Evaluate integral by using spherical coordinates

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Homework Help Overview

The discussion revolves around evaluating a triple integral using spherical coordinates. The original poster presents an integral with specific limits and transformations but expresses uncertainty about determining the limits of integration, particularly in visualizing the geometric constraints of the problem.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the transformation of variables into spherical coordinates and the need to identify the limits of integration. There are questions about the geometric interpretation of the surfaces involved, specifically a cone and a sphere, and how they intersect.

Discussion Status

Some participants have provided guidance on identifying the limits of integration based on the geometric shapes involved. There is ongoing exploration of how to visualize the first octant and the implications of the surfaces on the limits. Multiple interpretations of the limits are being considered, and there is a recognition of potential errors in the original setup.

Contextual Notes

Participants note constraints related to drawing the first octant and understanding the limiting surfaces, which complicates the determination of the integration limits. There is also mention of a potential typo in the expressions for the limits.

Fizic
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030sqrt(9-x2)sqrt(x2+y2)sqrt(18-x2-y2) (x2+y2+z2)dzdxdy

x=\rhosin\varphicosθ
y=\rhosin\varphisinθ
z=\rhocos\varphi

Change the integrand to \rho and integrate wrt d\rhodθd\varphi

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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Fizic said:
030sqrt(9-x2)sqrt(x2+y2)sqrt(18-x2-y2) (x2+y2+z2)dzdxdy

x=\rhosin\varphicosθ
y=\rhosin\varphisinθ
z=\rhocos\varphi

Change the integrand to \rho and integrate wrt d\rhodθd\varphi

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?
 
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.
 
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.
 
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.
 
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.
 
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.
 
Fizic said:
got 32.266.
I get 37.3. What expression do you get for the indefinite integral?
 

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