Evaluate integral by using spherical coordinates

In summary, the conversation discusses finding the limits of integration for a triple integral problem involving spherical coordinates and a given integrand. The participants discuss drawing a picture to determine the limits, with the cone and sphere intersecting to help define the domain. The final limits of integration are determined to be rho from 0 to sqrt18, theta from 0 to pi/2, and phi from 0 to pi/4. After integrating, the final result is found to be approximately 37.3.
  • #1
Fizic
23
0
030sqrt(9-x2)sqrt(x2+y2)sqrt(18-x2-y2) (x2+y2+z2)dzdxdy

x=[itex]\rho[/itex]sin[itex]\varphi[/itex]cosθ
y=[itex]\rho[/itex]sin[itex]\varphi[/itex]sinθ
z=[itex]\rho[/itex]cos[itex]\varphi[/itex]

Change the integrand to [itex]\rho[/itex] and integrate wrt d[itex]\rho[/itex]dθd[itex]\varphi[/itex]

I don't know how to find the limits of integration. Normally I would draw a picture and reason it out, but that doesn't seem plausible here.

Thanks.
 
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  • #2
Fizic said:
030sqrt(9-x2)sqrt(x2+y2)sqrt(18-x2-y2) (x2+y2+z2)dzdxdy

x=[itex]\rho[/itex]sin[itex]\varphi[/itex]cosθ
y=[itex]\rho[/itex]sin[itex]\varphi[/itex]sinθ
z=[itex]\rho[/itex]cos[itex]\varphi[/itex]

Change the integrand to [itex]\rho[/itex] and integrate wrt d[itex]\rho[/itex]dθd[itex]\varphi[/itex]

I don't know how to find the limits of integration. Normally I would draw a picture and reason it out, but that doesn't seem plausible here.

Thanks.

Sure it does. The ##z## limits are ##z=\sqrt{x^2+y^2}## and ##z=\sqrt{18-x^2-y^2}##. What are those two surfaces?
 
  • #3
LCKurtz said:
Sure it does. The ##z## limits are ##z=\sqrt{x^2+y^2}## and ##z=\sqrt{18-x^2-y^2}##. What are those two surfaces?

Z is limited on bottom is a cone and on top a sphere with radius sqrt18.
Y is limited on the left by the xz-plane and on the right by y=3.
X... I don't know.
 
  • #4
So where do the sphere and cone intersect? Have you drawn the first octant picture? Once you have that the ##\rho,\phi,\theta## limits are pretty obvious.
 
  • #5
LCKurtz said:
So where do the sphere and cone intersect? Have you drawn the first octant picture? Once you have that the ##\rho,\phi,\theta## limits are pretty obvious.

I don't know how to draw the first octant because I'm unsure of how to draw the limiting surface on X.

Z is a cone and sphere. Y is two planes.

I feel like the upper limit on X should be sqrt(9-y^2) instead of sqrt(9-x^2). In that case it would be easy to re-parametrize.
 
  • #6
Fizic said:
I don't know how to draw the first octant because I'm unsure of how to draw the limiting surface on X.

Z is a cone and sphere. Y is two planes.

I feel like the upper limit on X should be sqrt(9-y^2) instead of sqrt(9-x^2). In that case it would be easy to re-parametrize.

Yes, that is an obvious typo which I didn't notice until you pointed it out. Nevertheless, you have the cone and sphere given.

Draw the cone and sphere in the first octant. They are going to intersect where their z values are equal. Get that equation. That will tell you the xy domain. But you don't need that anyway. Once you have the picture you can read the spherical coordinate limits from it.
 
  • #7
LCKurtz said:
Yes, that is an obvious typo which I didn't notice until you pointed it out. Nevertheless, you have the cone and sphere given.

Draw the cone and sphere in the first octant. They are going to intersect where their z values are equal. Get that equation. That will tell you the xy domain. But you don't need that anyway. Once you have the picture you can read the spherical coordinate limits from it.

Ah... yes. Okay.

rho from 0 to sqrt18
theta from 0 to pi/2
phi from 0 to pi/4

integrating (rho)^3*sin(phi)

I got 149 but that's too big.

Edit: Woops. Did theta from 0 to 2pi in calculations. Divided by 4 and got 32.266.
 
  • #8
Fizic said:
got 32.266.
I get 37.3. What expression do you get for the indefinite integral?
 

1. What are spherical coordinates?

Spherical coordinates are a coordinate system that uses two angles and a distance to specify the position of a point in three-dimensional space. The angles are typically referred to as azimuth and inclination, and the distance is known as the radial distance.

2. How do you convert from Cartesian coordinates to spherical coordinates?

To convert from Cartesian coordinates (x, y, z) to spherical coordinates (ρ, θ, φ), you can use the following equations: ρ = √(x² + y² + z²), θ = arctan(y/x), and φ = arccos(z/ρ). However, be careful with the arctan function, as it can only give you a value between -π/2 and π/2. You may need to add or subtract π to get the correct value for θ.

3. How do you evaluate an integral using spherical coordinates?

To evaluate an integral using spherical coordinates, you need to first convert the integral limits and the integrand into spherical coordinates. Then, you can use the triple integral formula for spherical coordinates: ∫∫∫ f(ρ, θ, φ) ρ² sin(φ) dρ dθ dφ. Finally, you can solve the integral using your preferred method (e.g. integration by parts, substitution, etc.).

4. What are the advantages of using spherical coordinates?

One of the main advantages of using spherical coordinates is that they are well-suited for problems involving spherical symmetry, such as those involving spheres, cones, and other curved surfaces. They can also simplify calculations in three-dimensional space, as the integration limits and the integrand often become simpler when expressed in spherical coordinates.

5. Are there any limitations to using spherical coordinates?

Yes, there are some limitations to using spherical coordinates. For example, they are not suitable for problems involving objects with sharp edges or corners. Additionally, the conversion between Cartesian and spherical coordinates can be complicated for some points, and the triple integral formula for spherical coordinates can be more difficult to work with compared to other coordinate systems such as Cartesian or cylindrical coordinates.

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