MHB Calculation of a surface integral

[mathmari]

Hey!

I want to calculate the surface integral of $$F(x,y,z)=(0,0,z)$$ on the unit sphere with parametrization
$$x=\sin u \cos v, \ y=\sin u \sin v , \ z=\cos u \\ 0\leq u\leq \pi, \ 0\leq v\leq 2\pi$$
with positive direction the direction of $T_u\times T_v$. Could you give some hints how to calculate this? (Wondering)

I haven't really understood how we use the fact that the positive direction is the direction of $T_u\times T_v$...

Do we maybe calculate the following?

$$\int_{\Phi}F\cdot ds=\int_SF\cdot T_u\times T_v \ dvdu$$ ? (Wondering)

 

[I like Serena]

Hey!

I want to calculate the surface integral of $$F(x,y,z)=(0,0,z)$$ on the unit sphere with parametrization
$$x=\sin u \cos v, \ y=\sin u \sin v , \ z=\cos u \\ 0\leq u\leq \pi, \ 0\leq v\leq 2\pi$$
with positive direction the direction of $T_u\times T_v$. Could you give some hints how to calculate this? (Wondering)

I haven't really understood how we use the fact that the positive direction is the direction of $T_u\times T_v$...

Hey mathmari! (Smile)

The orientation of a surface is the direction that we give to the normal vector, meaning we have 2 choices.
When we pick $T_u\times T_v$ we get a normal vector that has a specific direction, which serves to identify the orientation. (Nerd)

Do we maybe calculate the following?

$$\int_{\Phi}F\cdot ds=\int_SF\cdot T_u\times T_v \ dvdu$$ ? (Wondering)

Yep! (Nod)

 

[mathmari]

The orientation of a surface is the direction that we give to the normal vector, meaning we have 2 choices.
When we pick $T_u\times T_v$ we get a normal vector that has a specific direction, which serves to identify the orientation. (Nerd)

Ah... I see... (Thinking) Is the formula of $T$ the following? $$T(u,v)=(\sin u \cos v, \sin u \sin v , \cos u)$$ ? (Wondering)

 

[I like Serena]

Ah... I see... (Thinking) Is the formula of $T$ the following? $$T(u,v)=(\sin u \cos v, \sin u \sin v , \cos u)$$ ? (Wondering)

Yes. (Nod)

 

[mathmari]

So, we have the following:

$$T_u=(\cos u \cos v, \cos u \sin v , -\sin u) \\ T_v=(-\sin u \sin v, \sin u \cos v , 0)$$

Therefore, $$T_u\times T_v=\left (\cos v \sin^2u, \sin^2u \sin v, \cos u \sin u\right )$$

So, $$\int_{\Phi}F\cdot ds=\int_SF\cdot T_u\times T_v \ dvdu=\int_0^{\pi} \int_0^{2\pi} \cos^2 u \sin u \ dvdu=2\pi \int_0^{\pi} \cos^2 u \sin u \ du=2\pi \left [-\frac{\cos^3u}{3}\right ]_0^{\pi}=\frac{4\pi}{3}$$

Is this correct? (Wondering)

 

[I like Serena]

So, $$\int_{\Phi}F\cdot ds=\int_SF\cdot T_u\times T_v \ dvdu=\int_0^{\pi} \int_0^{2\pi} \cos^2 u \sin u \ dvdu=2\pi \int_0^{\pi} \cos^2 u \sin u \ du=2\pi \left [-\frac{\cos^3u}{3}\right ]_0^{\pi}=\frac{4\pi}{3}$$

Is this correct? (Wondering)

Let's see... (Thinking)
... according to Gauss's theorem we have:
$$\oint_{\Phi}F\cdot ds = \int_{V(\Phi)} \nabla \cdot F dV = \int_{V(\Phi)} dV = \frac 43\pi
$$
Yes! I think it is correct! (Happy)

 

[mathmari]

Yes! I think it is correct! (Happy)

Great! Thanks a lot! (Happy)