Improper integral with spherical coordinates

In summary, the question asks to check the convergence of the integral x^2y^2z^2/r^(17/2) * f(x,y,z)dV, where f(x,y,z) is a continuous positive function in the domain D = {(x,y,z); x^2 + y^2 +z^2<=1} and r = sqrt(x^2 + y^2 + z^2). The attempt at a solution involves using spherical coordinates with the boundaries r between 0 and R (R=1), a between 0 and pi, and b between -pi and pi. After calculating the determinant as r^2sin(a), taking the limit m-->infinity with r between 1/m and
  • #1
Cyn
8
0

Homework Statement


I have a question.

I have a function f(x,y,z) which is a continuous positive function in D = {(x,y,z); x^2 + y^2 +z^2<=1}. And let r = sqrt(x^2 + y^2 + z^2). I have to check whether the following jntegral is convergent.

x^2y^2z^2/r^(17/2) * f(x,y,z)dV.

Homework Equations



Sphericak coordinates
x = rsin(a)cos(b)
y = rsin(a)sin(b)
z = rcos(a)

The Attempt at a Solution



Because you know that f continuous and positive is can you say that the integral of f is between m and M. But now, I have to know what the other integral is. I have to use sphericak coordinates.
x = rsin(a)cos(b)
y = rsin(a)sin(b)
z = rcos(a)

The determinant is r^2sin(a). But if I calculate this with the standard boundaries:
r between 0 and R (R=1)
a between 0 and pi
b between -pi and pi

And if I take r between 1/m and 1 and let the limit m-->infinity, then I find that the integral is 8/105 pi. But the answer is 16/105 pi.
Have I do something wrong or have I need to use different boundaries?

Thank you
 
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  • #2
Cyn said:

Homework Statement


I have a question.

I have a function f(x,y,z) which is a continuous positive function in D = {(x,y,z); x^2 + y^2 +z^2<=1}. And let r = sqrt(x^2 + y^2 + z^2). I have to check whether the following jntegral is convergent.

x^2y^2z^2/r^(17/2) * f(x,y,z)dV.

Homework Equations



Sphericak coordinates
x = rsin(a)cos(b)
y = rsin(a)sin(b)
z = rcos(a)

The Attempt at a Solution



Because you know that f continuous and positive is can you say that the integral of f is between m and M. But now, I have to know what the other integral is. I have to use sphericak coordinates.
x = rsin(a)cos(b)
y = rsin(a)sin(b)
z = rcos(a)

The determinant is r^2sin(a). But if I calculate this with the standard boundaries:
r between 0 and R (R=1)
a between 0 and pi
b between -pi and pi

And if I take r between 1/m and 1 and let the limit m-->infinity, then I find that the integral is 8/105 pi. But the answer is 16/105 pi.
Have I do something wrong or have I need to use different boundaries?

Thank you

Show your actual work; it looks like you might have made a simple error, but we cannot tell without seeing what you did. Also: you need to tell us what is the formula for the function f(x,y,z).

When you show your work, please do NOT post an image of handwritten material; take the time to type it out. (Note, for example, that you can type ##\int_a^b f(x) \, dx## as int_{x=a..b} f(x) dx in plain text, so it should not be too bad.)
 

1. What is an improper integral with spherical coordinates?

An improper integral with spherical coordinates is a type of integral that is evaluated over a region in three-dimensional space using the spherical coordinate system. This system uses the distance from the origin, the inclination angle, and the azimuth angle to define a point in space.

2. What are the advantages of using spherical coordinates for improper integrals?

Spherical coordinates offer advantages for evaluating integrals over regions with spherical symmetry, as they simplify the integration process by reducing the number of variables. They are also useful for solving problems involving three-dimensional shapes, such as spheres or cones.

3. How do you convert an improper integral from rectangular coordinates to spherical coordinates?

To convert an improper integral from rectangular coordinates to spherical coordinates, you can use the following equations:

  • r = √(x² + y² + z²)
  • θ = arccos(z / r)
  • φ = arctan(y / x)

where r is the distance from the origin, θ is the inclination angle, and φ is the azimuth angle.

4. What is the Jacobian determinant and how is it used in improper integrals with spherical coordinates?

The Jacobian determinant is a mathematical tool used to convert integrals from one coordinate system to another. In the case of spherical coordinates, the Jacobian determinant is equal to r²sinθ, which is used to account for the changes in volume when integrating over a spherical region.

5. What are some common applications of improper integrals with spherical coordinates?

Improper integrals with spherical coordinates are commonly used in physics and engineering applications, such as calculating the mass and center of mass of a three-dimensional object, determining the electric field strength at a point in space, and finding the volume of a solid of revolution.

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