Represent a 3d region and compute this triple integral

[DottZakapa]

Homework Statement

Let $E=\left\{ (x,y,z) \in R^3 : 1 \leq x^2+y^2+z^2 \leq 4, 3x^2+3y^2-z^2\leq 0, z\geq0 \right\}$
- Represent the region E in 3-dimensions
-represent the section of e in (x,z) plane
-compute $\int \frac {y^2} {x^2+y^2} \,dx \,dy \,dz$

Relevant Equations

integrals

Let $E=\left\{ (x,y,z) \in R^3 : 1 \leq x^2+y^2+z^2 \leq 4, 3x^2+3y^2-z^2\leq 0, z\geq0 \right\}$
- Represent the region E in 3-dimensions
-represent the section of e in (x,z) plane
-compute $\int \frac {y^2} {x^2+y^2} \,dx \,dy \,dz$

the domain is a sphere of radius 2 with an inner spherical hole of radius 1 which intersects a cone on the positive z-axis.

using spherical coordinates

$\begin{cases} x=r cos\theta sin \phi\\ y=rsin\theta sin\phi\\ z=rcos\phi\\ \end{cases}$

$\begin{cases} 1\leq r\leq 2\\ 0 \leq \theta \leq 2\pi\\ 0 \leq \phi \leq \frac {\pi} 6\\ \end{cases}$

the integral becomes

$\int_{1}^2 \int_{0}^{2\pi} \int_{0}^{\frac \pi 6} \frac {(rsin\theta sin\phi)^2} {(r cos\theta sin \phi)^2+(rsin\theta sin\phi)^2} r^2 \sin \phi \,d\phi \, d\theta \,dr$=

= $\int_{1}^2 \int_{0}^{2\pi} \int_{0}^{\frac \pi 6} r^2 sin\phi sin\theta^2 ,d\phi \, d\theta \,dr$
up to here is correct?

 

[vanhees71]

Looks good to me, assuming concerning the integration order of the final integral you mean
$$\int_1^2 \mathrm{d} r \int_0^{2 \pi} \mathrm{d} \theta \int_0^{\pi/6} \mathrm{d} \phi \; r^2 \sin \phi \sin^2 \theta.$$
I'm aware that many textbooks have the confusing notation with the "differentials" in the integrals at the end, but then you have a hard time to read which integral sign belongs to which of the integration variables.

 

[DottZakapa]

Looks good to me, assuming concerning the integration order of the final integral you mean
$$\int_1^2 \mathrm{d} r \int_0^{2 \pi} \mathrm{d} \theta \int_0^{\pi/6} \mathrm{d} \phi \; r^2 \sin \phi \sin^2 \theta.$$
I'm aware that many textbooks have the confusing notation with the "differentials" in the integrals at the end, but then you have a hard time to read which integral sign belongs to which of the integration variables.

yes it is like that, concerning the domain is all correct?

 

[vanhees71]

I think so.

 

[DottZakapa]

I think so.

thanks