# Surface integral $\iint_Y \textbf{F} \cdot \textbf{N} dS$ using Gauss theorem (divergence theorem)

Calculate the surface integer $$\iint_Y \textbf{F} \cdot \textbf{N} dS$$ Where Y: $$(x-z)^2+(y-z)^2=1-z^2$$ and $$0 \leq z \leq 1$$ (N is outward oriented). $$\textbf{F} = (y,x,1+x^2z)$$

If i shortly summarize what i have done:

I started with adding horizontal planes at $$z = 0$$ (call this plane B) and $$z = 1$$ (L), now the entire volume is restricted and we can use Gauss theorem.

$$\iint_Y \textbf{F} \cdot \textbf{N} dS + \iint_B \textbf{F} \cdot \textbf{N} dS + \iint_L \textbf{F} \cdot \textbf{N} dS = \iiint \text{div}\textbf{F}dxdydz$$

However, since we are orientating outwards we get that $$\textbf{N}_L = (0,0,1)$$ and $$\textbf{N}_B = (0,0,-1)$$, therefore we have that

$$\iint_Y \textbf{F} \cdot \textbf{N} dS = \iiint \text{div}\textbf{F}dxdydz$$ (Is this correct?)

From here i simply get div F$$= (0,0,x^2)$$ and then i calculate the tripple integer useing the following limits $$0 \leq z \leq 1 \Rightarrow 1 \leq x,y \leq 2$$ Giving me the final answer 6.

I dont have the key but I'm almost 100% sure I made atleast one mistake somewhere.

Thanks in advance.

## 1 Answer

Firstly, just a quick tip that you can type multiple integrals using \iint and \iiint

$$\iint \qquad \qquad \iiint$$

instead of doing \int\int or \int\int\int.

You have the right idea of trying to add in planes to make your domain closed in order to use the Divergence Theorem. However, note that at $$z=1$$ we have

$$(x-1)^2+(y-1)^2 = 0$$

which is just a single point $$(x,y)=(1,1)$$, so we don't actually need the $$z=1$$ plane. Indeed, we then have

$$\iint_Y \mathbf F \cdot \mathbf N \, dS + \iint_B \mathbf F \cdot \mathbf N \, dS = \iiint_R \nabla \cdot \mathbf F \, dV$$

However, $$B$$ is NOT the entire $$z=0$$ plane. It is only the portion of it that is inside $$Y$$. i.e.

$$B = \{ (x,y,z) \in \Bbb R^3 : x^2+y^2 \leq 1 \; , \; z = 0\}$$

$$R$$ is then the region bounded by $$Y$$ and $$B$$, i.e.

$$R = \{ (x,y,z) \in \Bbb R^3 : (x-z)^2+(y-z)^2 \leq 1-z^2 \; , \; 0 \leq z \leq 1 \}$$

I believe you know how to do the rest from this point onwards, but I shall leave you with the full solution anyways.

As you said, the outward unit normal at $$B$$ is $$(0,0,-1)$$, so

$$\iint_B \mathbf F \cdot \mathbf N \, dS = \iint_{x^2+y^2 \leq 1} \begin{pmatrix} y \\ x \\ 1\end{pmatrix} \cdot \begin{pmatrix} 0 \\ 0 \\ -1\end{pmatrix} \, dS = \iint_{x^2+y^2 \leq 1}(-1) \, dS = -\pi$$

As for the other integral, we need to parameterise $$R$$, which can be done using

$$x = z+r\cos \theta \quad y = z + r\sin \theta \quad z = z$$

where $$0\leq r \leq \sqrt{1-z^2}$$, $$0\leq \theta \leq 2\pi$$, $$0 \leq z \leq 1$$. The Jacobian is

$$J = \det \begin{pmatrix} \frac{\partial x}{\partial r} & \frac{\partial x}{\partial \theta} & \frac{\partial x}{\partial z} \\ \frac{\partial y}{\partial r} & \frac{\partial y}{\partial \theta} & \frac{\partial y}{\partial z} \\ \frac{\partial z}{\partial r} & \frac{\partial z}{\partial \theta} & \frac{\partial z}{\partial z} \end{pmatrix} = \det \begin{pmatrix} \cos \theta & -r\sin \theta & 1 \\ \sin\theta & r\cos \theta & 1 \\ 0 & 0 & 1 \end{pmatrix} = r$$

Moreover, we compute

$$\nabla \cdot \mathbf F = x^2$$

(note that taking the divergence of a vector gives you a scalar!). It follows that

$$\iiint_R \nabla \cdot \mathbf F \, dV = \int_{z=0}^{z=1}\int_{\theta=0}^{\theta=2\pi}\int_{r=0}^{r=\sqrt{1-z^2}}\big(z+r\cos\theta)^2 \cdot rdrd\theta dz = \frac{4\pi}{15}$$