Triple Integral Help in Spherical Coordinates

[sevag00]

Hello everyone. I need help regarding putting boundaries in triple integrals using spherical coordinates. The figure attached below is an ice cream cone. I want someone to explain to me how to put the boundaries in the order of dr dz dθ? I had understood this in class but unfortunately i forgot the explanation.

 

[DryRun]

Hi sevag00

Hello everyone. I need help regarding putting boundaries in triple integrals using spherical coordinates. The figure attached below is an ice cream cone. I want someone to explain to me how to put the boundaries in the order of dr dz dθ? I had understood this in class but unfortunately i forgot the explanation.

Your question contradicts itself. You want to express the integrals in spherical coordinates, but the boundaries in terms of drdθdz are cylindrical coordinates!

What is the whole question? Are you required to find the volume of the cone including the section of paraboloid on top?

 

[sevag00]

lol. I meant the volume in cylindrical coordinates.

 

[LCKurtz]

Hello everyone. I need help regarding putting boundaries in triple integrals using spherical coordinates. The figure attached below is an ice cream cone. I want someone to explain to me how to put the boundaries in the order of dr dz dθ? I had understood this in class but unfortunately i forgot the explanation.

There's no reason to use spherical coordinates on this problem. You want cylindrical coordinates. It is best not to integrate the dr variable first; better would be $r\, dz dr d\theta$. The dz limits will be z on the lower surface and z on the upper surface, both expressed in terms of r. Once you have those z values in terms of r, you can set them equal to see what r is at their intersection.

 

[sevag00]

Yeah, i know it's easy integrating in order dzdrdθ . But the question in the book is asking drdzdθ.

 

[HallsofIvy]

Well, "drdzd\theta" is the hard way! I recommend doing it as two separate integrals. For z from 0 to 1, we have the cone, z= r, so, for each z, r goes from 0 to z. For z= 1 to 2, we have the paraboloid, z= 2- r^2 so, for each z, r^2=2-z and r, being positive, goes from 0 to \sqrt{2- z}.

 

[DryRun]

The answer will be the sum of two triple integrals; volume of the cone + volume of paraboloid.

 

[sevag00]

Is this in order dzdrdθ?

 

[DryRun]

I recommend that you do these as two separate integrals- one over the cone from z= 0 up to z= 1 and the second over the paraboloid from z= 1 up to 2.

For the second triple integral, i think it should be from z=1 to $z=2-r^2$

The projection of both cone and paraboloid to the xy-plane is the circle with center at (0, 0) and radius 2. To cover that, in polar coordinates, \theta goes from 0 to 2\pi and r from 0 to 1.

The radius of the projection is r=1.

 

[LCKurtz]

Is this in order dzdrdθ?

Is what in that order? Use the quote button to reply to a message so we know what you are replying to.

 

[sevag00]

Actually, i know the answer. But i didn't get how r is from 0 to z + 0 to sqrt(2-z). Answer is attached.

 

[LCKurtz]

Actually, i know the answer. But i didn't get how r is from 0 to z + 0 to sqrt(2-z).

Doing dr first, on the top part (the paraboloid) r goes from 0 to the r on the paraboloid. The equation of the paraboloid is $z = 2-r^2$. Just solve that for $r$ to get $\sqrt{2-z}$.

 

[sevag00]

Isn't the dr supposed to be from the x-axis to y-axis?

 

[LCKurtz]

Isn't the dr supposed to be from the x-axis to y-axis?

That doesn't make any sense. Think of starting on the z axis up at the level of the paraboloid. r is perpendicular to the z axis and it goes from r = 0 out till it hits the paraboloid. What direction it goes depends on $\theta$.

 

[sevag00]

Aha. I think i get it.

EDIT: Yes, i get it. Thanks for the help.