A vector calculus problem..

Question

I am a beginner in vector calculus . It will be great if someone can guide me solve this problem . Thanks. I don't know how to proceed. I am studying by online sources only so I have no teacher it will be great if someone solves this problem for me

Answer 1

There are easier ways to do this, but this generalizes

1. Parametrize the surface. For the sphere, the easiest parametrization is spherical coordinates $$g(\phi,\theta) = (\cos(\phi)\sin(\theta),\sin(\phi)\sin(\theta),\cos(\theta))$$ where $\theta$ is the polar angle and $\phi$ is the azimuthal angle. We only want the first octant so $0<\phi<\pi/2$ and $0<\theta < \pi/2.$

2. Compute the surface element, given by $$dS = \left|\frac{\partial g}{\partial \phi}\times\frac{\partial g}{\partial \theta}\right|d\theta d\phi$$ where $\times$ is the cross product and $||$ is the length of the vector. When you compute this you should get $dS = \sin(\theta)d\theta d\phi.$ You could also reason this with a good picture. Note that the formula I gave is completely general and works for any parametrized surface where you call the parameters $\theta$ and $\phi.$

3. Put everything in terms of the parameters and write down the integral. Here, the only thing in the integral is $x$ which we know is given by $x=\cos(\phi)\sin(\theta).$ So we can write $$ \int xdS = \int_0^{\pi/2}\int_0^{\pi/2}\left(\cos(\phi)\sin(\theta)\right) \sin(\theta) d\theta d\phi$$

4. Do the double integral.

Answer 2

First parametrize your surface $S$. One way to parametrize is given by spherical coordinates: $$\phi:[0,\frac{\pi}{2}]\times[0,\frac{\pi}{2}]\to\mathbb{R}^3,\quad \phi(s,t)=(\cos s\cos t, \cos s\sin t,\sin s).$$ Here the domain is $[0,\frac{\pi}{2}]\times[0,\frac{\pi}{2}]$ because we want all three coordinates of $\phi(s,t)$ to be non-negative, as $S$ lies in the first octant.

In this case we have $x=x(s,t)=\cos s\cos t.$ Then find the cross product of the partial derivatives of $\phi$: $$\frac{\partial\phi}{\partial s}\times\frac{\partial\phi}{\partial t},$$ Then by definition of surface integral $$\int x dS=\int_{[0,\frac{\pi}{2}]\times[0,\frac{\pi}{2}]}x(s,t)\left\|\frac{\partial\phi}{\partial s}\times\frac{\partial\phi}{\partial t}\right\|dsdt.$$

I left the detailed calculations for you to fill in.