- #1

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- Homework Statement
- f = r ^3

A = (x^2, y^2, z^2)

Calculate div(fA)

- Relevant Equations
- Nabla operator rules such as

$$\nabla \cdot (\phi A) = (\nabla \phi) \cdot A + \phi \nabla \cdot A$$

Using the formula in 'relevant equations' I calculate

$$div(fA) = \nabla(fA) = (\nabla f) \cdot A + f \nabla \cdot A$$

$$3r^2 \cdot (x^2, y^2, z^2) + r^3 \cdot (2x + 2y + 2z)$$But the answer is

$$3r \cdot (x^3 + y^3 + z^3) + r^3 \cdot (2x + 2y + 2z)$$

I find no way of easily turning ##3r^2 \cdot (x^2, y^2, z^2)## into ##3r \cdot (x^3 + y^3 + z^3)## and even if I could I don't see why they would even bother obfuscating the answer like that. I assume f is a scalar, yet in the first exercise ##grad(fg) = {g = 1/r^2} = e_r##

Overall I am just throughly confusing by something so seemingly obvious.

$$div(fA) = \nabla(fA) = (\nabla f) \cdot A + f \nabla \cdot A$$

$$3r^2 \cdot (x^2, y^2, z^2) + r^3 \cdot (2x + 2y + 2z)$$But the answer is

$$3r \cdot (x^3 + y^3 + z^3) + r^3 \cdot (2x + 2y + 2z)$$

I find no way of easily turning ##3r^2 \cdot (x^2, y^2, z^2)## into ##3r \cdot (x^3 + y^3 + z^3)## and even if I could I don't see why they would even bother obfuscating the answer like that. I assume f is a scalar, yet in the first exercise ##grad(fg) = {g = 1/r^2} = e_r##

Overall I am just throughly confusing by something so seemingly obvious.