Verify Stokes' Theorem for $v=zi+xj+yk$ over the hemispherical surface $x^2+y^2+z^2=1$ and $z \gt 0$.

Question

Verify Stokes theorem if $v=zi+xj+yk$ (where $i,j,k$ are the identity vectors for the $x,y,z$ axis) is taken over the hemispherical surface $x^2+y^2+z^2=1$ and $z \gt 0$.

Stokes theorem being: $$\int\limits_C \vec{F} \cdot d\vec{r} = \iint\limits_S \mathrm{curl}\ \vec{F} \cdot d\vec{S}$$

According to the back of my textbook, both sides of the equation come to $\pi$, and I am unable to get these answers on either side. For the LHS I get 0, and for the RHS, I also get 0. I am in need of more guidance because I believe that there is something I am missing. I will show my work below. Apologies in advance for y sloppy writing. Thanks.

First, $$\int\limits_C \vec{F} \cdot d\vec{r}$$ $r(t)=costi+sintj$ $=\int (zi+xj+yk)(cost i+sintj)dt = \int costsint dt$ $=1/4 cost2t$ from $0$ to $2\pi$ $=1-1=0$.

Now, $$\iint\limits_S \mathrm{curl}\ \vec{F} \cdot d\vec{S}$$ $\mathrm{curl}\ \vec{F}=i+j+k$ and $n=(2xi+2yj+2zk)/\sqrt{4x^2+4y^2+4z^2}=xi+yj+zk$ So, $\iint\limits_S \mathrm{curl}\ \vec{F} \cdot d\vec{S}= \int (x+y+z) dS= \int (cost + sint)dt = sint-cost$ where $0 \leq t \leq 2\pi$, which $=0$.

Answer 1

For the first integral, you forgot to take the derivative of $r$ inside the integral. If you fix that you will quickly find $\pi$ as expected.

For the second integral, your expression of $\vec{n}$ is correct but then you go wrong after that. You need to parametrize the hemisphere (e.g. with spherical coordinates) and then find the correct expression of $\vec{dS} = \vec{n}dS$ and $\vec{F}$ in these coordinates. You can try spherical coordinates $x = \sin \theta \cos \phi$, $y = \sin \theta \sin \phi$, $z = \cos \theta$ where $\theta \in [0, \pi/2]$ and $\phi \in [0,2\pi]$ (colatitude and azimuth). In this case $dS = (\sin \theta) d\theta d\phi$, and it is actually quick to see that the integral $\iint \mathrm{curl}(\vec{F}) \cdot \vec{dS} = \int_{\theta = 0}^{\pi/2} \int_{\phi=0}^{2\pi} \vec{F} \cdot \vec{n}\, dS$ is equal to $\pi$.

Answer 2

For the integral $\int_C \vec{F}\cdot\mathrm d\vec{r}$, you represent $C$ correctly as$$\vec{r}=\cos(t)\vec{i}+\sin(t)\vec{j}+ 0\vec{k}$$ but then you have$$\int_C(z\vec{i}+ x\vec{j}+ y\vec{k})(\cos(t)\vec{i}+\sin(t)\vec{j}+0\vec{k})\,\mathrm dt,$$which is wrong. With $\vec{r}=\cos(t)\vec{i}+\sin(t)\vec{j}+ 0\vec{k}$, $\mathrm d\vec{r}$ is the derivative of $\vec{r}$:$$\mathrm d\vec r= \{-\sin(t)\hat i+\cos(t)\hat j\}\,\mathrm dt.$$

Answer 3

is what I've done. I did not use spherical coordinates in the second integration. Since I took projection on xy plane, I used polar transformation. Because after projection, it gives circle on xy plane. So why am I not getting the answer.

Answer 4

Using spherical coordinate transformation.