- #1
mathmari
Gold Member
MHB
- 5,049
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Hey!
Apply the divergence theorem over the region $1 \leq x^2+y^2+z^2 \leq 4$ for the vector field $\overrightarrow{F}=-\frac{\hat{i}x+\hat{j}y+\hat{k}z}{p^3}$, where $p=(x^2+y^2+z^2)^\frac{1}{2}$.
$\bigtriangledown \overrightarrow{F}=\frac{3}{p^5}\frac{x^2+y^2+z^2}{p^2}-\frac{3}{p^3}=\frac{3}{p^3}-\frac{3}{p^3}=0$
So $\int \int \int_D {\bigtriangledown \overrightarrow{F}}dV=0$To calculate $\int \int_S {\overrightarrow{F} \cdot \hat{n}}d \sigma$ we have to calculate first the $\hat{n}$.
Isn't it as followed?
$\hat{n}=\frac{\bigtriangledown f}{|\bigtriangledown f|}=\frac{\hat{i}x+\hat{j}y+\hat{k}z}{p}$
In my textbook it is: $\hat{n}=\pm \frac{\bigtriangledown f}{|\bigtriangledown f|}=\pm \frac{\hat{i}x+\hat{j}y+\hat{k}z}{p}$
$"+": p=2$
$"-" \text{ for } p=1$
Why is it so?
Apply the divergence theorem over the region $1 \leq x^2+y^2+z^2 \leq 4$ for the vector field $\overrightarrow{F}=-\frac{\hat{i}x+\hat{j}y+\hat{k}z}{p^3}$, where $p=(x^2+y^2+z^2)^\frac{1}{2}$.
$\bigtriangledown \overrightarrow{F}=\frac{3}{p^5}\frac{x^2+y^2+z^2}{p^2}-\frac{3}{p^3}=\frac{3}{p^3}-\frac{3}{p^3}=0$
So $\int \int \int_D {\bigtriangledown \overrightarrow{F}}dV=0$To calculate $\int \int_S {\overrightarrow{F} \cdot \hat{n}}d \sigma$ we have to calculate first the $\hat{n}$.
Isn't it as followed?
$\hat{n}=\frac{\bigtriangledown f}{|\bigtriangledown f|}=\frac{\hat{i}x+\hat{j}y+\hat{k}z}{p}$
In my textbook it is: $\hat{n}=\pm \frac{\bigtriangledown f}{|\bigtriangledown f|}=\pm \frac{\hat{i}x+\hat{j}y+\hat{k}z}{p}$
$"+": p=2$
$"-" \text{ for } p=1$
Why is it so?