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April 3rd, 2016, 05:18 AM   #1
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Area of hemisphere

So I have the top hemisphere of a sphere with radius 1 and spherical center at the origin of the xyz-coordinate system. I'm gonna find the area of the surface by integration.

Parameterization
$x = r \, cos(\theta)$, $y = r \, sin(\theta)$, $z = \sqrt{1 - x^2 - y^2} = \sqrt{1 - r^2}$

Surface function
$\vec{w} = r \, cos(\theta) \, \vec{i} + r \, sin(\theta) \, \vec{j} + \sqrt{1 - r^2} \, \vec{k}$

Absolute value of the cross product of the surface function
$\frac{d\vec{w}}{dr} \times \frac{d\vec{w}}{d\theta} = (\frac{r^2 \, cos(\theta)}{\sqrt{1 - r^2}}) \, \vec{i} + \frac{r^2 \, sin(\theta)}{\sqrt{1 - r^2}}) \, \vec{j} + r \, \vec{k}$

$|\frac{d\vec{w}}{dr} \times \frac{d\vec{w}}{d\theta}| = \frac{r}{\sqrt{1 - r^2}}$

Intregration
$\int_{0}^{2\pi} \int_{0}^{1} (\frac{r}{\sqrt{1 - r^2}}) \, r \, dr \, d\theta = \frac{(\pi)^2}{2}$

$\int_{0}^{2\pi} \int_{0}^{1} (\frac{r}{\sqrt{1 - r^2}}) \, dr \, d\theta = 2\pi$

Why is the last integral with just dr correct? I thought r*dr should always go together in an integral like this?
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April 3rd, 2016, 05:35 AM   #2
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First, you should NOT be integrating with respect to "r" at all because "r" is constant on the surface of a sphere! The problem is that your "parametrization" is for the sphere itself, not the surface of the sphere. The surface of the sphere is, of course, two dimensional and any parameterization should involve only two parameters, not three. Better is to use the parameterization for the sphere from spherical coordinates:




and "fix" [itex]\rho[/itex] at the radius of the sphere:




Now take the derivatives with respect to [itex]\theta[/itex] and [itex]\phi[/itex].
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April 3rd, 2016, 09:52 AM   #3
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Quote:
Originally Posted by Country Boy View Post
First, you should NOT be integrating with respect to "r" at all because "r" is constant on the surface of a sphere! The problem is that your "parametrization" is for the sphere itself, not the surface of the sphere. The surface of the sphere is, of course, two dimensional and any parameterization should involve only two parameters, not three. Better is to use the parameterization for the sphere from spherical coordinates:




and "fix" [itex]\rho[/itex] at the radius of the sphere:




Now take the derivatives with respect to [itex]\theta[/itex] and [itex]\phi[/itex].
I knew something was odd, especially since r is constantly equal to 1. Think I'll get it right now. Much appreciated!
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April 4th, 2016, 05:23 AM   #4
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Okay, so now I've tried again.

Parameterization
$x = \rho \, sin(\phi) \, cos(\theta) = (1) \, sin(\phi) \, cos(\theta)$
$y = \rho \, sin(\phi) \, sin(\theta) = (1) \, sin(\phi) \, sin(\theta)$
$z = \rho \, cos(\phi) = (1) \, cos(\phi)$

Surface function
$\vec{w}(\phi, \, \theta) = sin(\phi) \, cos(\theta) \, \vec{i} + sin(\phi) \, sin(\theta) \, \vec{j} + cos(\theta) \, \vec{k}$

Absolute value of the cross product of the surface function
$\frac{d\vec{w}}{d\phi} \times \frac{d\vec{w}}{d\theta} = (sin(\phi)^2 \, cos(\theta)) \, \vec{i} + (sin(\phi)^2 \, sin(\theta)) \, \vec{j} + (cos(\phi) \, sin(\phi)) \, \vec{k}$

$|\frac{d\vec{w}}{d\phi} \times \frac{d\vec{w}}{d\theta}| = sin(\phi)$

Intregration
$\int_{0}^{2\pi} \int_{0}^{\frac{\pi}{2}} (sin(\phi)) \, (\rho)^2 \, sin(\phi) \, d\phi \, d\theta = \int_{0}^{2\pi} \int_{0}^{\frac{\pi}{2}} (sin(\phi)^2) \, (1)^2 \, d\phi \, d\theta = \frac{\pi^2}{2}$

$\int_{0}^{2\pi} \int_{0}^{\frac{\pi}{2}} (sin(\phi)) \, d\phi \, d\theta = 2\pi$

Once again, if I include $(\rho)^2 \, sin(\phi)$ in the integral I get the wrong answer. When I exclude it, I get the correct answer. Why?
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April 4th, 2016, 08:11 AM   #5
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Quote:
Originally Posted by Country Boy View Post
First, you should NOT be integrating with respect to "r" at all because "r" is constant on the surface of a sphere! The problem is that your "parametrization" is for the sphere itself, not the surface of the sphere. The surface of the sphere is, of course, two dimensional and any parameterization should involve only two parameters, not three. Better is to use the parameterization for the sphere from spherical coordinates:




and "fix" [itex]\rho[/itex] at the radius of the sphere:




Now take the derivatives with respect to [itex]\theta[/itex] and [itex]\phi[/itex].
You need to change the inside of your brackets to MATH and /MATH, respectively, instead of tex and /tex, respectively. *


* Maybe math and /math work as well.
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