Tripple Integral Homework: Finding \int\int\int_D dV

  • Thread starter Thread starter sandy.bridge
  • Start date Start date
  • Tags Tags
    Integral
Click For Summary

Homework Help Overview

The problem involves evaluating the triple integral \(\int\int\int_D{e^{(x^2+y^2+z^2)^{3/2}/2}}dV\) over a specified region defined by the inequalities \(1\leq{x^2+y^2+z^2}\leq{3}\), \(z^2\geq{2}(x^2+y^2)\), and \(2x\leq{y}\leq3x\). The original poster is reviewing triple integrals and is struggling with determining the limits of integration, particularly for the angle \(\phi\).

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • Participants discuss visualizing the surface defined by \(z^2=2(x^2+y^2)\) and its implications for determining the angle \(\phi\). There is a suggestion that this surface is a cone, and participants explore how this affects the limits of integration. Questions arise about how to find the angle \(\phi\) given the curved nature of the surface.

Discussion Status

The discussion is active, with participants exploring different interpretations of the geometric shapes involved and their implications for the limits of integration. Some guidance has been offered regarding the relationship between the cone and the sphere, and how to determine the angle \(\phi\) based on their intersection.

Contextual Notes

Participants are working within the constraints of the problem statement and are attempting to visualize the geometric relationships between the surfaces defined by the inequalities. There is an emphasis on understanding the implications of these surfaces for the limits of integration in spherical coordinates.

sandy.bridge
Messages
797
Reaction score
1

Homework Statement



find the value of [tex]\int\int\int_D{e^{(x^2+y^2+z^2)^{3/2}/2}}dV[/tex] given [tex]1\leq{x^2+y^2+z^2}\leq{3}, z^2\geq{2}(x^2+y^2), 2x\leq{y}\leq3x[/tex]
I am reviewing tripple integrals and am having a bit of difficulty determining the limits for each part. I have,
[tex]arctan(2)\leq\theta\leq{arctan(3)}, 1\leq\rho\leq{\sqrt{3}}[/tex]

but I can't seem to visualize what is happening with
[tex]\phi[/tex]
Any suggestions?
 
Last edited:
Physics news on Phys.org
Can you visualize what the surface z^2=2*(x^2+y^2) looks like?
 
Kind of; paraboloid? If it is indeed a paraboloid, how does one determine the angle phi when the parabloid has a curved surface?
[tex]\phi{_1}\leq\phi\leq\pi/2[/tex]
 
sandy.bridge said:
Kind of; paraboloid? If it is indeed a paraboloid, how does one determine the angle phi when the parabloid has a curved surface?
[tex]\phi{_1}\leq\phi\leq\pi/2[/tex]

z=2*(x^2+y^2) is a paraboloid. z^2=2*(x^2+y^2) isn't. Think a little harder. I'll give you a hint. Sketch z^2=2*x^2 in the x-z plane.
 
Last edited:
It's a cone, I believe. Now, this cone is going to intersect the outer edge of the sphere, and therefore, plugging the equation for the cone into x^2+y^2+z^2=3, we get
[tex]3x^2+3y^2=3\rightarrow{x^2+y^2=1}[/tex]
Do I have the right idea so far?
 
sandy.bridge said:
It's a cone, I believe. Now, this cone is going to intersect the outer edge of the sphere, and therefore, plugging the equation for the cone into x^2+y^2+z^2=3, we get
[tex]3x^2+3y^2=3\rightarrow{x^2+y^2=1}[/tex]
Do I have the right idea so far?

You have the right idea that it's a cone. Doesn't the cone determine the range of ϕ? Isn't that what you were asking about?
 
Indeed, it does. However, is this not limited by the point that the cone intersects the outer edge of shell? This is why I combined the equations
[tex]x^2+y^2+z^2=3, z^2=2(x^2+y^2)[/tex]
This point of interesction can then be used to determine the angle from the xy-plane, which phi will merely be 90 less the angle measured.
 
Here is the drawing:
yes-4.jpg

So [tex]\phi[/tex] will go from the angle that it intersects to pi/2
 
Last edited:
sandy.bridge said:
Indeed, it does. However, is this not limited by the point that the cone intersects the outer edge of shell? This is why I combined the equations
[tex]x^2+y^2+z^2=3, z^2=2(x^2+y^2)[/tex]
This point of interesction can then be used to determine the angle from the xy-plane, which phi will merely be 90 less the angle measured.

Right, but I don't think there is any need to intersect z^2=2(x^2+y^2) with anything. ANY point on the cone that's above the x-y plane has the same value of phi.
 
  • #10
This is true. I got the right answer. Thanks
 

Similar threads

  • · Replies 105 ·
4
Replies
105
Views
11K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
Replies
11
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 27 ·
Replies
27
Views
3K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 9 ·
Replies
9
Views
2K