# Represent a 3d region and compute this triple integral

• DottZakapa
In summary, the region E in 3-dimensions is represented by a sphere with an inner spherical hole. The section of e in the (x,z) plane is computed using spherical coordinates and the integral becomes
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?

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.

vanhees71 said:
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?

I think so.

vanhees71 said:
I think so.
thanks

## 1. What is a 3D region?

A 3D region is a space in 3-dimensional coordinate system. It can be represented by a combination of x, y, and z coordinates.

## 2. How do you represent a 3D region?

A 3D region can be represented in various ways, such as with equations, graphs, or geometric shapes. It is important to understand the boundaries and limits of the region in order to accurately represent it.

## 3. What is a triple integral?

A triple integral is a mathematical tool used to calculate the volume of a 3D region. It involves integrating a function over a 3D space, taking into account the boundaries and limits of the region.

## 4. Why is it important to compute a triple integral?

Computing a triple integral allows us to calculate important quantities such as volume, mass, and center of mass in a 3D space. It is a crucial tool in many fields of science, including physics, engineering, and economics.

## 5. What are some common applications of computing triple integrals?

Some common applications of computing triple integrals include calculating the volume of irregularly shaped objects, determining the mass of a three-dimensional object, and finding the center of mass of a 3D system.

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