# Integral without Divergence Theorem

I wish to solve the following integral(Ans. $108\pi$)
$$\iint_S \vec F\cdot \hat{n}dS$$ Where
$$\vec F=4xz\hat{i}+xyz^2\hat{j}+3z\hat{k}$$
and $S$ is the surface of the region above the $xy$ plane bounded by the cone
$$z^2=x^2+y^2$$
and the plane $z=3$
My attempt:
Everything in Spherical Coordinates,
Here,
$$r\in[0,3\sqrt{2}]$$
$$\theta=\frac{\pi}{4}$$
$$\phi\in[0,2\pi]$$
$$dS=r\sin(\theta)d\phi dr=\frac{r}{\sqrt{2}}d\phi dr$$
$$\vec F=4r^2\sin(\theta)\cos(\phi)\cos(\theta)\hat{i}+r^4\sin^2(\theta)\cos(\phi)\sin(\phi)\cos^2(\theta)\hat{j}+3r\cos(\theta)\hat{k}$$
The $\hat{n}$ can be calculated using gradient
$$\vec{\nabla}(x^2+y^2-z^2)=2(x\hat{i}+y\hat{j}-z\hat{k})$$
$$|\vec{\nabla}(x^2+y^2-z^2)|=\sqrt{4(x^2+y^2+z^2)}=2r$$
$$\hat{n}=\frac{\vec{\nabla}(x^2+y^2-z^2)}{|\vec{\nabla}(x^2+y^2-z^2)|}=\frac{x}{r}\hat{i}+\frac{y}{r}\hat{j}-\frac{z}{r}\hat{k}=\sin(\theta)\cos(\phi)\hat{i}+\sin(\theta)\sin(\phi)\hat{j}-\cos(\theta)\hat{k}$$
Now,
$$\vec{F}\cdot\hat{n}=4r^2\sin^2(\theta)\cos^2(\phi)\cos(\theta)+r^4\sin^3(\theta)\cos(\phi)\sin^2(\phi)\cos^2(\theta)-3r\cos^2(\theta)$$
$$\vec{F}\cdot\hat{n}=\frac{4r^2\cos^2(\phi)}{(\sqrt{2})^3}+\frac{r^4\cos(\phi)\sin^2(\phi)}{(\sqrt{2})^5}-\frac{3r}{2}$$
$$\vec{F}\cdot\hat{n}dS=(r^3\cos^2(\phi)+\frac{r^5\cos(\phi)\sin^2(\phi)}{8}-\frac{3r^2}{2\sqrt{2}})drd\phi$$
$$\iint_S \vec F\cdot \hat{n}dS=\int_{0}^{2\pi}\int_{0}^{3\sqrt{2}}(r^3\cos^2(\phi)+\frac{r^5\cos(\phi)\sin^2(\phi)}{8}-\frac{3r^2}{2\sqrt{2}})drd\phi$$
$$\iint_S \vec F\cdot \hat{n}dS=\int_{0}^{2\pi}[\frac{r^4\cos^2(\phi)}{4}+\frac{r^6\cos(\phi)\sin^2(\phi)}{48}-\frac{r^3}{2\sqrt{2}}]_{0}^{3\sqrt{2}}d\phi$$
$$\iint_S \vec F\cdot \hat{n}dS=\int_{0}^{2\pi}[81\cos^2(\phi)+\frac{3^6\cos(\phi)\sin^2(\phi)}{6}-27]d\phi$$
$$\iint_S \vec F\cdot \hat{n}dS=81\int_{0}^{2\pi}\cos^2{\phi}d\phi+\frac{3^6}{6}\int_{0}^{2\pi}\cos(\phi)\sin^2(\phi)d\phi-27\int_{0}^{2\pi}d\phi$$
$$\iint_S \vec F\cdot \hat{n}dS=81\pi+0-54\pi=27\pi$$ What is wrong ??

EDIT: It looks like I forgot the top cap !! (Thanks to StackTD) For the upper cap we have
$$z=3$$
$$\hat{n}=\hat{k}$$
$$\vec{F}.\hat{n}=\vec{F}.\hat{k}=3z=9$$
$$\iint_S\vec{F}.\hat{n}dS=\iint_S9dS=9\iint_SdS=9\times area=9\times\pi(3)^2=81\pi$$

Total
$$27\pi+81\pi=108\pi$$

• Cylindrical coordinates looks to be a better choice here. – Paul Jul 14 '17 at 14:08
• There are 2 surfaces which have to be integrated separately also. – Paul Jul 14 '17 at 14:09
• Which is the second surface? Isn't this the only surface? – Nishant Garg Jul 14 '17 at 14:10
• The top of the surface, i.e. the "cap" of the cone, given by the plane $z=3$. At least, if $S$ is supposed to be this closed surface (that would lead to $108\pi$). – StackTD Jul 14 '17 at 14:16
• Ah, thanks!!! I solved it because of your help @StackTD – Nishant Garg Jul 14 '17 at 14:30