Calculating Surface Area of Sphere for r = 3, 0 < theta < pi/2, 0 < phi < pi/3

[geft]

For r = 3, 0 < theta < pi/2, 0 < phi < pi/3

$$\int_{\theta=0}^{\frac{\pi}{2}}\int_{\phi=0}^{\frac{\pi}{3}}r^2sin\theta d \theta d \phi$$
$$=R^2 [-cos\theta]_{0}^{\frac{\pi}{2}} [\phi]_{0}^{\frac{\pi}{3}}$$
$$=(3^2)(1)(\frac{\pi}{3}) = 3\pi$$

The above is my working. The answer should be the same as the volume since 1/3R^3 (in the volume equation) = R^2 when R = 3 but the answer given is 9pi.

 

[HallsofIvy]

For r = 3, 0 < theta < pi/2, 0 < phi < pi/3

$$\int_{\theta=0}^{\frac{\pi}{2}}\int_{\phi=0}^{\frac{\pi}{3}}r^2sin\theta d \theta d \phi$$
$$=R^2 [-cos\theta]_{0}^{\frac{\pi}{2}} [\phi]_{0}^{\frac{\pi}{3}}$$
$$=(3^2)(1)(\frac{\pi}{3}) = 3\pi$$

The above is my working. The answer should be the same as the volume since 1/3R^3 (in the volume equation) = R^2 when R = 3 but the answer given is 9pi.

What "volume" are you talking about? Yes, the volume of the entire sphere of radius 3 is
$$\frac{4}{3}\pi (3)^3= 4\pi (3)^2$$
the surface area of the sphere. But here you are working with only a portion of the sphere.

For the entire sphere, $\theta$ goes from 0 to $\pi$ and $\phi$ goes from 0 to $2\pi$ (you are using "engineering" notation which swaps $\theta$ and $\phi$ from "mathematics" notation).

With $\theta$ going from 0 to $\pi/2$, half its range, and $\phi$ going from 0 to $\pi/3$, 1/6 of its range, you are dealing with 1/12 of the entire sphere. The surface area of a sphere of radius 3 is $4\pi 3^2= 36\pi$ and the surface area of 1/12 of that sphere is $3\pi$, exactly what you get.

If the "area given" is $9\pi$ either the answer in the book is wrong or you are doing the wrong limits of integration. $9\pi$ would be 1/4 the surface area of a sphere of radius 3. you could get that by taking $\theta$ from 0 to $\pi/2$ as you have and then taking $\phi$ from 0 to $\pi$, not from 0 to $\pi/3$.

 

[geft]

That's what I was thinking, but this is what's written on the book.

http://i.imgur.com/rfvVp.jpg

I calculated the (constrained) volume earlier and got the answer correctly, but the surface area stumped me. Is there something I'm missing from the question? Do we have to limit phi from 0 to pi when calculating surface areas?