Surface area of the upper hemisphere

Homework Statement

For upper hemisphere S: x^2 + y^2 + z^2 = 1 , with z≥0, find the area element dS and unit normal vector N. Compute the total area of the hemisphere, ∫∫dS over S.

Homework Equations

Unit normal to surface f(x,y,z) = const

Surface area element of surface z = f(x,y)
dS = sqrt(1 + (df/dx)^2 + (df/dy)^2)*dx*dy

The Attempt at a Solution

N(hat) = (x/sqrt(x^2+y^2+z^2), y/sqrt(x^2+y^2+z^2), z/sqrt(x^2+y^2+z^2))
dS = dx*dy/sqrt(1-x^2-y^2)

So, I computed the gradient of f(x,y,z) and normalized it to get N(hat). I also used z = +sqrt(1-x^2-y^2) to find dS. Now that I have these things I do not know how to use them to integrate over the surface. I have a feeling I need to use Stokes Theorem to solve this, but I don't know how. Am I even on the right track? Thanks.

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vela
Staff Emeritus
Homework Helper

Homework Statement

For upper hemisphere S: x^2 + y^2 + z^2 = 1 , with z≥0, find the area element dS and unit normal vector N. Compute the total area of the hemisphere, ∫∫dS over S.

Homework Equations

Unit normal to surface f(x,y,z) = const

Surface area element of surface z = f(x,y)
dS = sqrt(1 + (df/dx)^2 + (df/dy)^2)*dx*dy

The Attempt at a Solution

N(hat) = (x/sqrt(x^2+y^2+z^2), y/sqrt(x^2+y^2+z^2), z/sqrt(x^2+y^2+z^2))
dS = dx*dy/sqrt(1-x^2-y^2)

So, I computed the gradient of f(x,y,z) and normalized it to get N(hat). I also used z = +sqrt(1-x^2-y^2) to find dS. Now that I have these things I do not know how to use them to integrate over the surface. I have a feeling I need to use Stokes Theorem to solve this, but I don't know how. Am I even on the right track? Thanks.
You have z=f(x,y). What region in the xy-plane corresponds to the hemisphere? That's the region you want to integrate over to calculate the surface area S = ∫ dS.

The unit circle corresponds to the hemisphere. I'm aware I would use this curve on the left side of Stokes theorem $\oint$ F*dr over C = ∫∫div(F)*N(hat)*dS over S, but I don't know what F is, or if the same N(hat) and dS above are those that I found originally. Do I use them here?

vela
Staff Emeritus
Homework Helper
No, you just want to evaluate
$$\int \frac{1}{\sqrt{1-x^2-y^2}}\,dx\,dy$$ with the appropriate limits.

I found the answer to be 2*pi. I parameterized the sphere via spherical coordinates, x = cos(t)sin(s) y = sin(t)sin(s) and z = cos(s). This is a stretching of the (s,t) domain of (0<t<2*pi)X(0<s<pi) into the whole sphere.

Using |Tt X Ts|, where Tt and Ts are tangent vectors on the surface. Taking their cross product provides a normal vector to the surface.

You find Tt with Tt = [dx/dt(t,s), dy/dt(t,s), dz/dt(t,s)] and similarly for Ts. Thus, you have your tangent vectors in terms of your parameters t and s, and can integrate ∫∫|Tt X Ts|dtds.
|Tt X Ts| reduces to sin(s), and the limits are the boundaries of the aforementioned domain. An easy integral to 4*pi. Taking half of this gives you half the sphere SA = 2*pi.

The problem is that I don't know if there was an easier way to do this via Stokes or perhaps Divergence theorem, or a better way to conceptualize this.

vela
Staff Emeritus
Homework Helper
The unit circle corresponds to the hemisphere. I'm aware I would use this curve on the left side of Stokes theorem $\oint$ F*dr over C = ∫∫div(F)*N(hat)*dS over S, but I don't know what F is, or if the same N(hat) and dS above are those that I found originally. Do I use them here?
Stokes theorem has the curl on the RHS:
$$\oint_C \vec{F}\cdot d\vec{r} = \int_S (\nabla \times \vec{F})\cdot \hat{n}\,dS$$ You might be confusing it with the divergence theorem:
$$\int_V (\nabla \cdot \vec{F})\,dV = \oint_S \vec{F} \cdot \hat{n}\,dS$$In either case, the normal vector and area element in these integrals are the ones you calculated in your OP if you're integrating over the surface of the hemisphere.
The problem is that I don't know if there was an easier way to do this via Stokes or perhaps Divergence theorem, or a better way to conceptualize this.
You could use either one to calculate ##\int dS## by using an appropriate vector field ##\vec{F}## to make the integrand of the RHS equal to 1, but I think it would be simplest just to evaluate the integral directly as you did.

Note with the integral you derived using Cartesian coordinates,
$$\int \frac{1}{\sqrt{1-x^2-y^2}}\,dx\,dy$$ you could easily evaluate it by changing to polar coordinates.

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