5
$\begingroup$

Given $\vec F=y\hat i+(x-2xz)\hat j-xy\hat k$ evaluate $$\iint_R(\nabla \times\vec F)\cdot \vec n dS $$ Where $S$ is surface represented by $x^2+y^2+z^2=a^2$ for $z\ge 0$

My attempt:

i found curl $$\nabla\times \vec F=\left|\begin{matrix} \hat i & \hat j & \hat k\\ \frac{\partial }{\partial x}&\frac{\partial }{\partial y}&\frac{\partial }{\partial z} \\ y& x-2xz&-xy\end{matrix}\right|=(x,y,-2z)$$ normal vector $$\vec n=\frac{\nabla g}{|\nabla g|}=\frac{(2x, 2y, 2z)}{2a}=(x/a, y/a, z/a)$$ used Stokes theorem andtaking projection on the XY plane: $z=0$ $$\iint_R(\nabla \times\vec F)\cdot \vec n dS=\iint (x,y,-2z)\cdot (x/a, y/a, z/a)dxdy$$ $$=\frac1a\int \int \left(x^2+y^2-2z^2\right)dxdy$$ putting limits of $x=-a$ to $x=a$ and $y=-\sqrt{a^2-x^2}$ to $y=\sqrt{a^2-x^2}$ and $z=0$ $$=\frac1a\int_{-a}^{a} \int_{\sqrt{a^2-x^2}}^{\sqrt{a^2-x^2}} \left(a^2-0\right)dxdy$$ $$=\pi a^3$$ I am not sure where I made mistake. please correct me and help solve this integration. Thanks

$\endgroup$
2
  • $\begingroup$ In the end you need to take $z=\sqrt{a^2-x^2-y^2}$ not zero. $\endgroup$ Feb 21, 2022 at 16:11
  • $\begingroup$ You have two errors. One is that $ds\ne dx dy$. The surface element is on the sphere, not in the $z=0$ plane. Then you mention Stokes, but I don't see where you are using it. It would transform your surface integral into a line integral. $\endgroup$
    – Andrei
    Feb 21, 2022 at 16:12

2 Answers 2

5
$\begingroup$

You can use Stokes' theorem twice: $$I=\iint_S(\nabla\times \vec F)\cdot \vec n dS=\oint_B\vec F\cdot d\vec l=\iint_{S_1}(\nabla\times \vec F)\cdot \vec n_1 dS_1$$ Here $S$ and $S_1$ share the same boundary $B$. In your problem $B$ is the circle of radius $a$, centered in the origin, in the $z=0$ plane. If you want $S_1$ to be the disk of radius $a$, centered in the origin, in the $z=0$ plane, then $\vec n_1=\hat k$. Then $$I=\iint_{S_1}(x,y,0)\cdot(0,0,1)dS_1=0$$

$\endgroup$
5
$\begingroup$

As a direct computation your flux should be $$\iint_S(\nabla \times\vec F)\cdot \vec n dS=\iint_{x^2+y^2\leq a^2} (x,y,-2f)\cdot (x/a, y/a, f/a)\sqrt{1+f_x^2+f_y^2}dxdy$$ where $z=f(x,y)=\sqrt{a^2-x^2-y^2}$.

Another way (easier): by applying Stokes' Theorem, we have $$\iint_S(\nabla \times\vec F)\cdot \vec n dS=\int_{\gamma}\vec F\cdot d\vec s=\int_0^{2\pi}(y(t)x'(t)+x(t)y'(t)+0)dt=[x(t)y(t)]_0^{2\pi}=0$$ where $\gamma$ is the circle $x^2+y^2=a^2$ in the plane $z=0$ counterclockwise oriented, i.e. $\gamma(t)=(a\cos(t),a\sin(t),0)$ with $t\in [0,2\pi]$.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .