Triple integral in spherical form

In summary: So the limits for z should be as you say, and the limits for theta should be 0 to pi/4. So the limits for ρ should be 3/\sqrt{2}\leqslant r\leqslant 3 and for theta 0\leqslant\theta\leqslant pi/4.In summary, the triple integral 1/(x^2+y^2+z^2)dxdydz is bounded above by the sphere z=(9-x^2-y^2)^1/2 and below by the cone z=(x^2+y^2)^1/2. The limits for ρ, the spherical coordinate, are 3/\sqrt{2
  • #1
naspek
181
0
consider this following triple integral
1/(x^2+y^2+z^2)dxdydz

bounded above by sphere z=(9-x^2-y^2)^1/2 and below by the cone z=(x^2+y^2)^1/2

what i have done:

z=Pcospi

P^2=x^2+y^2+z^2

9=x^2+y^2+z^2

P=0 to 3

pi=0 to pi/4

theta=0 to 2pi

is this the correct range?
 
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  • #2
I would first equate the z's to find that [itex]x^{2}+y^{2}=9/2[/itex], then this means that the radius is [itex]3/\sqrt{2}[/itex] and i think that you limits are: [itex]3/\sqrt{2}\leqslant r\leqslant 3[/itex] and all your limits are okay.
 
  • #3
naspek said:
consider this following triple integral
1/(x^2+y^2+z^2)dxdydz

bounded above by sphere z=(9-x^2-y^2)^1/2 and below by the cone z=(x^2+y^2)^1/2

what i have done:

z=Pcospi

P^2=x^2+y^2+z^2

9=x^2+y^2+z^2

P=0 to 3

pi=0 to pi/4

theta=0 to 2pi

is this the correct range?

hunt_mat said:
I would first equate the z's to find that [itex]x^{2}+y^{2}=9/2[/itex], then this means that the radius is [itex]3/\sqrt{2}[/itex] and i think that you limits are: [itex]3/\sqrt{2}\leqslant r\leqslant 3[/itex] and all your limits are okay.

I think his P is supposed to be the spherical coordinate ρ, so his limits are correct. naspek, don't forget the spherical dV element.
 
  • #4
I think you're right coming to think of it, the point being that the intersection (which is a point on the sphere) is a circle of radius [itex]3/\sqrt{2}[/itex].
 

Related to Triple integral in spherical form

1. What is a triple integral in spherical form?

A triple integral in spherical form is a mathematical tool used to calculate the volume of a three-dimensional region in space. It is represented by three integrals, each representing a different dimension.

2. How is a triple integral in spherical form different from other types of integrals?

A triple integral in spherical form is different from other types of integrals because it is specifically used to calculate the volume of a three-dimensional region, rather than just an area or a line. It also uses spherical coordinates, which take into account the distance from the origin and angle measurements.

3. What are the advantages of using spherical coordinates for a triple integral?

Using spherical coordinates for a triple integral can make calculations easier and more efficient. It can also be helpful when dealing with spherical or cylindrical objects, as their equations are often simpler in spherical coordinates.

4. Are there any limitations to using a triple integral in spherical form?

One limitation of using a triple integral in spherical form is that it can be more difficult to visualize the region being integrated over compared to other coordinate systems. It also may not be the most efficient method for certain types of regions, such as those with irregular shapes.

5. How can I apply a triple integral in spherical form to real-world problems?

Triple integrals in spherical form can be applied to a variety of real-world problems, such as calculating the volume of a spherical tank or determining the mass of a planet. They can also be used in physics and engineering for calculations involving forces, electric fields, and fluid dynamics.

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