Question

A spherical solid, centered at the origin, has radius 4 and mass density(x,y,z)=6-(x^2+y^2+z^2). Set up the triple integral and find its mass.

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Answer 1

There's something very off about this question.

In spherical coordinates,

x² + y² + z² = ρ²

so that

f(x, y, z) = 6 - (x² + y² + z²)

transforms to

g(ρ, θ, φ) = 6 - ρ²

When transforming to spherical coordinates, we also introduce the Jacobian determinant, so that

dV = dx dy dz = ρ² sin(φ) dρ dθ dφ

Since we integrate over a sphere with radius 4, the domain of integration is the set

E = {(ρ, θ, φ) : 0 ≤ ρ ≤ 4 and 0 ≤ θ ≤ 2π and 0 ≤ φ ≤ π}

so that the integral is

`\displaystyle \int_(\phi=0)^(\phi=\pi) \int_(\theta=0)^(\theta=2\pi) \int_(\rho=0)^(\rho=4) (6 - \rho^2) \rho^2 \sin(\phi) \, d\rho \, d\theta \, d\phi`

Computing the integral is simple enough.

`\displaystyle = \int_(\phi=0)^(\phi=\pi) \int_(\theta=0)^(\theta=2\pi) \int_(\rho=0)^(\rho=4) (6 \rho^2 - \rho^4) \sin(\phi) \, d\rho \, d\theta \, d\phi`

`\displaystyle = 2\pi \int_(\phi=0)^(\phi=\pi) \int_(\rho=0)^(\rho=4) (6 \rho^2 - \rho^4) \sin(\phi) \, d\rho \, d\phi`

`\displaystyle = 2\pi \left(\int_(\phi=0)^(\phi=\pi) \sin(\phi) \, d\phi\right) \left(\int_(\rho=0)^(\rho=4) (6 \rho^2 - \rho^4) \, d\rho\right)`

`\displaystyle = 2\pi \cdot 2 \cdot \left(-\frac{384}5\right) = \boxed{-\frac{1536\pi}5}`

but the mass can't be negative...

Chances are good that this question was recycled without carefully changing all the parameters. Going through the same steps as above, the mass of a spherical body with radius R and mass density given by

`\delta(x, y, z) = k - (x^2 + y^2 + z^2)`

for some positive number k is

`(4\pi r^3)/(15) \left(5k - 3r^2\right)`

so in order for the mass to be positive, we must have

5k - 3r² ≥ 0 ⇒ k ≥ 3r²/5

In this case, k = 6 and r = 4, but 3•4²/5 = 9.6.