{x^2+y^2}}, -1 \right)$

$ \displaystyle \vec F \cdot \vec n = \sqrt{x^2+y^2} - (x^2 + y^2)^2$

So the surface integral is,

$\displaystyle \iint_{x^2 + y^2 \leq 1} \left (\sqrt{x^2+y^2} - (x^2 + y^2)^2 \right) ~ dx ~ dy$

In polar coordinates,

$\displaystyle \int_0^{2\pi} \int_0^1 r \left(r - r^4 \right) ~ dr ~ d\theta = \frac{\pi}{3}$

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Alternatively, parametrize the surface as -

$\phi(r, \theta) = (r \cos\theta, r\sin\theta, r), 0 \leq r \leq 1, 0 \leq \theta \leq 2\pi$

$\phi_{\theta} \times \phi_r = (r\cos\theta, r\sin\theta, -r)$

$\vec F(\phi(r,\theta)) = (r\cos\theta, r\sin\theta, r^4)$

So the surface integral is,

$ \displaystyle \int_0^{2\pi} \int_0^1 (r\cos\theta, r\sin\theta, r^4) \cdot (r\cos\theta, r\sin\theta, -r) ~ dr ~ d\theta$

$ = \displaystyle \frac{\pi}{3}$

Answer 2

Your approach is not correct: Stokes' theorem is about the flux of $\text{curl}(\vec{F})$ (which in this case is identically zero) not of $\vec{F}$.

A direct computation: $$\begin{align} \iint_S \vec F\cdot d\vec S&=\iint_{x^2+y^2\leq 1}(x,y,(x^2+y^2)^2)\cdot\left(\frac{x}{\sqrt{x^2+y^2}},\frac{y}{\sqrt{x^2+y^2}},-1\right)dxdy\\ &=\iint_{x^2+y^2\leq 1}\left(\sqrt{x^2+y^2}-(x^2+y^2)^2\right)dxdy\\ &=2\pi\int_{0}^1(\rho-\rho^4)\rho d\rho=2\pi\left[\frac{\rho^3}{3}-\frac{\rho^6}{6}\right]_0^1=\frac{\pi}{3}. \end{align} $$ where $d\vec{S}=(f_x,f_y,-1)dxdy$ with $f(x,y)=\sqrt{x^2+y^2}$.

Another way, by using divergence theorem: $$\begin{align} \iint_S \vec F\cdot d\vec S&=\iiint_C \text{div}(\vec{F}) dxdydz- \iint_D \vec F\cdot d\vec S\\ &=\iiint_C (1+1+4z^3) dxdydz- \iint_D (x,y,1)\cdot (0,0,1) dxdy\\ &=2|C|+4\int_0^1 z^3\cdot \pi z^2 dz-|D|\\ &=\frac{2\pi}{3}+4\pi\left[\frac{z^6}{6}\right]_0^1-\pi=\frac{\pi}{3} \end{align} $$ where $C$ is the solid cone and $D$ is the disc $\{(x,y,1): x^2+y^2\leq 1\}$ upward oriented.