Calculate the flux of F across G where $\mathbf F(x, y, z) = 6x\mathbf i + 6y\mathbf j + 2\mathbf k$; G is the surface cut from the bottom of the paraboloid $z = x^2 + y^2$ by the plane z = 3
I found n but I dont know exactly how to find the endpoints for the integration. Thanks
I would employ the divergence theorem here. https://en.wikipedia.org/wiki/Divergence_theorem
$$ \iiint_\Omega (\nabla\cdot\mathbf{F})dV=\iint_{\partial\Omega}(\mathbf{F}\cdot\mathbf{n})dS. \tag 1 $$
Here $\Omega$ is the volume bounded by paraboloid $z=x^2+y^2$ and plane $z=3$. $\partial \Omega$ is the boundary of $\Omega$. $\partial\Omega$ consists of $G$ and the circle cut from the plane $z=3$ by paraboloid (let us call this circle $\partial\Omega_2$). $$ \iint_{\partial\Omega}(\mathbf{F}\cdot\mathbf{n})dS=\iint_{G}(\mathbf{F}\cdot\mathbf{n})dS+\iint_{\partial\Omega_2}(\mathbf{F}\cdot\mathbf{n})dS $$ What you need is $\iint_{G}(\mathbf{F}\cdot\mathbf{n})dS$. $$ \iint_{G}(\mathbf{F}\cdot\mathbf{n})dS = \iint_{\partial\Omega}(\mathbf{F}\cdot\mathbf{n})dS - \iint_{\partial\Omega_2}(\mathbf{F}\cdot\mathbf{n})dS. $$ Using $(1)$: $$ \iint_{G}(\mathbf{F}\cdot\mathbf{n})dS =\iiint_\Omega (\nabla\cdot\mathbf{F})dV-\iint_{\partial\Omega_2}(\mathbf{F}\cdot\mathbf{n})dS.\tag 2 $$ Let us calculate $\iiint_\Omega (\nabla\cdot\mathbf{F})dV$. $$ \nabla\cdot \mathbf{F}\equiv\frac{\partial F_x}{\partial x}+\frac{\partial F_y}{\partial y}+\frac{\partial F_z}{\partial z}=\frac{\partial(6x)}{\partial x}+\frac{\partial (6y)}{\partial y}+\frac{\partial (2)}{\partial z}=6+6+0=12. $$ So $$ \iiint_\Omega (\nabla\cdot\mathbf{F})dV=\iiint_\Omega 12dV=12V_\Omega. $$ Here $V_\Omega$ is the volume of $\Omega$. $\Omega$ is the stack of "pancakes" each with the radius $\sqrt{z}$ and thickness $dz$. Here $z$ varies from $0$ to $3$. So: $$ V_\Omega=\int_0^3\pi(\sqrt{z})^2dz=\pi\int_0^3zdz=\pi\frac{z^2}{2}\bigg|_0^3=\frac{9\pi}{2}. $$ $$ \iiint_\Omega (\nabla\cdot\mathbf{F})dV=12V_\Omega=12\cdot\frac{9\pi}{2}=54\pi.\tag 3 $$ Let us calculate $\iint_{\partial\Omega_2}(\mathbf{F}\cdot\mathbf{n})dS$. $\mathbf{i}$ and $\mathbf{j}$ components of $\mathbf{F}$ are parallel to $\partial \Omega_2$, so their flux through $\partial\Omega_2$ is simply $0$. $\mathbf{k}$ component of $\mathbf{F}$ is orthogonal to $\partial\Omega_2$ at all points of $\partial\Omega_2$, so $(\mathbf{F}\cdot\mathbf{n})_{\partial\Omega_2}=2$ and: $$ \iint_{\partial\Omega_2}(\mathbf{F}\cdot\mathbf{n})dS=2S_{\partial\Omega_2}. $$ Here $S_{\partial\Omega_2}$ is the area of $\partial\Omega_2$. Since $\partial\Omega_2$ is the circle of $\sqrt{3}$ radius, its area is $S_{\partial\Omega_2}=\pi(\sqrt{3})^2=3\pi$. So: $$ \iint_{\partial\Omega_2}(\mathbf{F}\cdot\mathbf{n})dS=2S_{\partial\Omega_2}=2\cdot3\pi=6\pi.\tag 4 $$ Substituting $(3)$ and $(4)$ to $(2)$, we get the final answer: $$ \iint_{G}(\mathbf{F}\cdot\mathbf{n})dS=54\pi-6\pi=48\pi.\tag 5 $$ Please note, that the divergence theorem considers the "positive" normal to surface $\partial\Omega$ (and therefore to $G$) to be pointed outwards of $\Omega$. If you need $G$ to be oriented the other way, just change the sign in the final answer. In fact, the orientation of $G$ (i.e. "positive" normal to it) should be specified in the task.