http://calc3.askbot.com/questions/Open source question and answer forum written in Python and DjangoenCopyright Askbot, 2010-2011.Fri, 25 Jul 2014 18:04:23 -0500http://calc3.askbot.com/question/183/setting-up-spherical-integrals/I was wondering if anybody had an easily explainable way of how they choose the integration terms for spherical integrals. I know in class he was saying to use you as the point on z? So does that mean that the z term will always go from 0 to a certain number? For example in the last question on the quiz today, would the z integral have gone from 0 to the top function? or from the bottom of the function he gave, back to the top? ** Sorry I can't remember the question exactly** I feel like my thinking is wrong on how to come up with the terms, I thought your $\rho$ integral would go from the bottom function to the top, the $\phi$ integral would range from 0 to $\pi$ depending on how much of the sphere you want, and then the $\theta$ integral would range from 0 to $2\pi$ again ranging on how much of the sphere you are looking at. So I guess if anybody can remember how they went about question number 3 and could help me out that would be fantastic! :)Fri, 25 Jul 2014 13:45:28 -0500http://calc3.askbot.com/question/183/setting-up-spherical-integrals/http://calc3.askbot.com/question/183/setting-up-spherical-integrals/?answer=185#post-id-185Ok, so here's my whack at it while building off of Christina's input. Like Christina said $\phi$ is the angle that is measured from the $+z$-axis, having no correlation with the $xy$-plane. And also like Christina said, $\theta$ is measured off of the $+x$-axis when looking onto the $xy$-plane. The question from what I remember stated to set up the integral in spherical coordinates with $x^2+y^2+z^2=1$ and $y>0$ of the function:$$\int\int\int yDV$$ And as Professor McClure talked about in class:$$x^2+y^2+z^2=\rho^2$$$$x=\rho\cos(\theta)\sin(\phi)$$$$y=\rho\sin(\theta)\sin(\phi)$$$$z=\rho\cos(\phi)$$Setting $x^2+y^2+z^2=\rho^2$ we get $\rho^2=1$, which is $\rho=\sqrt{1}$ or $\rho=1$. Knowing that $\rho=1$ and that there is no inner restriction $\rho$ has to go from $0$ to $1$. And seeing as $\phi$ is only relevant to any restrictions on the $z$-axis, $\phi$ has to go from $0$ to $\pi$. Then for $\theta$ we can note that it stated that $y>0$, meaning that $\sin(\theta)$ should never be negative (At least this is how I interpreted it, I could be wrong). So referring back to the ancient unit circle we can see that $\sin(\theta)$ is positive from $0$ to $\pi$ and we get that we need to integrate with respect to $\theta$ from $0$ to $\pi$. Also to note that the Cartesian $y$ is equal to $\rho\sin(\theta)\sin(\phi)$, having an end result of:$$\int_0^{\pi}\int_0^{\pi}\int_0^1 \rho\sin(\theta)\sin(\phi)\rho^2\sin(\phi)\delta \rho \delta \phi \delta \theta$$ Or at least this is how I understand it to be.Fri, 25 Jul 2014 18:04:23 -0500http://calc3.askbot.com/question/183/setting-up-spherical-integrals/?answer=185#post-id-185http://calc3.askbot.com/question/183/setting-up-spherical-integrals/?answer=184#post-id-184I don't really know if my answer will be easily explainable but I will try to explain it as I understand it. With spherical coordinates we use what he called an "ordered triplet" associated with some point in space. I think you have a good understanding of $\varphi$ and $\theta$. $\varphi$ is the angle that is swept out from the positive z axis and $\theta$ is the angle that rotates around the z axis. The role of $\rho$ is to represent the **distance** from our point to the **origin**, thus $0\leq\rho$. So I am pretty sure that $\rho$ will always be integrated from zero to some positive value. I hope this helps and if I am incorrect about any of the finer points, someone will let us know:)Fri, 25 Jul 2014 16:06:35 -0500http://calc3.askbot.com/question/183/setting-up-spherical-integrals/?answer=184#post-id-184