Quiz 1 Review

Spherical and cylindrical problems

Wed, 23 Jul 2014 12:03:06 -0500

I am having a lot of trouble visualizing and understanding how to set up spherical/cylindrical integrals. In particular, I am having trouble with #4 on the homework sheet. "Let $D$ denote the three-dimensional domain above the cone $z=\sqrt{x^2+y^2}$ and inside the sphere $x^2+y^2+z^2<=4$. Evaluate $\int\int\int(x^2+y^2+z^2)dV$." By looking at this, I can guess that spherical coordinates would work well because $x^2+y^2+z^2=P^2$ but past that I am lost. Please help!

Answer by SpaceManSpiff for Spherical and cylindrical problems

Wed, 23 Jul 2014 14:45:30 -0500

For starters, here's a groovy picture. ![image description](/upfiles/calc3.askbot.com/14061389454389333.png) That image shows a sphere of radius 2 centered at the origin intersecting with the given cone. Our integral with respect to theta will be from $0$ to $2 \pi$ since the rotation about the $z$-axis will have to be all the way around the section of the sphere above the cone. Our integral with respect to $\rho$ will be from $0$ to $2$ since the sphere is of radius two. So next you have to determine the angle from the "top" of the sphere to the inside surface of the cone. If you think about points that are on the cone in the $x$-$z$ plane a pattern becomes pretty apparent. Points like $(1,0,1)$ and $(0,0,0)$ indicate that the surface of the cone is sloped at $45^{o}$ relative to the $x$-axis. Here's the same groovy plot including the parametrized line through the origin, $ \langle t,0,t \rangle.$ ![image description](/upfiles/calc3.askbot.com/14061439752158272.png) You can see that the line runs along the surface of the cone confirming the $45^{o}$ angle hypothesis. So your integral is $$\int_{0}^{2 \pi} \int_{0}^{\frac{\pi}{4}} \int_{0}^{2} \rho ^{2} \rho^2 d \rho d \phi d \theta$$ $$=\int_{0}^{2 \pi} \int_{0}^{\frac{\pi}{4}} \int_{0}^{2} \rho ^{4} d \rho d \phi d \theta$$ Hopefully I haven't confused $\phi$ for $\theta$. Sorry if I did.

Answer by Justin for Spherical and cylindrical problems

Wed, 23 Jul 2014 19:01:28 -0500

I agree with Mr. Spiff, except for one thing. I believe that the integral should be: $$ \int_0^{2\pi} \int_0^{\pi/4} \int_0^2 \rho^4 \sin(\varphi) \ d\rho \ d\varphi \ d\theta $$ Note the extra $ \sin(\varphi) $. I also have a more rigorous reasoning for the $\pi/4$ radian angle: We know that $z = \sqrt{x^2 + y^2}$, which means that in cylindrical coordinates, $z = r$. By transforming $z$ and $r$ into their spherical form ($z = \rho \cos(\varphi), r = \rho \sin (\varphi) $), we obtain: $$ \rho \cos(\varphi) = \rho \sin (\varphi) $$ Assuming that $ \rho > 0 $, we can divide by $\rho$ on both sides. $$ \cos(\varphi) = \sin (\varphi) $$ If you can't tell what $\varphi$ is yet, it may be helpful to isolate it by dividing by $\cos(\varphi)$ and then taking the arctangent on both sides of the equation. It is easier to just watch the process than to explain: $$ \frac{\cos(\varphi)}{\cos(\varphi)} = \frac{\sin (\varphi)}{\cos(\varphi)} $$ $$ 1 = \tan(\varphi) $$ $$ \varphi = \tan^{-1}(1) $$ Using the restriction $0 \leq \varphi \leq \pi$, we conclude that $\varphi = \frac{\pi}{4}$ a.k.a. $45^{\circ}$.