Evaluate triple integral of 16z dv where G is the upper half of the sphere x^2+y^2+z^2=1.
Thank you very much.
Recall $\displaystyle \iiint_G f(x,y,z)~dV = \int_{c}^{d} \int_{\alpha}^{\beta} \int_{a}^{b} f(\rho \sin \phi \cos \theta , \rho \sin \phi \sin \theta , \rho \cos \phi) \rho ^2 \sin \phi~d \rho d \theta d \phi$
where G is a spherical wedge given by $\displaystyle G = \{ (\rho , \theta , \phi) | a \leq \rho \leq b, \alpha \leq \theta \leq \beta , c \leq \phi \leq d \}$
so our integral here is:
$\displaystyle \int_{0}^{\frac {\pi}{2}} \int_{0}^{2 \pi} \int_{0}^{1} 16 \rho^3 \cos \phi \sin \phi~d \rho ~d \theta ~d \phi$
any questions? do you know what $\displaystyle \rho, \theta \mbox { and } \phi$ represent?
Hi perfecthacker and Jhevon,
Thank you very much.
@perfecthacker
"How are you doing in your multivariable Calculus class?"
It is just okay. I am not a good math student.
@Jhevon
" any questions? do you know what represent?"
Please explain it to me. Thank you very much.
I am just begin to read the chapter of Spherical coordinates in Triple integral. Still working on getting the main idea of this chapter.
also could you please explain to me how you arrive the limits for the above three variables?
well, read the chapter and then if you have specific questions you can ask.
just keep in mind that $\displaystyle \rho$ is the distance from the origin to the surface of the solid you are dealing with. here, we were dealing with a sphere of radius 1 centered at the origin. so the distance from the origin to the surface of the sphere was, of course, 1. since the sphere was solid (meaning, there was no shape "carved" out of it), the distance $\displaystyle \rho$ ranged from 0 to 1.
$\displaystyle \theta$ is the angle in the xy-plane over which we are dealing with. we were dealing with a sphere. it's trace in the xy-plane is a circle. now the question is, do we want a whole circle, or just some sector of the circle. we wanted the upper-half plane, so it's trace in the xy-plane was indeed a full circle. thus $\displaystyle 0 \leq \theta \leq 2 \pi$
$\displaystyle \phi$ describes the angle formed between the z-axis and the surface. it opens down from the z-axis and continues as long as the surface is present. so here, the angle opened down from the z-axis, but it stopped when it hit the xy-plane, since we are dealing with the upper-half part of the sphere, there is no sphere in the negative z-region. the angle between the z-axis and the xy-plane is $\displaystyle \frac {\pi}{2}$, thus, $\displaystyle 0 \leq \phi \leq \frac {\pi}{2}$. it is worthy to note that $\displaystyle \phi$ does not exceed $\displaystyle \pi$ radians. if it goes right around and crosses the negative z-axis, it means we could have measure the angle from the other side and obtained a much smaller angle