Solving spherical surface area

[geft]

For r = 3, 0 < theta < pi/2, 0 < phi < pi/3, find the surface area of the sphere. The answer given is 9pi, but I can't seem to work it out. Below is my working:

$$2\int_{r=0}^{3}\int_{\phi=0}^{\frac{\pi}{3}}r\sin\theta dr d \phi + <br /> 2\int_{r=0}^{3}\int_{\theta=0}^{\frac{\pi}{2}}r dr d \theta + <br /> \int_{\theta=0}^{\frac{\pi}{2}}\int_{\phi=0}^{\frac{\pi}{3}}r^2\sin{\theta} d \theta d \phi$$

$$= 2 [\frac{1}{2}R^2]_0^3[sin \frac {\pi}{2}][\phi]_0^{\frac{\pi}{3}}+ <br /> 2 [\frac{1}{2}R^2]_0^3[\theta]_0^{\frac{\pi}{2}} + <br /> R^2 [-\cos{\theta}]_{0}^{\frac{\pi}{2}} [\phi]_{0}^{\frac{\pi}{3}}$$

$$= (9)(1)(\frac{\pi}{3})+(9)(\frac{\pi}{2})+(9)(1)(\frac{\pi}{3})$$

$$= 3\pi + 4.5\pi + 3\pi = 10.5\pi$$

 

[HallsofIvy]

I have no clue at all what you are doing! For one thing, in the first integral you are integrating with respect to r- but r is fixed at 3. And integrating a function of $\theta$ with respect to r and $\phi$ will NOT give a number! The middle integral appears to be the area of a section of a circle. What circle? And the third integral, except that it is missing the integration with respect to r, is a "volume"! Why are you doing three different integrals instead of just one?

The surface area of a sphere or radius 3, with $0\le \theta\le \pi/3$ and $0\le\phi\le \pi/2$ is given by
$$9\int_{\phi= 0}^{\pi/3}\int_{\theta= 0}^{\pi/2} sin(\theta)d\theta d\phi$$

Another way to answer this is to note that since $\theta$ runs from 0 to $\pi/2$, half of its range from 0 to $\pi$ to cover the entire sphere, and $\phi$ runs from 0 to $\pi/3$, 1/6 of its range from 0 to $2\pi$, the area must be 1/12 of the entire surface area of the sphere of radius 3.

However, the area of the given surface is NOT $9\pi$.

(This problem is using "Engineering notation" which swaps the roles of $\theta$ and $\phi$ from "Mathematics notation".)

 

[geft]

Sorry, I forgot to mention that I'm using the http://en.wikipedia.org/wiki/Spherical_coordinate_system" (the red/yellow/blue diagram) for the calculation of electric force. I'm supposed to find the surface area enclosed by the parameters. When these limits are imposed, the sphere is reduced to something resembling a pizza slice. The area of each side is then added together.

I can include pictures from my book if they can be of help.

 

[vela]

The factor of two you have in front of the first integral shouldn't be there. There's only one wedge that lies in the xy-plane.

 

[geft]

Thanks!