What is the most efficient way to set up a density integral for a conical solid?

[1d20]

I’ve got two questions about two problems. First I just want to confirm that I’m setting up this density integral properly:

“Find the mass of the conical solid bounded by z = \sqrt{x^2 + y^2} and x^2 + y^2 + z^2 = 4 if the density at any point is proportional to the distance to the origin.

I’m taking that last part to mean that Density = ρ, so when I convert to spherical coordinates and set up the integral I get:

\int{0_2\pi} \int{0_\pi/4} \int{0_2} ρ^3 dρ d∅ dθ

Is that ρ^3 right? Or should it be (ρ^2 + ρ)?

 

[1d20]

Okay, next question.

“For the solid bounded by z = 4 - y^2, y = x, x = 0, z = 0 find the volume using a double integral.”

I did this using a triple, but I’m stuck trying to use a double. Converting to cylindrical coordinates, z goes from 0 to 4 - r^2sin^2θ while θ goes from ∏/4 to ∏/2. All well and good.

But when I set up the integral:

\int{\pi/2_\pi/4} \int{4 - r^2sin^2θ_0} r dr dθ

It’s clear that I’m going to end up with several unintegrated r terms. Gah, what am I missing?

 

[HallsofIvy]

I’ve got two questions about two problems. First I just want to confirm that I’m setting up this density integral properly:

“Find the mass of the conical solid bounded by z = \sqrt{x^2 + y^2} and x^2 + y^2 + z^2 = 4 if the density at any point is proportional to the distance to the origin.

I’m taking that last part to mean that Density = ρ, so when I convert to spherical coordinates and set up the integral I get:

[ itex ] \int{0_2\pi} \int{0_\pi/4} \int{0_2} ρ^3 dρ d∅ dθ[ /itex ]

Use instead
[ itex ] \int_0^{2\pi} \int_0^{\pi/4} \int_0^2 \rho^3 d\rho d\phi d\theta [/ itex ]
to get
\int_0^{2\pi} \int_0^{\pi/4} \int_0^2 \rho^3 d\rho d\phi d\theta
as far as the LaTeX goes.

Is that ρ^3 right? Or should it be (ρ^2 + ρ)?

Have you thought about what you are doing? The mass of any object with density function \delta(x,y,z) is \int\int\int \delta(x,y,z)dV. I cannot imagine why you would think of adding.

However, you have the differential of volume wrong. For spherical coordinates it is
\rho^2 sin(\phi) d\rho d\phi d\theta

Also, saying that the density is proportional to the radius does not mean it is equal to the radius. It means it is equal to some constant times the radius.

 

[HallsofIvy]

Okay, next question.

“For the solid bounded by z = 4 - y^2, y = x, x = 0, z = 0 find the volume using a double integral.”

I did this using a triple, but I’m stuck trying to use a double. Converting to cylindrical coordinates, z goes from 0 to 4 - r^2sin^2θ while θ goes from ∏/4 to ∏/2. All well and good.

But when I set up the integral:

\int{\pi/2_\pi/4} \int{4 - r^2sin^2θ_0} r dr dθ

It’s clear that I’m going to end up with several unintegrated r terms. Gah, what am I missing?

I cannot see any good reason to convert to cylindrical coordinates. There is no circular symmetry that woud make cylindrical coordinates simpler. In Cartesian coordinates, the integral for volume would be
\int_{y= -2}^2\int_{z= 0}^{4- y^2}\int_{x= 0}^y dxdzdy

Can you convert that to a double integral?

 

[1d20]

Use instead
[ itex ] \int_0^{2\pi} \int_0^{\pi/4} \int_0^2 \rho^3 d\rho d\phi d\theta [/ itex ]

Thanks; this forum's code is very frustrating.

Have you thought about what you are doing?

Of course I've thought about it; that's why I'm here. If I hadn't thought about it, I'd have integrated according to my first instinct. Which in this case happened to be mostly right, but obviously I didn't know that.

I cannot see any good reason to convert to cylindrical coordinates. There is no circular symmetry that woud make cylindrical coordinates simpler. In Cartesian coordinates, the integral for volume would be
\int_{y= -2}^2\int_{z= 0}^{4- y^2}\int_{x= 0}^y dxdzdy

Can you convert that to a double integral?

I've had no practice with this so let's find out...

\int_0^{2} \int_0^{y} (4 - y^2) dx dy