Another triple integral problem

In summary, the problem is to calculate the integral of the given function over the region D, defined as a sphere centered at (0,0,A) with radius R. Using spherical coordinates, the most logical change of variables is x = ρ cosθ sinφ, y = ρ sinθ sinφ, and z = ρ cosφ + A. After setting the boundaries for ρ and θ, the integral can be solved using cylindrical coordinates, with an integral over r as the first step.
  • #1
Gianmarco
42
3

Homework Statement


Calculate
[tex]
\int_D \frac{dxdydz}{\sqrt{x^2+y^2+(z-A)^2}}, \: A>R
[/tex]
on ## D = {(x,y,z)\: s.t. x^2+y^2+z^2 \leq R^2}##.

Homework Equations


In spherical coordinates:
[tex]
x=\rho cos\theta sin\phi\\y=\rho sin\theta sin\phi\\ z=cos\phi\\dxdydz=\rho^2sin\phi d\theta d\rho d\phi
[/tex]

The Attempt at a Solution


Since the integrand is ##\frac{1}{\rho}## of a sphere centered in ##(0,0,A)##, I think the most logical change of coordinates is:
[tex]
x=\rho cos\theta sin\phi\\y=\rho sin\theta sin\phi\\ z=\rho cos\phi + A
[/tex]
I am centered at a point ##(0,0,A)##, so nothing changed with respect to the xy-plane, therefore ##0 \leq \theta \leq 2\pi##; that much is obvious. And, since A>R, the boundaries for ##\rho## are ##A-R \leq \rho \leq A+R##. I found those geometrically.
Changing the variables to##\rho, \theta, \phi## in D we get:
##\rho^2 + 2\rho Acos\phi + A^2 \leq R^2##
And at this point I have no idea. Solving for ##\phi## would give me an upper limit, but it would be a an arccos of the form ##arccos(\frac{a}{\rho} - b\rho)## which I'd have to integrate in ##\rho## afterwards. Any ideas?
 
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  • #2
I have got the impression that the cylindrical coordinate is more helpful instead. In this case, the integral will be
$$
\int_0^{2\pi} \int_{-R}^R \int_0^{r(z)} \frac{rdrd\phi dz}{\sqrt{r^2+(z-A)^2}}
$$
Performing the integral over ##r## first should be easy.
 
  • Like
Likes Gianmarco
  • #3
I was able to solve it, thank you for the hint blue leaf! :)
 

Related to Another triple integral problem

1. What is a triple integral?

A triple integral is a mathematical concept that involves integrating a three-dimensional function over a three-dimensional region. It is used to calculate the volume of a solid or the mass of an object with variable density.

2. How do you set up a triple integral?

To set up a triple integral, you need to determine the limits of integration for each variable. This is typically done by visualizing the region of integration and breaking it down into smaller, simpler shapes. Then, you can use the appropriate formula to integrate over each variable.

3. What is the purpose of using a triple integral?

The purpose of using a triple integral is to calculate the volume or mass of a three-dimensional object that cannot be easily calculated using other methods. It is also used in physics and engineering to solve problems involving three-dimensional systems.

4. What are some common applications of triple integrals?

Triple integrals have many applications in various fields, such as calculating the moment of inertia of an object, finding the center of mass of a three-dimensional object, and determining the probability of an event in three-dimensional space. They are also used in fluid dynamics, electromagnetism, and quantum mechanics.

5. How do I solve a triple integral problem?

To solve a triple integral problem, you need to follow a specific set of steps. First, determine the limits of integration for each variable. Then, set up the triple integral using the appropriate formula. Next, evaluate the integral using integration techniques such as substitution or integration by parts. Finally, simplify the result to get the final answer.

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