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...

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