Spherical Coordinates - Help me find my bounds

[RJLiberator]

Homework Statement

A vase is filled to the top with water of uniform density f = 1. The side profile of the barrel is given by the surface of revolution obtained by revolving the graph of g(z) = 2 + cos(z) over the z-axis, and bounded by 0 ≤ z ≤ π. Find the mass of the vase.

Homework Equations

The Attempt at a Solution

So I know that Mass = the triple integral of density dV.
I need to find my bounds.

I would like to use spherical coordinates as it seems to make sense with this problem.

Theta goes from 0 to 2pi.
phi goes from 0 to pi ? since it is bounded by 0 and pi? Or am I wrong here?
and rho goes from 0 to something involving the g(z) equation, but I am not sure how to manipulate it :/.

Any tips to get me started or confirmation on the phi bounds would be great.

 

[Dick]

Homework Statement

A vase is filled to the top with water of uniform density f = 1. The side profile of the barrel is given by the surface of revolution obtained by revolving the graph of g(z) = 2 + cos(z) over the z-axis, and bounded by 0 ≤ z ≤ π. Find the mass of the vase.

Homework Equations

The Attempt at a Solution

So I know that Mass = the triple integral of density dV.
I need to find my bounds.

I would like to use spherical coordinates as it seems to make sense with this problem.

Theta goes from 0 to 2pi.
phi goes from 0 to pi ? since it is bounded by 0 and pi? Or am I wrong here?
and rho goes from 0 to something involving the g(z) equation, but I am not sure how to manipulate it :/.

Any tips to get me started or confirmation on the phi bounds would be great.

Cylindrical coordinates make much more sense since you have the radial limit as an explicit function of z.

 

[Svein]

g(z) = 2 + cos(z) over the z-axis, and bounded by 0 ≤ z ≤ π

Therefore, in cylindrical coordinates:

Theta goes from 0 to 2pi.

Yes, and r=g(z).

 

[RJLiberator]

Ah, so I assume the wrong coordinates. I guess this makes sense since a barrel is a cylinder... Sigh. 0_o. Let me try some work now.

 

[RJLiberator]

So I am getting:

Theta: from 0 to 2pi.
r from 0 to 2+cos(z)
z from 0 to pi.

This would mean we integrate in the order dr dz dtheta

 

[RJLiberator]

So I performed the triple integral with the bounds:
Theta: from 0 to 2pi.
r from 0 to 2+cos(z)
z from 0 to pi.

using cylindrical coordinates and received the answer of (9pi^2)/2

It seems to be right, my one concern is with the bounds on r. It's clear to me that it starts at 0, but why am I allowed to put 2+cos(z) as the upper limit. So z=r in cylindrical coordinates, and the function was g(z)=2+cos(z) so it was that simple?

 

[Dick]

So I performed the triple integral with the bounds:
Theta: from 0 to 2pi.
r from 0 to 2+cos(z)
z from 0 to pi.

using cylindrical coordinates and received the answer of (9pi^2)/2

It seems to be right, my one concern is with the bounds on r. It's clear to me that it starts at 0, but why am I allowed to put 2+cos(z) as the upper limit. So z=r in cylindrical coordinates, and the function was g(z)=2+cos(z) so it was that simple?

Sound right. I'm not sure what you are saying with 'z=r in cylindrical coordinates'. The function g(z) defines the largest value of r for a given value of z, isn't that what the problem description implies?

 

[RJLiberator]

Hm. I see.

The function g(z) definitely defines the largest value for r, that makes more sense to me. r is the distance from the z-axis in cylindrical coordinates and that is what is implied by the question (bounds).

That makes sense to me. The other bounds are rather obvious, so that seems to work.

Thank you for your guidance.

 

[LCKurtz]

You could also have done it by cross sections as a single integral since it is a solid of revolution.