Triple integral to find volume of ice cream cone

[PhysicsMark]

Homework Statement

Use a triple integral in rectangular coordinates to find the volume of the ice cream cone defined as follows

The region R in the xy-plane is the circle of radius 1 with center at the origin.
The sides of the cone are defined by the function z= \sqrt{x^2+y&2}
The top of the cone is a portion of the sphere with radius \sqrt{2} and center at the origin.
Repeat using triple integrals in cylindrical and spherical coordinates

Homework Equations

The problem with a picture can be found here http://www.docbenton.com/MAT240-56707/maple/MAPLE-6.pdf

The Attempt at a Solution

I need some help in starting. I am not sure how to set up the triple integral. I really need the practice on this problem.

 

[LCKurtz]

Some questions to get you started on spherical first since that is the most natural for this problem. Do you see why?

1. What is the equation of the sphere in spherical coordinates.
2. Can you figure out the central angle of the cone? This has to do with the limits for \phi
3. Do you know the spherical element of volume dV?
4. Can you see what the limits for \rho would be? Ditto \phi and \theta?

Show us which of these you do/don't know and we can go from there.

 

[PhysicsMark]

Hi. Thank you for taking the time to respond.

1. What is the equation of the sphere in spherical coordinates? \rho=\sqrt{2}

2.Can you figure out the central angle of the cone? I'm a little confused. What is the "central angle of the cone"? I believe the limits of \phi are 0\leq\phi\leq\pi/4

3. Do you know the spherical element of volume dV? I do not.

4. Can you see what the limits for rho would be? Ditto for phi and theta? The limit for theta should be 0\leq\theta\leq2\pi

Sorry for the mix of latex and text. I am running out of time at the moment. I am not sure of the answer to question 3. Are you talking about the volume of sphere being 4/3\pir^{2}?

 

[LCKurtz]

Hi. Thank you for taking the time to respond.

1. What is the equation of the sphere in spherical coordinates? \rho=\sqrt{2}

Correct.

2.Can you figure out the central angle of the cone? I'm a little confused. What is the "central angle of the cone"? I believe the limits of \phi are 0\leq\phi\leq\pi/4

Yes, I meant the angle from the vertical axis to the slant side. You have the \phi limits correct.

3. Do you know the spherical element of volume dV? I do not.

Well, of course you need to know dV. It should be in your book:
dV = \rho^2\sin(\phi)d\rho d\phi d\theta

4. Can you see what the limits for rho would be? Ditto for phi and theta? The limit for theta should be 0\leq\theta\leq2\pi

Sorry for the mix of latex and text. I am running out of time at the moment. I am not sure of the answer to question 3. Are you talking about the volume of sphere being 4/3\pir^{2}?

No, we aren't doing the volume of a sphere. Since \rho is going to go from 0 to \sqrt 2 and you know the other limits and dV now, just put the proper limits on

V = \int\int\int 1\ dV

and integrate.

 

[PhysicsMark]

Wow, thank you. I have another question. Am I integrating "1" or am I integrating \rho^{2}sin(\phi).

Its 1 isn't it? Nevermind, I think that is obvious. I'm going to post this question just in case. Thanks again for all the help.

 

[LCKurtz]

Wow, thank you. I have another question. Am I integrating "1" or am I integrating \rho^{2}sin(\phi).

Its 1 isn't it? Nevermind, I think that is obvious. I'm going to post this question just in case. Thanks again for all the help.

Remember that in spherical coordinates:

<br /> dV = \rho^2\sin(\phi)d\rho d\phi d\theta<br />

The fact that you are integrating 1 dV is what gives you the volume. If you were, for example, integrating a mass density \int\int\int_V\delta(x,y,z)\ dV you would be calculating the total mass. The \rho^2\sin(\phi) is built into the dV.

 

[PhysicsMark]

I have worked out the problem and gotten a solution. I have the volume as 1.0017 units. Can anyone confirm this?


My last step looked like this:

(2/3)^(3/2) \int-cos(\pi/4)+1 d\theta

as theta goes from 0 to 2\pi

 

[LCKurtz]

I have worked out the problem and gotten a solution. I have the volume as 1.0017 units. Can anyone confirm this?


My last step looked like this:

(2/3)^(3/2) \int-cos(\pi/4)+1 d\theta

as theta goes from 0 to 2\pi

That first factor should be:
\frac {2^{\frac 3 2}}{3}
The rest looks right. If you were in my class I would have told you to give the answer in exact radical form, not a decimal.