April 3rd, 2016, 05:18 AM  #1 
Newbie Joined: Oct 2014 From: Norway Posts: 15 Thanks: 1  Area of hemisphere
So I have the top hemisphere of a sphere with radius 1 and spherical center at the origin of the xyzcoordinate 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? 
April 3rd, 2016, 05:35 AM  #2 
Math Team Joined: Jan 2015 From: Alabama Posts: 3,261 Thanks: 894 
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]. 
April 3rd, 2016, 09:52 AM  #3  
Newbie Joined: Oct 2014 From: Norway Posts: 15 Thanks: 1  Quote:
 
April 4th, 2016, 05:23 AM  #4 
Newbie Joined: Oct 2014 From: Norway Posts: 15 Thanks: 1 
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? 
April 4th, 2016, 08:11 AM  #5  
Banned Camp Joined: Jun 2014 From: Earth Posts: 945 Thanks: 191  Quote:
* Maybe math and /math work as well.  

Tags 
area, hemisphere 
Thread Tools  
Display Modes  

Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
Solid hemisphere resting on an inclined plane  john Cartwell  Applied Math  0  October 28th, 2014 12:41 PM 
Calculating volume of hemisphere and surface area of hemisphere  john Cartwell  Calculus  1  October 28th, 2014 12:11 PM 
Estimated Volume of a Solid Hemisphere  jaredbeach  Calculus  9  January 12th, 2012 09:49 PM 
Area of square with convex parts  area of circle  gus  Algebra  1  April 17th, 2011 05:25 PM 
Centre of gravity in a hemisphere  Algebra  2  March 19th, 2009 01:54 PM 