Evaluate $\int_{M}(F\cdot N)\,dS$

Question

Evaluate $\int_{M}(F\cdot N)\,dS$ where $F=(e^{\cos z},x^2,z)$ and $M=\{(x,y,z):z=1-(x^2+y^2),x^2+y^2\leq 1\}$

I try to use Stokes' theorem:

We will evaluate it on the curve $(\cos\theta,\sin\theta,0)$.

Therefore $$\int_0^{2\pi}(e,\cos^2\theta,0)\cdot (-\sin\theta,\cos\theta,0)d\theta =\int_0^{2\pi}-e\sin\theta+\cos^3\theta d\theta=0$$

Is it correct?

Answer 1

I was waiting for Matthew's solution, but since he hasn't posted, here's a quick answer for you.

You can parametrize the surface and compute the surface integral directly. Polar coordinates is the easiest way to go. Let $\vec x(r,\theta) = (r\cos\theta,r\sin\theta,1-r^2)$, $0\le r\le 1$, $0\le\theta\le 2\pi$. Then \begin{align*} \int_S F\cdot N\,dS &= \int_0^{2\pi}\int_0^1 (e^{1-r^2},r^2\cos^2\theta,1-r^2)\cdot\left(\frac{\partial\vec x}{\partial r}\times\frac{\partial\vec x}{\partial\theta}\right)dr\,d\theta\\&=\int_0^{2\pi}\int_0^1 (e^{1-r^2},r^2\cos^2\theta,1-r^2)\cdot r(2r\cos\theta,2r\sin\theta,1)\,dr\,d\theta \\ &= \int_0^{2\pi}\int_0^1 r\big(e^{1-r^2}2r\cos\theta+ 2r^3\cos^2\theta\sin\theta+1-r^2\big)dr\,d\theta \\ &=\pi/2. \end{align*} (Note it's easier to actually do the $\theta$ integral first, as the first two terms integrate to $0$ by symmetry.) I've assumed that the surface $S$ is oriented with $N$ pointing upward, but you didn't state so explicitly in your question.

Much easier is to apply the Divergence Theorem by closing up the surface with the disk $D$ on the bottom. Now, since $N$ points upward on the paraboloid, it must point downward on $D$. I will emphasize this by writing $D^-$ to denote the downward orientation. Then $$\int_S F\cdot N\,dS + \int_{D^-} F\cdot N\,dS = \int_V \text{div}\,F\,dV = \int_V 1\,dV.$$ This means that $$\int_S F\cdot N\,dS = \int_V 1\,dV + \int_{D^+} F\cdot N\,dS.$$ This makes sense. What enters the bottom disk plus the net sources inside $V$ tell you what leaves the paraboloid. In this case, $F\cdot N = z = 0$ on $D$, so we just get $$\int_S F\cdot N\,dS = \int_V 1\,dV = \int_0^{2\pi}\int_0^1\int_0^{1-r^2}r\,dz\,dr\,d\theta = \frac{\pi}2;$$ indeed, this is precisely the same integral we had in the direct computation.

Answer 2

Define $\mathbb{D}:=\{(x,y,z)\in\mathbb{R}^3|z=0,x^2+y^2\leq1\}$ and take $E$ as the solid $$E=\{(x,y,z)\in\mathbb{R}^3|0\leq z\leq1-x^2-y^2\}$$ From the divergence theorem we can say $$\int_{E}(\nabla\cdot F)dV=\int_M(F\cdot N)dS+\int_{\mathbb{D}}(F\cdot N)dS$$ However, since $$\nabla \cdot F=\frac{\partial}{\partial x}\bigg[e^{\cos(z)}\bigg]+\frac{\partial}{\partial y}\bigg[{x^2}\bigg]+\frac{\partial}{\partial z}\bigg[{z}\bigg]=1$$ and $$\int_E (\nabla \cdot F)dV=\int_{-1}^{1}\int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}}\int_{0}^{1-x^2-y^2}dV=\frac{\pi}{2}$$ we can say that $$\int_{M}(F\cdot N)dS=\frac{\pi}{2}-\int_{\mathbb{D}}(F\cdot N)dS$$ In order to calculate $\int_{\mathbb{D}}(F\cdot N)dS$ we can parameterize $\mathbb{D}$ on $[0,1]\times[0,2\pi)$ by $$\vec{r}(u,v):=(u\sin(v),u\cos(v),0)$$ Notice our parameterization of $\mathbb{D}$ induces an outward pointing normal vector of $E$ which is required in order to apply the divergence theorem. Moreover, $$\int_{\mathbb{D}}(F\cdot N)dS=\int_{0}^{2\pi}\int_{0}^{1}\big[(F\circ\vec{r})(u,v)\cdot (\vec{r}_u\times\vec{r}_v)(u,v)\big]dudv=\int_{0}^{2\pi}\int_{0}^{1}\big[(e,u^2\sin^2(v),0)\cdot (0,0,-u)\big]dudv=0$$ Finally $$\int_{M}(F\cdot N)dS=\frac{\pi}{2}$$